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How to Calculate Acceleration

Last Updated: March 20, 2023 Fact Checked

This article was co-authored by Sean Alexander, MS . Sean Alexander is an Academic Tutor specializing in teaching mathematics and physics. Sean is the Owner of Alexander Tutoring, an academic tutoring business that provides personalized studying sessions focused on mathematics and physics. With over 15 years of experience, Sean has worked as a physics and math instructor and tutor for Stanford University, San Francisco State University, and Stanbridge Academy. He holds a BS in Physics from the University of California, Santa Barbara and an MS in Theoretical Physics from San Francisco State University. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,884,973 times.

Sean Alexander, MS

Calculating Acceleration from a Force

Step 1 Define Newton’s Second Law of Motion.

  • Newton’s law can be represented by the equation F net = m x a , where F net is the total force acting on the object, m is the object’s mass, and a is the acceleration of the object.
  • When using this equation, keep your units in the metric system . Use kilograms (kg) for mass, newtons (N) for force, and meters per second squared (m/s 2 ) for acceleration.

Step 2 Find the mass...

  • For this equation, you will want to convert the mass into kilograms. If the mass you have is in grams simply divide that mass by 1000 to convert to kilograms.

Step 3 Calculate the net...

  • For example: Let’s say you and your big brother are playing tug-of-war. You pull the rope to the left with a force of 5 newtons while your brother pulls the rope in the opposite direction with a force of 7 newtons. The net force on the rope is 2 newtons to the right, in the direction of your brother.
  • In order to properly understand the units, know that 1 newton (N) is equal to 1 kilogram X meter/second squared (kg X m/s 2 ). [5] X Research source

Step 4 Rearrange the equation F = ma to solve for acceleration.

  • Force is directly proportional to the acceleration, meaning that a greater force will lead to a greater acceleration.
  • Mass is inversely proportional to acceleration, meaning that with a greater mass, the acceleration will decrease.

Step 5 Use the formula to solve for acceleration.

  • For example: A 10 Newton force acts uniformly on a mass of 2 kilograms. What is the object’s acceleration?
  • a = F/m = 10/2 = 5 m/s 2

Calculating Average Acceleration from Two Velocities

Step 1 Define the equation...

  • Acceleration is a vector quantity, meaning it has both a magnitude and a direction. [9] X Research source The magnitude is the total amount of acceleration whereas the direction is the way in which the object is moving. If it is slowing down the acceleration will be negative.

Step 2 Understand the variables....

  • Because acceleration has a direction, it is important to always subtract the initial velocity from the final velocity. If you reverse them, the direction of your acceleration will be incorrect.
  • Unless otherwise stated in the problem, the starting time is usually 0 seconds.

Step 3 Use the formula to find acceleration.

  • If the final velocity is less than the initial velocity, acceleration will turn out to be a negative quantity or the rate at which an object slows down.
  • Write the equation: a = Δv / Δt = (v f - v i )/(t f - t i )
  • Define the variables: v f = 46.1 m/s, v i = 18.5 m/s, t f = 2.47 s, t i = 0 s.
  • Solve: a = (46.1 – 18.5)/2.47 = 11.17 meters/second 2 .
  • Define the variables: v f = 0 m/s, v i = 22.4 m/s, t f = 2.55 s, t i = 0 s.
  • Solve: a = (0 – 22.4)/2.55 = -8.78 meters/second 2 .

Confirming Your Understanding

Step 1 Understand the Direction of Acceleration.

  • Example Problem: A toy boat with mass 10kg is accelerating north at 2 m/s 2 . A wind blowing due west exerts a force of 100 Newtons on the boat. What is the boat's new northward acceleration?
  • Solution: Because the force is perpendicular to the direction of motion, it does not have an effect on motion in that direction. The boat continues to accelerate north at 2 m/s 2 .

Step 3 Understand Net Force.

  • Example Problem: April is pulling a 400 kg container right with a force of 150 newtons. Bob stand on the left of the container and pushes with a force of 200 newtons. A wind blowing left exerts a force of 10 newtons. What is the acceleration of the container?
  • Solution: This problem uses tricky language to try and catch you. Draw a diagram and you'll see the forces are 150 newtons right, 200 newtons right, and 10 newtons left. If "right" is the positive direction, the net force is 150 + 200 - 10 = 340 newtons. Acceleration = F / m = 340 newtons / 400 kg = 0.85 m/s 2 .

Calculator, Practice Problems, and Answers

how to solve acceleration problems

Expert Q&A

Sean Alexander, MS

You Might Also Like

Calculate Instantaneous Velocity

  • ↑ Sean Alexander, MS. Academic Tutor. Expert Interview. 14 May 2020.
  • ↑ https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/newtons-laws-of-motion/
  • ↑ https://www.livescience.com/46560-newton-second-law.html
  • ↑ http://www.physicsclassroom.com/Class/newtlaws/u2l2d.cfm
  • ↑ http://study.com/academy/lesson/what-is-a-newton-units-lesson-quiz.html
  • ↑ https://byjus.com/physics/newtons-second-law-of-motion-and-momentum/
  • ↑ http://physics.info/acceleration/
  • ↑ http://www.physicsclassroom.com/Class/1DKin/U1L1e.cfm#vttable
  • ↑ https://pressbooks.online.ucf.edu/osuniversityphysics/chapter/3-3-average-and-instantaneous-acceleration/

About This Article

Sean Alexander, MS

To calculate acceleration, use the equation a = Δv / Δt, where Δv is the change in velocity, and Δt is how long it took for that change to occur. To calculate Δv, use the equation Δv = vf - vi, where vf is final velocity and vi is initial velocity. To caltulate Δt, use the equation Δt = tf - ti, where tf is the ending time and ti is the starting time. Once you've calculated Δv and Δt, plug them into the equation a = Δv / Δt to get the acceleration. To learn how to calculate acceleration from a force, read the article! Did this summary help you? Yes No

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  • Physics Formulas

Acceleration Formula

One may have perceived that pushing a terminally ill bus can give it a sudden start. That’s because lift provides an upward push when it starts. Here velocity changes and this is acceleration! Henceforth, the frame accelerates. Acceleration is described as the rate of change of velocity of an object. A body’s acceleration is the final result of all the forces being applied to the body, as defined by Newton’s second law. Acceleration is a vector quantity that is described as the frequency at which a body’s velocity changes.

Formula of Acceleration

Acceleration is the rate of change in velocity to the change in time. It is denoted by symbol a and is articulated as-

acceleration formula 1

The  S.I  unit for acceleration is meter per second square or m/s 2 .

velocity in terms of acceleration

  • Final Velocity is v
  • Initial velocity is u
  • Acceleration is a
  • Time taken is t
  • Distance traveled is s

Acceleration Solved Examples

Underneath we have provided some sample numerical based on acceleration which might aid you to get an idea of how the formula is made use of:

Problem 1:  A toy car accelerates from 3 m/s to 5 m/s in 5 s. What is its acceleration? Answer:

Given: Initial Velocity u = 3  m/s, Final Velocity v = 5m/s, Time taken t = 5s.

Acceleration formula 5

Problem 2:  A stone is released into the river from a bridge. It takes 4s for the stone to touch the river’s water surface. Compute the height of the bridge from the water level.

(Initial Velocity) u = 0 (because the stone was at rest), t = 4s (t is Time taken) a = g = 9.8 m/s 2 , (a is Acceleration due to gravity) distance traveled by stone = Height of bridge  = s The distance covered is articulated by

Acceleration formula 9

s = 0 + 1/2 × 9.8 × 4 = 19.6 m/s 2

Therefore, s = 19.6 m/s 2

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Physics Problems with Solutions

Physics Problems with Solutions

how to solve acceleration problems

  • Acceleration: Tutorials with Examples

Examples with explanations on the concepts of acceleration of moving object are presented. More problems and their solutions can also be found in this website.

Average Acceleration

An object with initial velocity v 0 at time t 0 and final velocity v at time t has an average acceleration between t 0 and t given by

Examples with soltutions

What is the acceleration of an object that moves with uniform velocity? Solution: If the velocity is uniform, let us say V, then the initial and final velocities are both equal to V and the definition of the acceleration gives

A car accelerates from rest to a speed of 36 km/h in 20 seconds. What is the acceleration of the car in m/s 2 ? Solution: The initial velocity is 0 (from rest) and the final velocity is 36 km/h. Hence

More References and links

  • Velocity and Speed: Tutorials with Examples
  • Velocity and Speed: Problems with Solutions
  • Uniform Acceleration Motion: Problems with Solutions
  • Uniform Acceleration Motion: Equations with Explanations

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

  • (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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2.5: Motion with Constant Acceleration (Part 1)

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Learning Objectives

  • Identify which equations of motion are to be used to solve for unknowns.
  • Use appropriate equations of motion to solve a two-body pursuit problem.

You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’s displacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement. In this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body pursuit problems .

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is \(\Delta\)t = t f − t 0 , taking t 0 = 0 means that \(\Delta\)t = t f , the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, x 0 is the initial position and v 0 is the initial velocity . We put no subscripts on the final values. That is, t is the final time , x is the final position , and v is the final velocity . This gives a simpler expression for elapsed time, \(\Delta\)t = t. It also simplifies the expression for x displacement, which is now \(\Delta\)x = x − x 0 . Also, it simplifies the expression for change in velocity, which is now \(\Delta\)v = v − v 0 . To summarize, using the simplified notation, with the initial time taken to be zero,

\[\Delta t = t\]

\[\Delta x = x - x_{0}\]

\[\Delta v = v - v_{0},\]

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.

We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that is,

\[\bar{a} = a = constant \ldotp\]

Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor does it degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can describe motion accurately by assuming a constant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, motion can be considered in separate parts, each of which has its own constant acceleration.

Displacement and Position from Velocity

To get our first two equations, we start with the definition of average velocity:

\[\bar{v} = \frac{\Delta x}{\Delta t} \ldotp\]

Substituting the simplified notation for \(\Delta\)x and \(\Delta\)t yields

\[\bar{v} = \frac{x - x_{0}}{t} \ldotp\]

Solving for x gives us

\[x = x_{0} + \bar{v} t,\label{3.10}\]

where the average velocity is

\[\bar{v} = \frac{v_{0} + v}{2} \ldotp \label{3.11}\]

The equation \(\bar{v} = \frac{v_{0} + v}{2}\) reflects the fact that when acceleration is constant, v is just the simple average of the initial and final velocities. Figure \(\PageIndex{1}\) illustrates this concept graphically. In part (a) of the figure, acceleration is constant, with velocity increasing at a constant rate. The average velocity during the 1-h interval from 40 km/h to 80 km/h is 60 km/h:

\[\bar{v} = \frac{v_{0} + v}{2} = \frac{40\; km/h + 80\; km/h}{2} = 60\; km/h \ldotp\]

In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the average velocity is greater than in part (a).

Graph A shows velocity in kilometers per hour plotted versus time in hour. Velocity increases linearly from 40 kilometers per hour at 1 hour, point vo, to 80 kilometers per hour at 2 hours, point v. Graph B shows velocity in kilometers per hour plotted versus time in hour. Velocity increases from 40 kilometers per hour at 1 hour, point vo, to 80 kilometers per hour at 2 hours, point v. Increase is not linear – first velocity increases very fast, then increase slows down.

Solving for Final Velocity from Acceleration and Time

We can derive another useful equation by manipulating the definition of acceleration:

\[a = \frac{\Delta v}{\Delta t} \ldotp\]

Substituting the simplified notation for \(\Delta\)v and \(\Delta\)t gives us

\[a = \frac{v - v_{0}}{t}\; (constant\; a) \ldotp\]

Solving for v yields

\[v = v_{0} + at\; (constant\; a) \ldotp \label{3.12}\]

Example 3.7: Calculating Final Velocity

An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s 2 for 40.0 s. What is its final velocity?

First, we identify the knowns: v 0 = 70 m/s, a = −1.50 m/s 2 , t = 40 s.

Second, we identify the unknown; in this case, it is final velocity v f .

Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown in terms of the knowns. We calculate the final velocity using Equation \ref{3.12}, v = v 0 + at.

Substitute the known values and solve:

\[v = v_{0} + at = 70.0\; m/s + (-1.50\; m/s^{2})(40.0\; s) = 10.0\; m/s \ldotp\]

Figure \(\PageIndex{2}\) is a sketch that shows the acceleration and velocity vectors.

Figure shows airplane at two different time periods. At t equal zero seconds it has velocity of 70 meters per second and acceleration of -1.5 meters per second squared. At t equal 40 seconds it has velocity of 10 meters per second and acceleration of -1.5 meters per second squared.

Significance

The final velocity is much less than the initial velocity, as desired when slowing down, but is still positive (see figure). With jet engines, reverse thrust can be maintained long enough to stop the plane and start moving it backward, which is indicated by a negative final velocity, but is not the case here.

In addition to being useful in problem solving, the equation v = v 0 + at gives us insight into the relationships among velocity, acceleration, and time. We can see, for example, that

  • Final velocity depends on how large the acceleration is and how long it lasts
  • If the acceleration is zero, then the final velocity equals the initial velocity (v = v 0 ), as expected (in other words, velocity is constant)
  • If a is negative, then the final velocity is less than the initial velocity

All these observations fit our intuition. Note that it is always useful to examine basic equations in light of our intuition and experience to check that they do indeed describe nature accurately.

Solving for Final Position with Constant Acceleration

We can combine the previous equations to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

\[v = v_{0} + at \ldotp\]

Adding v 0 to each side of this equation and dividing by 2 gives

\[\frac{v_{0} + v}{2} = v_{0} + \frac{1}{2} at \ldotp\]

Since \(\frac{v_{0} + v}{2} = \bar{v}\) for constant acceleration, we have

\[\bar{v} = v_{0} + \frac{1}{2} at \ldotp\]

Now we substitute this expression for \(\bar{v}\) into the equation for displacement, x = x 0 + \(\bar{v}\)t, yielding

\[x = x_{0} + v_{0}t + \frac{1}{2} at^{2}\; (constant\; a) \ldotp \label{3.13}\]

Example 3.8: Calculating Displacement of an Accelerating Object

Dragsters can achieve an average acceleration of 26.0 m/s 2 . Suppose a dragster accelerates from rest at this rate for 5.56 s Figure \(\PageIndex{3}\). How far does it travel in this time?

Picture shows a race car with smoke coming off of its back tires.

First, let’s draw a sketch Figure \(\PageIndex{4}\). We are asked to find displacement, which is x if we take x 0 to be zero. (Think about x 0 as the starting line of a race. It can be anywhere, but we call it zero and measure all other positions relative to it.) We can use the equation \(x = x_{0} + v_{0}t + \frac{1}{2} at^{2}\) when we identify v 0 , a, and t from the statement of the problem.

Figure shows race car with acceleration of 26 meters per second squared.

First, we need to identify the knowns. Starting from rest means that v 0 = 0 , a is given as 26.0 m/s 2 and t is given as 5.56 s.

Second, we substitute the known values into the equation to solve for the unknown:

\[x = x_{0} + v_{0}t + \frac{1}{2} at^{2} \ldotp\]

Since the initial position and velocity are both zero, this equation simplifies to

\[x = \frac{1}{2} at^{2} \ldotp\]

Substituting the identified values of a and t gives

\[x = \frac{1}{2} (26.0\; m/s^{2})(5.56\; s)^{2} = 402\; m \ldotp\]

If we convert 402 m to miles, we find that the distance covered is very close to one-quarter of a mile, the standard distance for drag racing. So, our answer is reasonable. This is an impressive displacement to cover in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this. If the dragster were given an initial velocity, this would add another term to the distance equation. If the same acceleration and time are used in the equation, the distance covered would be much greater.

What else can we learn by examining the equation \(x = x_{0} + v_{0}t + \frac{1}{2} at^{2}\)? We can see the following relationships:

  • Displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3.8, the dragster covers only one-fourth of the total distance in the first half of the elapsed time.
  • If acceleration is zero, then initial velocity equals average velocity (v 0 = \(\bar{v}\)) , and \(x = x_{0} + v_{0}t + \frac{1}{2} at^{2}\) becomes x = x 0 + v 0 t.

Solving for Final Velocity from Distance and Acceleration

A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve v = v 0 + at for t, we get

\[t = \frac{v - v_{0}}{a} \ldotp\]

Substituting this and \(\bar{v} = \frac{v_{0} + v}{2}\) into \(x = x_{0} + \bar{v} t\), we get

\[v^{2} = v_{0}^{2} + 2a(x - x_{0})\; (constant\; a) \ldotp \label{3.14}\]

Example 3.9: Calculating Final Velocity

Calculate the final velocity of the dragster in Example 3.8 without using information about time.

The equation \(v^{2} = v_{0}^{2} + 2a(x - x_{0})\) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

First, we identify the known values. We know that v 0 = 0, since the dragster starts from rest. We also know that x − x 0 = 402 m (this was the answer in Example 3.8). The average acceleration was given by a = 26.0 m/s 2 . Second, we substitute the knowns into the equation \(v^{2} = v_{0}^{2} + 2a(x - x_{0})\) and solve for v:

\[v^{2} = 0 + 2(26.0\; m/s^{2})(402\; m) \ldotp\]

\[v^{2} = 2.09 \times 10^{4}\; m/s^{2}\]

\[v = \sqrt{2.09 \times 10^{4}\; m^{2}/s^{2}} = 145\; m/s \ldotp\]

A velocity of 145 m/s is about 522 km/h, or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

An examination of the equation \(v^{2} = v_{0}^{2} + 2a(x - x_{0})\) can produce additional insights into the general relationships among physical quantities:

  • The final velocity depends on how large the acceleration is and the distance over which it acts.
  • For a fixed acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance. It takes much farther to stop. (This is why we have reduced speed zones near schools.)

Acceleration Calculator

Table of contents

Our acceleration calculator is a tool that helps you to find out how fast the speed of an object is changing . It works in three different ways, based on:

  • Difference between velocities at two distinct points in time.
  • Distance traveled during acceleration.
  • The mass of an accelerating object and the force that acts on it.

If you're asking yourself what is acceleration , what is the acceleration formula, or what are the units of acceleration, keep reading, and you'll learn how to find acceleration. Acceleration is strictly related to the motion of an object, and every moving object possesses specific energy.

To keep things clear, we also prepared some acceleration examples that are common in physics. You can find there:

  • Centripetal acceleration and tangential acceleration.
  • Angular acceleration.
  • Acceleration due to gravity .
  • Particle accelerator.

Acceleration always occurs whenever there is a non-zero net force acting on an object. You can feel it in an elevator when you become a little heavier (accelerating) or lighter (decelerating) or when you're riding down a steep slope on your sled in the snow. What's more, from the general theory of relativity, we know that the entire Universe is not only expanding, but it is even an accelerated expansion! That means that the distance between two points is constantly becoming greater and greater, but we can't feel that on an everyday basis because every scale in the world expands too.

What is acceleration? — acceleration definition

Acceleration is the rate of change of an object's speed; in other words, it's how fast velocity changes. According to Newton's second law , acceleration is directly proportional to the summation of all forces that act on an object and inversely proportional to its mass. It's all common sense – if several different forces are pushing an object, you need to work out what they add up to (they may be working in different directions) and then divide the resulting net force by your object's mass.

This acceleration definition says that acceleration and force are, in fact, the same thing. When the force changes, acceleration changes too, but the magnitude of its change depends on the mass of an object (see our magnitude of acceleration calculator for more details). This is not true in a situation when the mass also changes, e.g., in rocket thrust, where burnt propellants exit from the rocket's nozzle. See our rocket thrust calculator to learn more.

We can measure acceleration experienced by an object directly with an accelerometer . If you hang an object on the accelerometer, it will show a non-zero value. Why is that? Well, it's because of gravitational forces that act on every particle that has mass. And where is a net force, there is an acceleration. An accelerometer at rest thus measures the acceleration of gravity, which on the Earth's surface is about 31.17405 ft/s² (9.80665 m/s²) . In other words, this is the acceleration due to gravity that any object gains in free fall when in a vacuum.

Speaking of vacuums, have you ever watched Star Wars or another movie that takes place in space? The epic battles of spaceships, the sounds of blasters, engines, and explosions. Well, it's a lie. Space is a vacuum, and no sound can be heard there (sound waves require matter to propagate). Those battles should be soundless! In space, no one can hear you scream.

How to find acceleration? – acceleration calculator

The acceleration calculator on this site considers only a situation in which an object has a uniform (constant) acceleration. In that case, the acceleration equation is, by definition, the ratio of the change in velocity over a particular time.

Here, you can learn how to find acceleration in two more ways. Let's see how to use our calculator (you can find acceleration equations in the section after):

Depending on what data you have, you may calculate acceleration in three different ways. First of all, select an appropriate option at the top ("Speed difference", "Distance traveled" or "Mass and force").

For speed difference — Enter the initial v i and final v f speeds of the object and how much time Δt it took for the speed to change (see our speed calculator if required).

For distance traveled — Enter initial speed v i , distance traveled Δd and time Δt passed during acceleration. Here, you don't need to know the final speed.

For mass and force — Enter the mass m of the object and the net force F acting on this object. This is an entirely different set of variables that arises from Newton's second law of motion (another definition of acceleration).

Read the resulting acceleration from the last field. You can also perform calculations in the other way if you know what acceleration is, for example, to estimate distance Δd . Just provide the rest of the parameters in this window.

Acceleration formula — three acceleration equations

In the 17th century, Sir Isaac Newton , one of the most influential scientists of all time, published his famous book Principia . In it, he formulated the law of universal gravitation, which states that any two objects with mass will attract each other with a force exponentially dependent on the distance between these objects (specifically, it is inversely proportional to the distance squared). The heavier the objects are, the greater the gravitational force. It explains, for example, why planets orbit around the very dense Sun.

In Principia , Newton also includes three laws of motion which are central to understanding the physics of our world. The acceleration calculator is based on three various acceleration equations, where the third is derived from Newton's work:

  • a = (v f − v i ) / Δt ;
  • a = 2 × (Δd − v i × Δt) / Δt² ; and
  • a = F / m .
  • a — Acceleration;
  • v i and v f are, respectively, the initial and final velocities;
  • Δt — Acceleration time;
  • Δd — Distance traveled during acceleration;
  • F — Net force acting on an object that accelerates; and
  • m — Mass of this object.

Now you know how to calculate acceleration! In the next paragraph, we discuss the units of acceleration (SI and Imperial).

Acceleration units

If you already know how to calculate acceleration, let's focus on the units of acceleration. You can derive them from the equations we listed above. All you need to know is that speed is expressed in feet per second (imperial/US system) or in meters per second (SI system) and time in seconds. Therefore, if you divide the speed by time (as we do in the first acceleration formula), you'll get acceleration unit ft/s² or m/s² depending on which system you use.

Alternatively, you can use the third equation. In this case, you need to divide force (poundals in US and newtons in SI) by mass (pounds in US and kilograms in SI), obtaining pdl/lb or N/kg . They both represent the same thing, as poundal is pdl = lb·ft/s² and the newton is N = kg·m/s² . When you substitute it and reduce the units, you'll get (lb·ft/s²) / lb = ft/s² or (kg·m/s²) / kg = m/s² .

There is also a third option that is, in fact, widely used. You can express acceleration by standard acceleration due to gravity near the surface of the Earth, which is defined as g = 31.17405 ft/s² = 9.80665 m/s² . For example, if you say that an elevator is moving upwards with the acceleration of 0.2g , it means that it accelerates with about 6.2 ft/s² or 2 m/s² (i.e., 0.2 × g ). We rounded the above expressions to two significant figures.

Acceleration examples

Centripetal acceleration and tangential acceleration

Acceleration is generally a vector, so you can always decompose it into components. Usually, we have two parts that are perpendicular to each other: the centripetal and the tangential . Centripetal acceleration changes the direction of the velocity , and therefore the shape of the track, but doesn't affect the value of the velocity. On the other hand, tangential acceleration is always parallel to the trajectory of motion. It changes the value of velocity only, and not its direction.

In a circular motion (the leftmost picture below), where an object moves around the circumference of a circle, there is only the centripetal component. An object will keep its speed at a constant value; think of the Earth, which has centripetal acceleration due to the gravity of the Sun (in fact, its speed changes a bit during a year).

When both components are present, the object's trajectory looks like the right picture. What happens if there is only tangential acceleration? Then linear motion occurs. This is similar to when you press down on the gas pedal in a car on a straight part of the freeway.

Centripetal and tangential acceleration components in a circular motion.

Angular acceleration

Angular acceleration plays a vital role in the description of rotational motion. However, don't confuse it with the previously mentioned centripetal or tangential accelerations. This physical quantity corresponds to the rate of change of angular velocity. In other words, it tells you how fast an object's rotations accelerate – the object spins faster and faster (or slower and slower if angular acceleration is less than zero). Check out our angular acceleration calculator for more information.

Did you know that we can find an analogy between this and Newton's law of dynamics in rotational motion? In his second law, if you can switch acceleration with angular acceleration, force with torque, and mass with moment of inertia, you'll end up with the angular acceleration equation. You might notice that some physical laws, like this one, are universal, which makes them really important in physics.

Gravitational acceleration

We mentioned acceleration due to gravity a few times earlier. It arises from the gravitational force that exists between every two objects that have mass (note that the gravity equation isn't dependent on an object's volume – only mass is essential here). It may sound weird at first, but according to the third Newton's law of motion, you act with the same force on the Earth as the Earth acts on you . However, the mass of the Earth is much bigger than a human mass (~10²² times bigger), so our impact on the Earth is pretty much zero. It's analogous to all the bacteria (~10¹⁸ times lighter than a human) living on your hand; you can't even notice them! On the other hand, we can feel the influence of our planet, and that's the acceleration due to gravity.

Standard gravity is, by definition, 31.17405 ft/s² (9.80665 m/s²), so if a human weighs 220 lb (about 100 kg), he is subjected to the gravitational force of about 7000 pdl (1000 N). Let's enter this value into window #3 of our calculator along with the mass of the Earth (1.317 × 10²⁵ lbs or 5.972 × 10²⁴ kg in scientific notation). What is the calculated acceleration? It is so small that our calculator considers it to be zero . We mean nothing compared to the planet!

Particle accelerator

After talking about huge objects in space, let's move to the microscopic world of particles. Although we can't see them with our eyes, we have harnessed high-energy particles, like electrons and protons, and use them regularly in particle accelerators, common in physics, chemistry, and medicine. We use them to kill cancer cells while sparing the surrounding healthy tissue or investigate a material's structure at the atomic scale. Recently, cancer is one of the diseases of affluence that probably result from the increasing wealth in society.

You probably know about the Large Hadron Collider (CERN) , the most powerful particle accelerator in the world. It allows us to take a step further to understand how the universe works and develop technologies that may have many essential applications in the future. However, to achieve such high energies, we have to accelerate particles to speeds that are close to the speed of light. Briefly, we can do it using magnetic or electric fields.

Is acceleration a vector?

Yes , acceleration is a vector as it has both magnitude and direction . The magnitude is how quickly the object is accelerating, while the direction is if the acceleration is in the direction that the object is moving or against it. This is acceleration and deceleration, respectively.

How does mass affect acceleration?

If the force the object is being pushed with stays the same, the acceleration will decrease as the mass increases . This is because F/m = a, so as the mass increases, the fraction becomes smaller and smaller.

Can acceleration be negative?

Yes , acceleration can be negative, which is known as deceleration . Two objects with equal but opposite acceleration will accelerate by the same amount, just in two opposite directions.

How do you find average acceleration?

  • Work out the change in velocity for your given time.
  • Calculate the change in time for the period you are considering.
  • Divide the change in velocity by the change in time.
  • The result is the average acceleration for that period.

How do I find the magnitude of acceleration?

  • Convert the magnitude of the force into Newtons.
  • Change the mass of the object to kilograms.
  • Divide both values together to find the acceleration in m/s².

What is the difference between acceleration and velocity?

Velocity is the speed with which an object is moving in a particular direction, while acceleration is how the speed of that object changes with time. Both have a magnitude and a direction, but their units are m/s and m/s², respectively.

How do you find angular acceleration?

To find the angular acceleration:

Use the angular acceleration equations, which is ε = Δω / Δt .

Find the initial and final angular velocity in radians/s.

Subtract the initial angular velocity from the final angular velocity to get the change in angular velocity .

Find the initial and final time for the period being considered.

Subtract the initial time from the final time to get the change in time .

Divide the change in angular velocity by the change in time to get the angular acceleration in radians/s².

Omni's not-flat Earth calculator helps you perform three experiments that prove the world is round.

Make the best pizza choice with our Pizza Size Calculator – compare sizes and prices for the perfect order!

Calculate the root mean square speed of an ideal gas.

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ol{padding-top:0px;}.css-ykr2zs ul:not(:first-child),.css-ykr2zs ol:not(:first-child){padding-top:4px;} Speed difference

Initial speed

Final speed

Time passed during acceleration.

Acceleration

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Solved Speed, Velocity, and Acceleration Problems

Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial. 

Speed and velocity Problems: 

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution : Speed is defined in physics  as the total distance divided by the elapsed time,  so the rocket's speed is \[\text{speed}=\frac{8000}{13}=615.38\,{\rm m/s}\]

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution : Average speed is the ratio of the total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\ \\ &=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\ \\ &=4000\,{\rm s}\end{align*} To convert it to hours, it must be divided by $3600\,{\rm s}$ which gives $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes, and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed. 

Solution : First find its total distance traveled ($D$) by summing all distances in each section, which gets $D=100+200+50=350\,{\rm m}$. Now, by definition of average speed, divide it by the total time elapsed $T=5+7+4=16$ minutes.

But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity. 

Solution : Average velocity , $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement . 

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ (due north) and in the second part $\Delta x_2=250\,\hat{i}$ (due east). The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of  \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\ \\&=790.5\,{\rm m}\end{align*} In addition, the total elapsed time is $t=12\times 60$ seconds. Therefore, the magnitude of the average velocity is \[\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}\]

Problem (5): An object moves along a straight line. First, it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.

Here, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula, i.e., $x=vt$. Thus, displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\ \\&=\frac{60+60}{5+3}\\ \\&=\boxed{15\,{\rm m/s}}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance at $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): To reach a park located south of his jogging path, Henry runs along a 15-kilometer route. If he completes the journey in 1.5 hours, determine his speed and velocity.

Solution:  Henry travels his route to the park without changing direction along a straight line. Therefore, the total distance traveled in one direction equals the displacement, i.e, \[\text{distance traveled}=\Delta x=15\,\rm km\]Velocity is displacement divided by the time of travel \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{15\,\rm km}{1.5\,\rm h} \\\\ &=\boxed{10\,\rm km/h}\end{align*} and by definition, its average speed is \begin{align*} \text{speed}&=\frac{\text{distance covered}}{\text{time interval}}\\\\&=\frac{15\,\rm km}{1.5\,\rm h}\\\\&=\boxed{10\,\rm km/h}\end{align*} Thus, Henry's velocity is $10\,\rm km/h$ to the south, and its speed is $10\,\rm km/h$. As you can see, speed is simply a positive number, with units but velocity specifies the direction in which the object is moving. 

Problem (8): In 15 seconds, a football player covers the distance from his team's goal line to the opposing team's goal line and back to the midway point of the field having 100-yard-length. Find, (a) his average speed, and (b) the magnitude of the average velocity.

Solution:  The total length of the football field is $100$ yards or in meters, $L=91.44\,\rm m$. Going from one goal's line to the other and back to the midpoint of the field takes $15\,\rm s$ and covers a distance of $D=100+50=150\,\rm yd$. 

average speed and velocity at football field

Distance divided by the time of travel gets the average speed, \[\text{speed}=\frac{150\times 0.91}{15}=9.1\,\rm m/s\] To find the average velocity, we must find the displacement of the player between the initial and final points. 

The initial point is her own goal line and her final position is the midpoint of the field, so she has displaced a distance of $\Delta x=50\,\rm yd$ or $\Delta x=50\times 0.91=45.5\,\rm m$. Therefore, her velocity is calculated as follows \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time elapsed}} \\\\ &=\frac{45.5\,\rm m}{15\,\rm s} \\\\&=\boxed{3.03\quad \rm m/s}\end{align*} Contrary to the previous problem, here the motion is not in one direction, hence, the displacement is not equal to the distance traveled. Accordingly, the average speed is not equal to the magnitude of the average velocity.

Problem (9): You begin at a pillar and run towards the east (the positive $x$ direction) for $250\,\rm m$ at an average speed of $5\,\rm m/s$. After that, you run towards the west for $300\,\rm m$ at an average speed of $4\,\rm m/s$ until you reach a post. Calculate (a) your average speed from pillar to post, and (b) your average velocity from pillar to post. 

Solution : First, you traveled a distance of $L_1=250\,\rm m$ toward east (or $+x$ direction) at $5\,\rm m/s$. Time of travel in this route is obtained as follows \begin{align*} t_1&=\frac{L_1}{v_1}\\\\ &=\frac{250}{5}\\\\&=50\,\rm s\end{align*} Likewise, traveling a distance of $L_2=300\,\rm m$ at $v_2=4\,\rm m/s$ takes \[t_2=\frac{300}{4}=75\,\rm s\]  (a) Average speed is defined as the distance traveled (or path length) divided by the total time of travel \begin{align*} v&=\frac{\text{path length}}{\text{time of travel}} \\\\ &=\frac{L_1+L_2}{t_1+t_2}\\\\&=\frac{250+300}{50+75} \\\\&=4.4\,\rm m/s\end{align*} Therefore, you travel between these two pillars in $125\,\rm s$ and with an average speed of $4.4\,\rm m/s$. 

(b) Average velocity requires finding the displacement between those two points. In the first case, you move $250\,\rm m$ toward $+x$ direction, i.e., $L_1=+250\,\rm m$. Similarly, on the way back, you move $300\,\rm m$ toward the west ($-x$ direction) or $L_2=-300\,\rm m$. Adding these two gives us the total displacement between the initial point and the final point, \begin{align*} L&=L_1+L_2 \\\\&=(+250)+(-300) \\\\ &=-50\,\rm m\end{align*} The minus sign indicates that you are generally displaced toward the west. 

Finally, the average velocity is obtained as follows: \begin{align*} \text{average velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{-50}{125} \\\\&=-0.4\,\rm m/s\end{align*} A negative average velocity indicating motion to the left along the $x$-axis. 

This speed problem better makes it clear to us the difference between average speed and average speed. Unlike average speed, which is always a positive number, the average velocity in a straight line can be either positive or negative. 

Problem (10): What is the average speed for the round trip of a car moving uphill at 40 km/h and then back downhill at 60 km/h? 

Solution : Assuming the length of the hill to be $L$, the total distance traveled during this round trip is $2L$ since $L_{up}=L_{down}=L$. However, the time taken for going uphill and downhill was not provided. We can write them in terms of the hill's length $L$ as $t=\frac L v$. 

Applying the definition of average speed gives us \begin{align*} v&=\frac{\text{distance traveled}}{\text{total time}} \\\\ &=\frac{L_{up}+L_{down}}{t_{up}+t_{down}} \\\\ &=\cfrac{2L}{\cfrac{L}{v_{up}}+\cfrac{L}{v_{down}}} \end{align*} By reorganizing this expression, we obtain a formula that is useful for solving similar problems in the AP Physics 1 exams. \[\text{average speed}=\frac{2v_{up} \times v_{down}}{v_{up}+v_{down}}\] Substituting the numerical values into this, yields \begin{align*} v&=\frac{2(40\times 60)}{40+60} \\\\ &=\boxed{48\,\rm m/s}\end{align*} What if we were asked for the average velocity instead? During this round trip, the car returns to its original position, and thus its displacement, which defines the average velocity, is zero. Therefore, \[\text{average velocity}=0\,\rm m/s\]

Acceleration Problems

Problem (9): A car moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate? 

Solution : The car is initially at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \] 

Problem (10): A car moving at a velocity of $15\,{\rm m/s}$, uniformly slows down. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution:  Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$.   Average acceleration is the difference in velocities divided by the time taken, so we have: \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-15}{10}\\\\ &=\boxed{-1.5\,{\rm m/s^2}}\end{align*}The minus sign indicates the direction of the acceleration vector, which is toward the $-x$ direction.

Problem (11): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car.

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=4\,{\rm s}$.  Average acceleration is defined as the difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{20-0}{4}\\\\&=5\,{\rm m/s^2}\end{align*} In above, we converted $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get \[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\] 

Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.

Solution:  A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration. Known: $v_i=10\,{\rm m/s}$, $v_f = 30\,{\rm m/s}$, $\Delta t=2\,{\rm s}$. Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (13): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution:  Known: $v_i=10\,{\rm m/s}$, $v_f=-20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$. Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\\\&=\frac{(-20)-10}{2}\\\\ &=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.

Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?

Solution : At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.

When the ball is tossed upward, the only external force that acts on it is the gravity force. 

Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$. 

Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground? 

Solution : The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!

Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Assume each trip is done in one dimension without changing direction. Thus, displacements associated with each segment are the same as the distance traveled in that direction and is calculated as below: \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\ \\ \Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\ \\ \Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity, $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$, to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\ \\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ \\&=\frac{40+60+100}{4+2+4}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}

Problem (17): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$.

Solution: In this velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note : whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

  • Distances and times are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}\]
  • Velocities and times are known: \[\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}\]
  • Distances and velocities are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}\]

Problem (18): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (19): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with a constant velocity of $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known, so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\\ \Rightarrow \frac{t_2}{t_1}&=4\end{align*} 

Read more related articles:  

Kinematics Equations: Problems and Solutions

Position vs. Time Graphs

Velocity vs. Time Graphs

In the following section, some sample AP Physics 1 problems on acceleration are provided.

Problem (20): An object moves with constant acceleration along a straight line. If its velocity at instant of $t_1 = 3\,{\rm s}$ is $10\,{\rm m/s}$ and at the moment of $t_2 = 8\,{\rm s}$ is $20\,{\rm m/s}$, then what is its initial speed?

Solution: Let the initial speed at time $t=0$ be $v_0$. Now apply average acceleration definition in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ and equate them.\begin{align*}\text{average acceleration}\ \bar{a}&=\frac{\Delta v}{\Delta t}\\\\\frac{v_1 - v_0}{t_1-t_0}&=\frac{v_2-v_0}{t_2-t_0}\\\\ \frac{10-v_0}{3-0}&=\frac{20-v_0}{8-0}\\\\ \Rightarrow v_0 &=4\,{\rm m/s}\end{align*} In the above, $v_1$ and $v_2$ are the velocities at moments $t_1$ and $t_2$, respectively. 

Problem (21): For $10\,{\rm s}$, the velocity of a car that travels with a constant acceleration, changes from $10\,{\rm m/s}$ to $30\,{\rm m/s}$. How far does the car travel?

Solution: Known: $\Delta t=10\,{\rm s}$, $v_1=10\,{\rm m/s}$ and $v_2=30\,{\rm m/s}$. 

Method (I) Without computing the acceleration: Recall that in the case of constant acceleration, we have the following kinematic equations for average velocity and displacement:\begin{align*}\text{average velocity}:\,\bar{v}&=\frac{v_1+v_2}{2}\\\text{displacement}:\,\Delta x&=\frac{v_1+v_2}{2}\times \Delta t\\\end{align*}where $v_1$ and $v_2$ are the velocities in a given time interval. Now we have \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\\&=\frac{10+30}{2}\times 10\\&=200\,{\rm m}\end{align*}

Method (II) with computing acceleration: Using the definition of average acceleration, first determine it as below \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{30-10}{10}\\\\&=2\,{\rm m/s^2}\end{align*} Since the velocities at the initial and final points of the problem are given so use the below time-independent kinematic equation to find the required displacement \begin{align*} v_2^{2}-v_1^{2}&=2\,a\Delta x\\\\ (30)^{2}-(10)^{2}&=2(2)\,\Delta x\\\\ \Rightarrow \Delta x&=\boxed{200\,{\rm m}}\end{align*}

Problem (22): A car travels along a straight line with uniform acceleration. If its velocity at the instant of $t_1=2\,{\rm s}$ is $36\,{\rm km/s}$ and at the moment $t_2=6\,{\rm s}$ is $72\,{\rm km/h}$, then find its initial velocity (at $t_0=0$)?

Solution: Use the equality of definition of average acceleration $a=\frac{v_f-v_i}{t_f-t_i}$ in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ to find the initial velocity as below \begin{align*}\frac{v_2-v_0}{t_2-t_0}&=\frac{v_1-v_0}{t_1-t_0}\\\\ \frac{20-v_0}{6-0}&=\frac{10-v_0}{2-0}\\\\ \Rightarrow v_0&=\boxed{5\,{\rm m/s}}\end{align*}

All these kinematic problems on speed, velocity, and acceleration are easily solved by choosing an appropriate kinematic equation. Keep in mind that these motion problems in one dimension are of the uniform or constant acceleration type. Projectiles are also another type of motion in two dimensions with constant acceleration.

Author:   Dr. Ali Nemati

Date Published: 9/6/2020

Updated: Jun 28,  2023

© 2015 All rights reserved. by Physexams.com

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1D Kinematics Problem Solving

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The equations of 1D Kinematics are very useful in many situations. While they may seem minimal and straightforward at first glance, a surprising amount of subtlety belies these equations. And the number of physical scenarios to which they can be applied is vast. These problems may not be groundbreaking advances in modern physics, but they do represent very tangible everyday experiences: cars on roads, balls thrown in the air, hockey pucks on ice, and countless more examples can be modeled with these three relatively simple equations.

Equation Review

1d kinematics problems: easy, 1d kinematics problems: medium.

The three fundamental equations of kinematics in one dimension are:

\[v = v_0 + at,\]

\[x = x_0 + v_0 t + \frac12 at^2,\]

\[v^2 = v_0^2 + 2a(x-x_0).\]

The first gives the change in velocity under a constant acceleration given a change in time, the second gives the change in position under a constant acceleration given a change in time, and the third gives the change in velocity under a constant acceleration given a change in distance.

Here, the subscript "0" always refers to "initial". So, \(v_0\) is the initial velocity, and \(x_0\) is the initial position. Letters with no subscript indicate the quantity value after some time, \(t\). So, in the first equation, \(v\) is the velocity of an object that began at velocity \(v_0\) and has moved with constant acceleration \(a\) for an amount of time \(t\).

Very often, rather than using the initial and final positions, we simply want to know the total change in position, the distance traveled. This change in position is always merely the initial position subtracted from the final position: \(x-x_0\), often called \(d\) for distance. In many problems, this simplifies things and makes it simpler to see what is being asked. With this change, the second and third equations are sometimes rewritten:

\[d = v_0 t + \frac12 at^2,\]

\[v^2 = v_0^2 + 2ad.\]

A ball is dropped from rest off a cliff of height \(100 \text{ m}\). Assuming gravity accelerates masses uniformly on Earth's surface at \(g = 9.8 \text{ m}/\text{s}^2\), how fast is the ball going when it hits the ground? How long does it take to hit the ground? Solution: The third kinematics equation gives the final speed as: \[v_f^2 = 2( 9.8 \text{ m}/\text{s}^2)(100 \text{ m}) \implies v_f \approx 44.3 \text{ m}/\text{s}.\] The first kinematical equation gives the time to accelerate up to this speed: \[t = \frac{v}{a} = \frac{44.3 \text{ m}/\text{s}}{9.8 \text{ m}/\text{s}^2} \approx 4.5 \text{ s}.\]
A soccer ball is kicked from rest at the penalty spot into the net \(11 \text{ m}\) away. It takes \(0.4 \text{ s}\) for the ball to hit the net. If the soccer ball does not accelerate after being kicked, how fast was it traveling immediately after being kicked? Solution: This is a straightforward application of the second equation of motion with \(a = 0\), i.e \(d = vt\): \[v = \frac{d}{t} = \frac{11 \text{ m}}{0.4 \text{ s}} = 27.5 \text{ m}/\text{s}.\]
A continuously accelerating car starts from rest as it zooms over a span of \(100 \text{ m}\). If the final velocity of the car is \(30 \text{ m}/\text{s}\), what is the acceleration of the car? Solution: Applying the third kinematical equation with \(v_0 = 0\), \[v^2 = 2ad \implies a = \frac{v^2}{2d} = \frac{900}{200} \text{ m}/\text{s}^2 = 4.5 \text{ m}/\text{s}^2.\]

A basketball is dropped from a height of \(10 \text{ m}\) above the surface of the moon, accelerating downwards at \(1.6 \text { m}/\text{s}^2\). How long does it take to hit the surface, in seconds to the nearest tenth?

A train traveling at \(40 \text{ m}/\text{s}\) is heading towards a station \(400 \text{ m}\) away. If the train must slow down with constant deceleration \(a\) into the station, how long does it take to come to a complete stop, in seconds? Answer to the nearest integer.

Sometimes kinematics problems require multiple steps of computation, which can make them more difficult. Below, some more challenging problems are explored.

A projectile is launched with speed \(v_0\) at an angle \(\theta\) to the horizontal and follows a trajectory under the influence of gravity. Find the range of the projectile. Solution: The projectile begins with velocity in the vertical direction of \(v_0 \sin \theta\). To reach the apex of its trajectory, where the projectile is at rest, thus requires a time: \[t = \frac{v_0 \sin \theta}{g}.\] The time that it takes to fall back to the ground is therefore double this time, \[t = \frac{2v_0 \sin \theta}{g}.\] The range is the total distance in the horizontal direction traveled during this time. This is just the velocity in the x-direction times the time: \[R = v_x t = v_0 \cos \theta t = \frac{2v_0^2 \sin \theta \cos \theta}{g} = \frac{v_0^2 \sin 2 \theta}{g}.\]
A package is dropped from a cargo plane which is traveling at an altitude of \(10000 \text{ m}\) with a horizontal velocity of \(250 \text{ m}/\text{s}\) and no vertical component of the velocity. The package is initially at rest with respect to the plane. On the ground, a man is speeding along parallel to the plane in a \(5 \text{ m}\) wide car traveling \(40 \text{ m}/\text{s}\) trying to catch the package. The car starts a distance \(X \text{ m}\) ahead of the plane. What does \(X\) need to be for the man to succeed in catching the package? Solution: First, compute how long it takes for the package to hit the ground: \[d = \frac12gt^2 \implies t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{20000 \text{ m}}{9.8 \text{ m}/\text{s}^2}} = 45.2 \text{ s}.\] How far does the package travel horizontally during that time? \[d_{\text{package}} = v_x t = (250 \text{ m}/\text{s})(45.2 \text{ s}) = 11300 \text{ m}.\] How far does the car travel during that time? \[d_{\text{car}} = v_x t = (40 \text{ m}/\text{s})(45.2 \text{ s}) = 1808 \text{ m}.\] If the package is caught, then \(d_{\text{car}} + X = d_{\text{package}}\). This requires: \[X = (11300 - 1808) \text{ m} = 9492 \text{ m},\] or nearly \(10\) kilometers! To be exact, the above quantity for \(X\) can be shifted by up to \(2.5 \text{ m}\) and still make contact with the car, because of the nonzero width of the car, but this is a negligible correction; \(X\) is very large in comparison.

A pitcher throws a baseball towards home plate, a distance of \(18 \text{ m}\) away, at \(v = 40 \text{ m}/\text{s}\). Suppose the batter takes \(.2 \text {s}\) to react before swinging. In swinging, the batter accelerates the end of the bat from rest through \(2 \text{ m}\) at some constant acceleration \(a\). Assuming that the end of the bat hits the ball if it crosses the plate within \(. 05 \text{ s}\) of the ball crossing the plate, what is the minimum required \(a\) in \(\text{m}/\text{s}^2\) to the nearest tenth for the batter to hit the ball?

SpaceX is trying to land their next reusable rocket back on a drone ship. The drone ship is traveling due west in an ocean current at a constant speed of \(5 \text{ m}/\text{s}\). The rocket is \(2000 \text{ m}\) east of the drone ship and \(5000 \text{ m}\) vertically above it, traveling vertically downwards at \(100 \text{ m}/\text{s}\). If the rocket can apply vertical and horizontal thrusts to change the acceleration of the rocket in the vertical and horizontal directions, and must accelerate constantly in both directions, find the magnitude of the net acceleration (vector) required to land on the ship with no vertical velocity. Answer in \(\text{ m}/\text{s}^2\) to the nearest tenth.

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  • 21.4 DC Voltmeters and Ammeters
  • 21.5 Null Measurements
  • 21.6 DC Circuits Containing Resistors and Capacitors
  • Introduction to Magnetism
  • 22.1 Magnets
  • 22.2 Ferromagnets and Electromagnets
  • 22.3 Magnetic Fields and Magnetic Field Lines
  • 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
  • 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
  • 22.6 The Hall Effect
  • 22.7 Magnetic Force on a Current-Carrying Conductor
  • 22.8 Torque on a Current Loop: Motors and Meters
  • 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
  • 22.10 Magnetic Force between Two Parallel Conductors
  • 22.11 More Applications of Magnetism
  • Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
  • 23.1 Induced Emf and Magnetic Flux
  • 23.2 Faraday’s Law of Induction: Lenz’s Law
  • 23.3 Motional Emf
  • 23.4 Eddy Currents and Magnetic Damping
  • 23.5 Electric Generators
  • 23.6 Back Emf
  • 23.7 Transformers
  • 23.8 Electrical Safety: Systems and Devices
  • 23.9 Inductance
  • 23.10 RL Circuits
  • 23.11 Reactance, Inductive and Capacitive
  • 23.12 RLC Series AC Circuits
  • Introduction to Electromagnetic Waves
  • 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
  • 24.2 Production of Electromagnetic Waves
  • 24.3 The Electromagnetic Spectrum
  • 24.4 Energy in Electromagnetic Waves
  • Introduction to Geometric Optics
  • 25.1 The Ray Aspect of Light
  • 25.2 The Law of Reflection
  • 25.3 The Law of Refraction
  • 25.4 Total Internal Reflection
  • 25.5 Dispersion: The Rainbow and Prisms
  • 25.6 Image Formation by Lenses
  • 25.7 Image Formation by Mirrors
  • Introduction to Vision and Optical Instruments
  • 26.1 Physics of the Eye
  • 26.2 Vision Correction
  • 26.3 Color and Color Vision
  • 26.4 Microscopes
  • 26.5 Telescopes
  • 26.6 Aberrations
  • Introduction to Wave Optics
  • 27.1 The Wave Aspect of Light: Interference
  • 27.2 Huygens's Principle: Diffraction
  • 27.3 Young’s Double Slit Experiment
  • 27.4 Multiple Slit Diffraction
  • 27.5 Single Slit Diffraction
  • 27.6 Limits of Resolution: The Rayleigh Criterion
  • 27.7 Thin Film Interference
  • 27.8 Polarization
  • 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
  • Introduction to Special Relativity
  • 28.1 Einstein’s Postulates
  • 28.2 Simultaneity And Time Dilation
  • 28.3 Length Contraction
  • 28.4 Relativistic Addition of Velocities
  • 28.5 Relativistic Momentum
  • 28.6 Relativistic Energy
  • Introduction to Quantum Physics
  • 29.1 Quantization of Energy
  • 29.2 The Photoelectric Effect
  • 29.3 Photon Energies and the Electromagnetic Spectrum
  • 29.4 Photon Momentum
  • 29.5 The Particle-Wave Duality
  • 29.6 The Wave Nature of Matter
  • 29.7 Probability: The Heisenberg Uncertainty Principle
  • 29.8 The Particle-Wave Duality Reviewed
  • Introduction to Atomic Physics
  • 30.1 Discovery of the Atom
  • 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
  • 30.3 Bohr’s Theory of the Hydrogen Atom
  • 30.4 X Rays: Atomic Origins and Applications
  • 30.5 Applications of Atomic Excitations and De-Excitations
  • 30.6 The Wave Nature of Matter Causes Quantization
  • 30.7 Patterns in Spectra Reveal More Quantization
  • 30.8 Quantum Numbers and Rules
  • 30.9 The Pauli Exclusion Principle
  • Introduction to Radioactivity and Nuclear Physics
  • 31.1 Nuclear Radioactivity
  • 31.2 Radiation Detection and Detectors
  • 31.3 Substructure of the Nucleus
  • 31.4 Nuclear Decay and Conservation Laws
  • 31.5 Half-Life and Activity
  • 31.6 Binding Energy
  • 31.7 Tunneling
  • Introduction to Applications of Nuclear Physics
  • 32.1 Diagnostics and Medical Imaging
  • 32.2 Biological Effects of Ionizing Radiation
  • 32.3 Therapeutic Uses of Ionizing Radiation
  • 32.4 Food Irradiation
  • 32.5 Fusion
  • 32.6 Fission
  • 32.7 Nuclear Weapons
  • Introduction to Particle Physics
  • 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
  • 33.2 The Four Basic Forces
  • 33.3 Accelerators Create Matter from Energy
  • 33.4 Particles, Patterns, and Conservation Laws
  • 33.5 Quarks: Is That All There Is?
  • 33.6 GUTs: The Unification of Forces
  • Introduction to Frontiers of Physics
  • 34.1 Cosmology and Particle Physics
  • 34.2 General Relativity and Quantum Gravity
  • 34.3 Superstrings
  • 34.4 Dark Matter and Closure
  • 34.5 Complexity and Chaos
  • 34.6 High-temperature Superconductors
  • 34.7 Some Questions We Know to Ask
  • A | Atomic Masses
  • B | Selected Radioactive Isotopes
  • C | Useful Information
  • D | Glossary of Key Symbols and Notation

Learning Objectives

By the end of this section, you will be able to:

  • Establish the expression for centripetal acceleration.
  • Explain the centrifuge.

We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration ( a c a c ); centripetal means “toward the center” or “center seeking.”

The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii r r and Δ s Δ s are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds v 1 = v 2 = v v 1 = v 2 = v . Using the properties of two similar triangles, we obtain

Acceleration is Δ v / Δ t Δ v / Δ t , and so we first solve this expression for Δ v Δ v :

Then we divide this by Δ t Δ t , yielding

Finally, noting that Δ v / Δ t = a c Δ v / Δ t = a c and that Δ s / Δ t = v Δ s / Δ t = v , the linear or tangential speed, we see that the magnitude of the centripetal acceleration is

which is the acceleration of an object in a circle of radius r r at a speed v v . So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that a c a c is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that a c a c is greater for tighter turns, as you have probably noticed.

It is also useful to express a c a c in terms of angular velocity. Substituting v = rω v = rω into the above expression, we find a c = rω 2 / r = rω 2 a c = rω 2 / r = rω 2 . We can express the magnitude of centripetal acceleration using either of two equations:

Recall that the direction of a c a c is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.

A centrifuge (see Figure 6.8 b) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity ( g ) ( g ) ; maximum centripetal acceleration of several hundred thousand g g is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth’s gravity.

Example 6.2

How does the centripetal acceleration of a car around a curve compare with that due to gravity.

What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 6.8 (a).

Because v v and r r are given, the first expression in a c = v 2 r ;  a c = rω 2 a c = v 2 r ;  a c = rω 2 is the most convenient to use.

Entering the given values of v = 25 . 0 m/s v = 25 . 0 m/s and r = 500 m r = 500 m into the first expression for a c a c gives

To compare this with the acceleration due to gravity ( g = 9 . 80 m/s 2 ) ( g = 9 . 80 m/s 2 ) , we take the ratio of a c / g = 1 . 25 m/s 2 / 9 . 80 m/s 2 = 0 . 128 a c / g = 1 . 25 m/s 2 / 9 . 80 m/s 2 = 0 . 128 . Thus, a c = 0 . 128 g a c = 0 . 128 g and is noticeable especially if you were not wearing a seat belt.

Example 6.3

How big is the centripetal acceleration in an ultracentrifuge.

Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at 7.5 × 10 4 rev/min. 7.5 × 10 4 rev/min. Determine the ratio of this acceleration to that due to gravity. See Figure 6.8 (b).

The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity ω ω . Because r r is given, we can use the second expression in the equation a c = v 2 r ; a c = rω 2 a c = v 2 r ; a c = rω 2 to calculate the centripetal acceleration.

To convert 7 . 50 × 10 4 rev / min 7 . 50 × 10 4 rev / min to radians per second, we use the facts that one revolution is 2π rad 2π rad and one minute is 60.0 s. Thus,

Now the centripetal acceleration is given by the second expression in a c = v 2 r ;  a c = rω 2 a c = v 2 r ;  a c = rω 2 as

Converting 7.50 cm to meters and substituting known values gives

Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of a c a c to g g yields

This last result means that the centripetal acceleration is 472,000 times as strong as g g . It is no wonder that such high ω ω centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.

Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In Centripetal Force , we will consider the forces involved in circular motion.

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physicsgoeasy

How to solve kinematics problems

PhysicsGoeasy

  • April 1, 2023
  • Kinematics , Mechanics

how to solve acceleration problems

Table of Contents

How to Solve Kinematics Problems

In this comprehensive article learn How to solve kinematics problems. Kinematics is the study of motion without considering the forces that cause it. In this article, we will discuss how to solve kinematics problems for one-dimensional, two-dimensional, and three-dimensional motion involving constant acceleration.

Understanding Kinematic Variables

Before studying the kinematic equations , it’s important to understand the variables involved in solving these problems. Here are the kinematic variables you’ll encounter frequently:

  • Initial Position $(x_0 \text{ or } y_0)$: The starting point or location of an object before any motion occurs, measured along the $x$ or $y$ axis.
  • Final Position $(x \text{ or } y)$: The ending point or location of an object after motion has taken place, measured along the $x$ or $ y$ axis.
  • Initial Velocity $(v_0 \text{ or } u)$: The speed and direction of an object at the beginning of a motion, usually denoted as $v_0$ (or $u$ in some cases).
  • Final Velocity $(v)$: The speed and direction of an object at the end of a motion.
  • Acceleration $(a)$: The rate at which an object’s velocity changes over time, indicating how quickly the object speeds up, slows down, or changes direction.
  • Time $(t)$: The duration or interval during which the motion occurs, usually measured in seconds.

These variables are the foundation for solving problems involving constant acceleration .

Understanding Acceleration

Definition of acceleration.

Acceleration is the rate of change of velocity over time. It can be represented as a vector, with positive and negative components corresponding to the direction of motion. Positive acceleration is in the direction of travel, while negative acceleration is opposite the direction of travel.

There are several types of acceleration , including:

types of acceleration - How to solve kinematics problems.

Acceleration Vectors and Direction of Travel

When solving problems, it’s important to consider the direction of motion. The direction can be represented by acceleration vectors. Remember that positive acceleration is in the positive direction of travel, while negative acceleration is in the negative direction.

This distinction can be crucial when analyzing motion in the vertical and horizontal directions. The acceleration vector $(\vec a)$ can be represented as: $$\vec a=a_x\hat i+a_y\hat j+a_z\hat k$$

Centripetal and Tangential Acceleration

In circular motion, there are two types of acceleration: centripetal acceleration $(a_c​)$ and tangential acceleration $(a_t​)$. Centripetal acceleration is directed towards the center of the circle and is responsible for maintaining circular motion. Tangential acceleration is perpendicular to centripetal acceleration and is responsible for changes in the object’s speed. The equations for centripetal and tangential acceleration are:

  • $a_c=\frac{v^2}{r}$
  • $a_t​=r\alpha$

where $v$ is the object’s speed, $r$ is the radius of the circle, and $\alpha$ is the angular acceleration.

Deceleration

Deceleration is the process of an object slowing down, which corresponds to negative acceleration. When the deceleration is constant, it is referred to as constant deceleration . This is a special case of motion with constant acceleration, where the acceleration has a negative value.

Problems involving constant deceleration can be solved using the same kinematic equations mentioned earlier, but with a negative acceleration value.

Average Velocity and Average Acceleration

Average velocity $(v_{avg}​)$ is the total displacement $(d=x-x_0)$ divided by the time interval $(t)$, while average acceleration $(a_{avg}​)$ is the change in velocity $(\Delta v)$ divided by the time interval. Understanding these concepts can help you analyze motion in both horizontal and vertical directions. The equations for average velocity and average acceleration are:

  • $v_{avg​}=d/t​$
  • $a_{avg}​=\frac{\Delta v}{t}$

Uniform Acceleration

Uniform acceleration occurs when the acceleration is constant throughout the motion. This is the case in most kinematics problems. When dealing with uniform acceleration, remember to consider the forces acting on the object, such as gravity or friction.

Kinematic Equations for Constant Acceleration

The equation of motion is a mathematical representation of an object’s motion. When dealing with constant acceleration , there are four key equations of motions to remember:

  • $x=x_0+v_0t+\frac{1}{2}at^2$
  • $v^2=v_0^2+2a(x-x_0)$
  • $x-x_0=\frac{1}{2}(v_0+v)t$

The four kinematic equations mentioned above are examples of equations of motion. They can be used to determine the unknown variable(s) in a problem, depending on the given information.

Kinematics equations - How to solve kinematics problems

These equations only work for motion with constant acceleration. Most of the problems you encounter would involve motion with constant acceleration.

We also have an article on constant acceleration problems along with their solutions. You can visit the link to know more.

One-Dimensional Motion

When solving one-dimensional motion problems, follow these steps:

  • Write down the given physical quantities (initial position, initial velocity, acceleration, and time).
  • Determine the unknown variable(s) you need to find.
  • Identify the appropriate kinematic equation(s) to use.
  • Solve the equation algebraically or calculate to find the unknown variable(s).

Example: Finding Position Equation

Suppose you are given the initial position, initial velocity, and acceleration of an object. You can use the first kinematic equation to find the position equation : $x=x_0+v_0t+\frac{1}{2}at^2$

Sample Problem: One-Dimensional Motion

A car accelerates uniformly from rest for a distance of $200$ meters in $10 $ seconds. Calculate the final velocity of the car and the acceleration.

Step 1: Identify the given information and what needs to be found. Initial velocity $(v_0​) = 0 m/s$ (since the car is initially at rest) Initial position $(x_0​) = 0 m$ (assuming the starting point) Final position $(x) = 200 m$ Time $(t) = 10 s$ We need to find the final velocity $(v)$ and the acceleration $(a)$.

Step 2: Choose the appropriate kinematic equation. We can use the equation: $x=x_0​+v_0​t+\frac{1}{2}​at^2$

Step 3: Solve the equation for acceleration. Since the car is initially at rest and starts from the origin, the equation simplifies to: $200m=0+0+\frac{1}{2}​​a(10s)^2$ Now, solve for a: $200m=50s^2⋅a$ $a=\frac{200m}{50s^2}​$ $a=4m/s^2$ So, the acceleration of the car is $4m/s^2$.

Step 4: Determine the final velocity of the car. We can use the equation: $v=v_0​+at$ Substitute the known values: $v=0+4m/s^2⋅10s$ Now, solve for $v$: $v=40m/s$ So, the final velocity of the car is $40m/s$.

In this example, we used one-dimensional kinematic equations to find the final velocity and acceleration of a car that accelerates uniformly from rest.

Two-Dimensional Motion

For two-dimensional motion , follow these steps:

  • Resolve the initial velocity vector into x and y components .
  • Find the acceleration in each direction (horizontal and vertical).
  • Solve the equation of trajectory by relating the motion in the x and y-directions with respect to time.

Remember that if the acceleration is only in the vertical direction, the horizontal motion can be treated as a constant-velocity problem.

Projectile Motion

Projectile motion is a type of two-dimensional motion where an object moves under the influence of gravity while having an initial velocity in the horizontal direction. This motion can be analyzed by breaking it down into horizontal and vertical components.

The horizontal motion is constant-velocity motion as their is no force in the horizontal direction. The vertical motion is constant-acceleration motion with the acceleration being the acceleration due to gravity (approximately -9.81 m/s²).

When solving projectile motion problems, it’s important to consider the initial velocity components, the time of flight, the maximum height reached, and the range of the projectile.

Sample Problem: Two-Dimensional Motion

A projectile is launched from the ground at an angle of $60$ degrees with a velocity of $50 m/s$. Determine the maximum height reached, the horizontal range, and the total time of flight.

Step 1: Identify the given information and what needs to be found. Initial velocity $(v_0​) = 50 m/s$ Launch angle $ = 60$ degrees Acceleration in the vertical direction $(a_y​) = -9.8 m/s^2$ (downward due to gravity) Acceleration in the horizontal direction $(a_x​) = 0$ (no horizontal acceleration) We need to find the maximum height $(h)$, horizontal range $(R)$, and the total time of flight $(t)$.

Step 2: Resolve the initial velocity vector into x and y components. $v_{0x}​=v_0​\cos(60^0)=50\cos(60^0)=25m/s$ $v_{0y}​=v_0​\sin(60^0)=50\sin(60^0) \approx 43.3m/s$

Step 3: Determine the time of flight. Use the vertical motion equation: $v_y​=v_{0y}​+a_y​t$ At the maximum height, the vertical velocity $(v_y​)$ will be $0$: $0=43.3m/s−9.8m/s^2⋅t$ Solve for $t$: $t\approx \frac{43.3m/s}{9.8m/s^2} \approx 4.42s$ The total time of flight is twice the time to reach the maximum height: $t_{total​} \approx 2\cdot 4.42s\approx 8.84s$

Step 4: Calculate the maximum height. Use the vertical motion equation: $y=v_{0y}​t−\frac{1}{2}​a_y​t^2$ Substitute the known values: $h\approx 43.3m/s\cdot 4.42s−\frac{1}{2}\cdot 9.8m/s^2 \cdot (4.42s)^2$ $h\approx 95.7m$

Step 5: Calculate the horizontal range. Use the horizontal motion equation: $x=v_{0x}​t$ Substitute the known values: $R\approx 25m/s \cdot 8.84s$ $R\approx 221m$

In this example, we used two-dimensional kinematic equations to find the maximum height, horizontal range, and total time of flight for a projectile launched at an angle.

Three-Dimensional Motion

For three-dimensional motion , the process is similar to two-dimensional motion:

  • Resolve the initial velocity vector into x, y, and z components.
  • Find the acceleration in each direction (x, y, and z).
  • Solve the equations for each direction, relating the motion with respect to time.

Quadratic Equations and Quadratic Formula

In some kinematics problems, you may encounter quadratic equations . These equations involve a variable raised to the second power, such as $ax^2+bx+c=0$. To solve them, you can use the quadratic formula : $$x=\frac{-b \pm\sqrt{b^2-4ac}}{2a}$$ This formula can be applied to problems involving position, velocity, and acceleration to find the unknown variables.

Let’s explore an example problem to understand how to solve quadratic equations in kinematics.

Sample Problem

A ball is thrown vertically upward with an initial velocity of $20m/s$ from the ground. Find the maximum height the ball reaches and the time it takes to reach the maximum height.

Step 1: Identify the given information and what needs to be found. Initial velocity $(v_0​) =20m/s$ Acceleration $(a) = −9.8m/s^2$ (downward due to gravity) Final velocity at maximum height $(v) = 0m/s$ (as the ball momentarily stops) We need to find the maximum height $(h)$ and the time it takes to reach the maximum height $(t)$.

Step 2: Choose the appropriate kinematic equation. We can use the equation: $v^2=v_0^2​+2a(x−x_0​)$, where $x$ represents the maximum height and $x_0$​ is the initial height (in this case, $0$).

Step 3: Solve the equation for the maximum height. Substitute the known values: $0^2=(20m/s)^2+2(−9.8m/s^2)(x−0)$ Solve for $x$: $0=400m^2/s^2−19.6m/s^2\cdot x$ Now, divide both sides by $−19.6m/s^2$: $x=\frac{19.6m/s^2}{400m2/s^2}​$ $x\approx 20.4m$ So, the maximum height the ball reaches is approximately $20.4m$.

Step 4: Determine the time it takes to reach the maximum height. We can use the equation: $v=v_0​+at$ Substitute the known values: $0m/s=20m/s−9.8m/s^2⋅t$ Now, solve for $t$: $9.8m/s^2⋅t=20m/s$ $t\approx \frac{20m/s}{9.8m/s^2}​$ $t\approx 2.04s$ So, it takes approximately $2.04s$ for the ball to reach the maximum height.

In this example, we used a kinematic equation that involved a quadratic term to find the maximum height and the time it takes to reach the maximum height.

Uniform Circular Motion

  • Draw a simple, neat diagram of the system.
  • Firstly consider the origin of the forces acting on each object. To do this find out the field forces acting on each object. Wherever contact is in the available account the contact force carefully
  • Find out the force acting on the body. The resultant force should provide the required centripet al required for Circular motion
  • Centripetal force \(=\frac{mv^2}{R}\) will give the velocity accordingly

Tips for Solving Kinematics Problems

  • Always draw a diagram to visualize the problem and identify the given variables.
  • Choose a coordinate system and resolve the vectors into components.
  • Write down all the known variables and the unknown variable(s) you need to find.
  • Select the appropriate kinematic equations to relate the variables.
  • Solve the equations algebraically to obtain the unknown variable(s).

Additional reading : Acceleration Formula

Related Posts:

Negative acceleration

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Title: an inexact bregman proximal point method and its acceleration version for unbalanced optimal transport.

Abstract: The Unbalanced Optimal Transport (UOT) problem plays increasingly important roles in computational biology, computational imaging and deep learning. Scaling algorithm is widely used to solve UOT due to its convenience and good convergence properties. However, this algorithm has lower accuracy for large regularization parameters, and due to stability issues, small regularization parameters can easily lead to numerical overflow. We address this challenge by developing an inexact Bregman proximal point method for solving UOT. This algorithm approximates the proximal operator using the Scaling algorithm at each iteration. The algorithm (1) converges to the true solution of UOT, (2) has theoretical guarantees and robust regularization parameter selection, (3) mitigates numerical stability issues, and (4) can achieve comparable computational complexity to the Scaling algorithm in specific practice. Building upon this, we develop an accelerated version of inexact Bregman proximal point method for solving UOT by using acceleration techniques of Bregman proximal point method and provide theoretical guarantees and experimental validation of convergence and acceleration.

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  3. Acceleration Formula With Solved Examples

    Acceleration is a vector quantity that is described as the frequency at which a body's velocity changes. Formula of Acceleration. Acceleration is the rate of change in velocity to the change in time. It is denoted by symbol a and is articulated as-The S.I unit for acceleration is meter per second square or m/s 2.

  4. How to Solve for Acceleration (Easy)

    A video tutorial explaining how to solve for acceleration using the a= Vf-Vi/t equation.

  5. Acceleration: Tutorials with Examples

    Examples with explanations on the concepts of acceleration of moving object are presented. More problems and their solutions can also be found in this website.. Average Acceleration The average acceleration is a vector quantity (magnitude and direction) that describes the rate of change (with the time) of the velocity of a moving object.. An object with initial velocity v 0 at time t 0 and ...

  6. Kinematic Equations: Sample Problems and Solutions

    Kinematic equations relate the variables of motion to one another. Each equation contains four variables. The variables include acceleration (a), time (t), displacement (d), final velocity (vf), and initial velocity (vi). If values of three variables are known, then the others can be calculated using the equations. This page demonstrates the process with 20 sample problems and accompanying ...

  7. Acceleration and velocity (practice)

    You might need: Calculator. A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m s 2 rightward. After 3 s , what will be the velocity of the rocket ship? Answer using a coordinate system where rightward is positive. m s.

  8. Acceleration Problems

    Acceleration Problems On this page I put together a collection of acceleration problems to help you understand acceleration better. The required equations and background reading to solve these problems is given on the kinematics page. Problem # 1 A particle is moving in a straight line with a velocity given by 5t 2, where t is time. Find an ...

  9. 2.5: Motion with Constant Acceleration (Part 1)

    We start with. v = v0 + at. Adding v 0 to each side of this equation and dividing by 2 gives. v0 + v 2 = v0 + 1 2at. Since v0 + v 2 = ˉv for constant acceleration, we have. ˉv = v0 + 1 2at. Now we substitute this expression for ˉv into the equation for displacement, x = x 0 + ˉv t, yielding. x = x0 + v0t + 1 2at2(constant a).

  10. What is acceleration? (article)

    So a velocity might be "20 m/s, downward". The speed is 20 m/s, and the direction is "downward". Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.

  11. Motion with constant acceleration review

    The strategy they showed in the last 2 videos were as follows. We have 5 variables, acceleration; final velocity; initial velocity; time; displacement. Now what we will do is fill in all the information into these 5 variables. For example the question provides us only the information for: final velocity; initial velocity; time and asks us to find out the displacement.

  12. Acceleration Calculator

    Acceleration is the rate of change of an object's speed; in other words, it's how fast velocity changes. According to Newton's second law, acceleration is directly proportional to the summation of all forces that act on an object and inversely proportional to its mass.It's all common sense - if several different forces are pushing an object, you need to work out what they add up to (they may ...

  13. 3.4 Motion with Constant Acceleration

    Displacement and Position from Velocity. To get our first two equations, we start with the definition of average velocity: v- = Δx Δt. v- = x − x0 t. v- = v0 + v 2. The equation v- = v0+v 2 reflects the fact that when acceleration is constant, v- is just the simple average of the initial and final velocities.

  14. Solved Speed, Velocity, and Acceleration Problems

    Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial. Speed and velocity Problems: Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

  15. Basic Physics: Solving 3 Acceleration Problems: Guided Practice

    More practice solving basic physics acceleration problems. In this video the EXPLAINER shows students how to set up and solve acceleration problems using the...

  16. 1D Kinematics Problem Solving

    Equation Review. The three fundamental equations of kinematics in one dimension are: \ [v = v_0 + at,\] \ [x = x_0 + v_0 t + \frac12 at^2,\] \ [v^2 = v_0^2 + 2a (x-x_0).\] The first gives the change in velocity under a constant acceleration given a change in time, the second gives the change in position under a constant acceleration given a ...

  17. 6.2 Centripetal Acceleration

    Substituting v = rω v = rω into the above expression, we find ac = (rω)2/r = rω2 a c = rω 2 / r = rω 2. We can express the magnitude of centripetal acceleration using either of two equations: ac = v2 r ; ac = rω2. a c = v 2 r ; a c = rω 2. 6.17. Recall that the direction of ac a c is toward the center.

  18. How to solve for ACCELERATION (tagalog)

    A lesson on how to solve for acceleration with sample problem solving. #acceleration #speed #velocity #positiveacceleration #negativeacceleration#formulafofa...

  19. How to solve kinematics problems

    Step 2: Choose the appropriate kinematic equation. We can use the equation: x= x0+v0t+ 1 2 at2 x = x 0 + v 0 t + 1 2 a t 2. Step 3: Solve the equation for acceleration. Since the car is initially at rest and starts from the origin, the equation simplifies to: 200m =0+0+ 1 2 a(10s)2 200 m = 0 + 0 + 1 2 a ( 10 s) 2.

  20. Gravitational Acceleration Physics Problems, Formula & Equations

    This physics video explains how to solve gravitational acceleration problems. This video provides all of the formulas & equations that will help you to calc...

  21. What are the kinematic formulas? (article)

    1. v = v 0 + a t. 2. Δ x = ( v + v 0 2) t. 3. Δ x = v 0 t + 1 2 a t 2. 4. v 2 = v 0 2 + 2 a Δ x. Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing.

  22. Position, velocity, acceleration problems and solutions

    Position, velocity, acceleration problems and solutions. When solving a Physics problem in general and one of Kinematics in particular, it is important that you follow an order. Get used to being organized when you solve problems, and you will see how it gives good results. It is worth spending a little time on the previous analysis of a ...

  23. An inexact Bregman proximal point method and its acceleration version

    The Unbalanced Optimal Transport (UOT) problem plays increasingly important roles in computational biology, computational imaging and deep learning. Scaling algorithm is widely used to solve UOT due to its convenience and good convergence properties. However, this algorithm has lower accuracy for large regularization parameters, and due to stability issues, small regularization parameters can ...

  24. Setting up problems with constant acceleration

    Setting up problems with constant acceleration. Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m s 2 . We want to know how many seconds the basketball is in the air before it hits the ground. We can ignore air resistance. Which kinematic formula would be most useful to solve ...