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Kinematic Equations: Explanation, Review, and Examples

  • The Albert Team
  • Last Updated On: April 29, 2022

problem solving with kinematics equation 1

Now that you’ve learned about displacement, velocity, and acceleration, you’re well on your way to being able to describe just about any motion you could observe around you with physics. All that’s left is to learn how these values really play into each other. We know a few ways to move between them, but they’re all pretty limited. What happens if you need to find displacement, but only know acceleration and time? We don’t have a way to combine all of those values yet. Enter the four kinematic equations. 

What We Review

The Kinematic Equations

The following four kinematic equations come up throughout physics from the earliest high school class to the highest level college course:

Don’t let all of these numbers and symbols intimidate you. We’ll talk through each one – what they mean and when we use them. By the end of this post, you’ll be a master of understanding and implementing each of these physics equations. Let’s start with defining what all of those symbols mean. 

The First Kinematic Equation

This physics equation would be read as “the final velocity is equal to the initial velocity plus acceleration times time”. All it means is that if you have constant acceleration for some amount of time, you can find the final velocity. You’ll use this one whenever you’re looking at changing velocities with a constant acceleration.

The Second Kinematic Equation

This one is read as “displacement equals final velocity plus initial velocity divided by two times time”. You’ll use this one whenever you don’t have an acceleration to work with but you need to relate a changing velocity to a displacement.

The Third Kinematic Equation

This one may look a bit scarier as it is longer than the others, but it is read as “displacement equals initial velocity times time plus one half acceleration times time squared”. All it means is that our displacement can be related to our initial velocity and a constant acceleration without having to find the final velocity. You’ll use this one when final velocity is the only value you don’t know yet.

It is worth noting that this kinematic equation has another popular form: x=x_{0}+v_{0}t+\frac{1}{2}at^{2} . While that may seem even more intimidating, it’s actually exactly the same. The only difference here is that we have split up \Delta x into x-x_{0} and then solved to get x on its own. This version can be particularly helpful if you’re looking specifically for a final or initial position rather than just an overall displacement.

The Fourth Kinematic Equation

Our last kinematic equation is read as “final velocity squared equals initial velocity squared plus two times acceleration times displacement”. It’s worth noting that this is the only kinematic equation without time in it. Many starting physicists have been stumped by reaching a problem without a value for time. While staring at an equation sheet riddled with letters and numbers can be overwhelming, remembering you have this one equation without time will come up again and again throughout your physics career.

It may be worth noting that all of these are kinematic equations for constant acceleration. While this may seem like a limitation, we learned before that high school physics courses generally utilize constant acceleration so we don’t need to worry about it changing yet. If you do find yourself in a more advanced course, new physics equations will be introduced at the appropriate times.

How to Approach a Kinematics Problem

So now that we have all of these different kinematic equations, how do we know when to use them? How can we look at a physics word problem and know which of these equations to apply? You must use problem-solving steps. Follow these few steps when trying to solve any complex problems, and you won’t have a problem.

Step 1: Identify What You Know

This one probably seems obvious, but skipping it can be disastrous to any problem-solving endeavor. In physics problems, this just means pulling out values and directions. If you can add the symbol to go with the value (writing t=5\text{ s} instead of just 5\text{ s} , for example), even better. It’ll save time and make future steps even easier.

Step 2: Identify the Goal

In physics, this means figuring out what question you’re actually being asked. Does the question want you to find the displacement? The acceleration? How long did the movement take? Figure out what you’re being asked to do and then write down the symbol of the value you’re solving for with a question mark next to it ( t=\text{?} , for example). Again, this feels obvious, but it’s also a vital step.

Step 3: Gather Your Tools

Generally, this means a calculator and an equation. You’ll want to look at all of the symbols you wrote down and pick the physics equation for all of them, including the unknown value. Writing everything down beforehand will make it easier to pull a relevant equation than having to remember what values you need while searching for the right equation. You can use the latter method, but you’re far more likely to make a mistake and feel frustrated that way.

Step 4: Put it all Together

Plug your values into your equation and solve for the unknown value. This will usually be your last step, though you may find yourself having to repeat it a few times for exceptionally complex problems. That probably won’t come up for quite a while, though. After you’ve found your answer, it’s generally a good idea to circle it to make it obvious. That way, whoever is grading you can find it easily and you can easily keep track of which problems you’ve already completed while flipping through your work.

Kinematic Equation 1: Review and Examples

To learn how to solve problems with these new, longer equations, we’ll start with v=v_{0}+at . This kinematic equation shows a relationship between final velocity, initial velocity, constant acceleration, and time. We will explore this equation as it relates to physics word problems. This equation is set up to solve for velocity, but it can be rearranged to solve for any of the values it contains. For this physics equation and the ones following, we will look at one example finding the variable that has already been isolated and one where a new variable needs to be isolated using the steps we just outlined. So, let’s jump into applying this kinematic equation to a real-world problem.

A car sits at rest waiting to merge onto a highway. When they have a chance, they accelerate at 4\text{ m/s}^2 for 7\text{ s} . What is the car’s final velocity?

problem solving with kinematics equation 1

We have a clearly stated acceleration and time, but there’s no clearly defined initial velocity here. Instead, we have to take this from context. We know that the car “sits at rest” before it starts moving. This means that our initial velocity in this situation is zero. Other context clues for an object starting at rest is if it is “dropped” or if it “falls”. Our other known values will be even easier to pull as we were actually given numerical values. Now it’s time to put everything into a list.

  • v_{0}=0\text{ m/s}
  • a=4\text{ m/s}^2
  • t=7\text{ s}

Our goal here was clearly stated: find the final velocity. We’ll still want to list that out so we can see exactly what symbols we have to work with on this problem.

We already know which of the kinematic equations we’re using, but if we didn’t, this would be where we search our equation sheet for the right one. Regardless, we’ll want to write that down too.

Step 4: Put it All Together

At this point, we’ll plug all of our values into our kinematic equation. If you’re working on paper, there’s no need to repeat anything we’ve put above. That being said, for the purposes of digital organization and so you can see the full problem in one spot, we will be rewriting things here.

Now let’s get a bit trickier with a problem that will require us to rearrange our kinematic equation.

A ball rolls toward a hill at 3\text{ m/s} . It rolls down the hill for 5\text{ s} and has a final velocity of 18\text{ m/s} . What was the ball’s acceleration as it rolled down the hill?

Just like before, we’ll make a list of our known values:

  • v_{0}=3\text{ m/s}
  • t=5\text{ s}
  • v=18\text{ m/s}

Again, our goal was clearly stated, so let’s add it to our list:

We already know which equation we’re using, but let’s pretend we didn’t. We know that we need to solve for acceleration, but if you look at our original list of kinematic equations, there isn’t one that’s set up to solve for acceleration:

This begs the question, how to find acceleration (or any value) that hasn’t already been solved for? The answer is to rearrange an equation. First, though, we need to pick the right one. We start by getting rid of the second equation in this list as it doesn’t contain acceleration at all. Our options are now:

  • \Delta x=v_{0}t+\dfrac{1}{2}at^{2}
  • v^{2}=v_{0}^{2}+2a\Delta x

Now we’ll need to look at the first list we made of what we know. We know the initial velocity, time, and final velocity. There’s only one equation that has all the values we’re looking for and all of the values we know with none that we don’t. This is the first kinematic equation:

In this case, we knew the kinematic equation coming in so this process of elimination wasn’t necessary, but that won’t often be the case in the future. You’ll likely have to find the correct equation far more often than you’ll have it handed to you. It’s best to practice finding it now while we only have a few equations to work with.

Like before, we’ll be rewriting all of our relevant information below, but you won’t need to if you’re working on paper.

Although you can plug in values before rearranging the equation, in physics, you’ll usually see the equation be rearranged before values are added. This is mainly done to help keep units where they’re supposed to be and to avoid any mistakes that could come from moving numbers and units rather than just a variable. We’ll be taking the latter approach here. Follow the standard PEMDAS rules for rearranging the equation and then write it with the variable we’ve isolated on the left. While that last part isn’t necessary, it is a helpful organizational practice:

For a review of solving literal equations, visit this post ! Now we can plug in those known values and solve:

Kinematic Equation 2: Review and Examples

Next up in our four kinematics equations is \Delta x=\dfrac{v+v_{0}}{2} t . This one relates an object’s displacement to its average velocity and time. The right-hand side shows the final velocity plus the initial velocity divided by two – the sum of some values divided by the number of values, or the average. Although this equation doesn’t directly show a constant acceleration, it still assumes it. Applying this equation when acceleration isn’t constant can result in some error so best not to apply it if a changing acceleration is mentioned.

A car starts out moving at 10\text{ m/s} and accelerates to a velocity of 24\text{ m/s} . What displacement does the car cover during this velocity change if it occurs over 10\text{ s} ?

  • v_{0}=10\text{ m/s}
  • v=24\text{ m/s}
  • t=10\text{ s}
  • \Delta x=\text{?}
  • \Delta x=\dfrac{v+v_{0}}{2} t

This time around we won’t repeat everything here. Instead, We’ll jump straight into plugging in our values and solving our problem:

problem solving with kinematics equation 1

A ball slows down from 15\text{ m/s} to 3\text{ m/s} over a distance of 36\text{ m} . How long did this take?

  • v_{0}=15\text{ m/s}
  • v=3\text{ m/s}
  • \Delta x=36\text{ m}

We don’t have a kinematic equation for time specifically, but we learned before that we can rearrange certain equations to solve for different variables. So, we’ll pull the equation that has all of the values we need and isolate the variable we want later:

Again, we won’t be rewriting anything, but we will begin by rearranging our equation to solve for time:

Now we can plug in our known values and solve for time.

Kinematic Equation 3: Review and Examples

Our next kinematic equation is \Delta x=v_{0}t+\frac{1}{2}at^{2} . This time we are relating our displacement to our initial velocity, time, and acceleration. The only odd thing you may notice is that it doesn’t include our final velocity, only the initial. This equation will come in handy when you don’t have a final velocity that was stated either directly as a number or by a phrase indicating the object came to rest. Just like before, we’ll use this equation first to find a displacement, and then we’ll rearrange it to find a different value.

A rocket is cruising through space with a velocity of 50\text{ m/s} and burns some fuel to create a constant acceleration of 10\text{ m/s}^2 . How far will it have traveled after 5\text{ s} ?

  • v_{0}=50\text{ m/s}
  • a=10\text{ m/s}^2
  • \Delta x=v_{0}t+\frac{1}{2}at^{2}

At this point, it appears that these problems seem to be quite long and take several steps. While that is an inherent part of physics in many ways, it will start to seem simpler as time goes on. This problem presents the perfect example. While it may have been easy to combine lines 4 and 5 mathematically, they were shown separately here to make sure the process was as clear as possible. While you should always show all of the major steps of your problem-solving process, you may find that you are able to combine some of the smaller steps after some time of working with these kinematic equations.

Later in its journey, the rocket is moving along at 20\text{ m/s} when it has to fire its thrusters again. This time it covers a distance of 500\text{ m} in 10\text{ s} . What was the rocket’s acceleration during this thruster burn?

  • v_{0}=20\text{ m/s}
  • \Delta x=500\text{ m}

As usual, we’ll begin by rearranging the equation, this time to solve for acceleration.

Now we can plug in our known values to find the value of our acceleration.

Kinematic Equation 4: Review and Examples

The last of the kinematic equations that we will look at is v^{2}=v_{0}^{2}+2a\Delta x . This one is generally the most complicated looking, but it’s also incredibly important as it is our only kinematic equation that does not involve time. It relates final velocity, initial velocity, acceleration, and displacement without needing a time over which a given motion occurred. For this equation, as with the others, let’s solve it as is and then rearrange it to solve for a different variable.

A car exiting the highway begins with a speed of 25\text{ m/s} and travels down a 100\text{ m} long exit ramp with a deceleration (negative acceleration) of 3\text{ m/s}^2 . What is the car’s velocity at the end of the exit ramp?

  • v_{0}=25\text{ m/s}
  • \Delta x=100\text{ m}
  • a=-3\text{ m/s}^2

Note that our acceleration here is a negative value. That is because our problem statement gave us a deceleration instead of an acceleration. Whenever you have a deceleration, you’ll make the value negative to use it as an acceleration in your problem-solving. This also tells us that our final velocity should be less than our initial velocity so we can add that to the list of what we know as well.

  • Final velocity will be less than initial.

Being able to know something to help check your answer at the end is what makes this subject a bit easier than mathematics for some students.

While we generally try to not have any operations going on for the isolated variable, sometimes it’s actually easier that way. Having your isolated variable raised to a power is generally a time to solve before simplifying. This may seem like an arbitrary rule, and in some ways it is, but as you continue through your physics journey you’ll come up with your own practices for making problem-solving easier.

Now that we have both sides simplified, we’ll take the square root to eliminate the exponent on the left-hand side:

If we remember back at the beginning, we said that our final velocity would have to be less than our initial velocity because the problem statement told us that we were decelerating. Our initial velocity was 25\text{ m/s} which is, indeed, greater than 5\text{ m/s} so our answer checks out.

problem solving with kinematics equation 1

A ghost is sliding a wrench across a table to terrify the mortal onlooker. The wrench starts with a velocity of 2\text{ m/s} and accelerates to a velocity of 5\text{ m/s} over a distance of 7\text{ m} . What acceleration did the ghost move the wrench with?

  • v_{0}=2\text{ m/s}
  • v=5\text{ m/s}
  • \Delta x=7\text{ m}

We can also make an inference about our acceleration here – that it will be positive. Not every problem will tell you clearly the direction of the acceleration, but if your final velocity is greater than your initial velocity, you can be sure that your acceleration will be positive.

  • Positive acceleration

You’ll get better at picking up on subtle hints like this as you continue your physics journey and your brain starts naturally picking up on some patterns. You’ll likely find this skill more and more helpful as it develops and as problems get more difficult.

We’ll start by rearranging our equation to solve for acceleration.

As usual, now that we’ve rearranged our equation, we can plug in our values.

Again, we can go back to the beginning when we said our acceleration would be a positive number and confirm that it is. 

Problem-Solving Strategies

At this point, you’re likely getting the sense that physics will be a lot of complex problem-solving. If so, your senses are correct. In many ways, physics is the science of explaining nature with mathematical equations. There’s a lot that goes into developing and applying these equations, but at this point in your physics career, you’ll find that the majority of your time will likely be spent on applying equations to word problems. If you feel that your problem-solving skills could still use some honing, check out more examples and strategies from this post by the Physics Classroom or through this video-guided tutorial from Khan Academy.

That was a lot of equations and examples to take in. Eventually, whether you’re figuring out how to find a constant acceleration or how to solve velocity when you don’t have a value for time, you’ll know exactly which of the four kinematic equations to apply and how. Just keep the problem-solving steps we’ve used here in mind, and you’ll be able to get through your physics course without any unsolvable problems.

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Module 1 Problem-Solving for Basic Kinematics

Applications.

There are four kinematic equations that describe the motion of objects without consideration of its causes.

Learning Objectives

Choose which kinematics equation to use in problems in which the initial starting position is equal to zero

Key Takeaways

  • The four kinematic equations involve five kinematic variables: [latex]\text{d}[/latex], [latex]\text{v}[/latex], [latex]\text{v}_0[/latex], [latex]\text{a}[/latex], and [latex]\text{t}[/latex].
  • Each equation contains only four of the five variables and has a different one missing.
  • It is important to choose the equation that contains the three known variables and one unknown variable for each specific situation.
  • kinematics : The branch of physics concerned with objects in motion.

Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. There are four kinematic equations when the initial starting position is the origin, and the acceleration is constant:

  • [latex]\text{v} = \text{v}_0 + \text{at}[/latex]
  • [latex]\text{d} = \frac{1}{2}(\text{v}_0 + \text{v})\text{t}[/latex] or alternatively [latex]\text{v}_{\text{average}} = \frac{\text{d}}{\text{t}}[/latex]
  • [latex]\text{d} = \text{v}_0\text{t} + (\frac{\text{at}^2}{2})[/latex]
  • [latex]\text{v}^2 = \text{v}_0^2 + 2\text{ad}[/latex]

Notice that the four kinematic equations involve five kinematic variables: [latex]\text{d}[/latex] , [latex]\text{v}[/latex] , [latex]\text{v}_0[/latex] , [latex]\text{a}[/latex] , and [latex]\text{t}[/latex]. Each of these equations contains only four of the five variables and has a different one missing. This tells us that we need the values of three variables to obtain the value of the fourth and we need to choose the equation that contains the three known variables and one unknown variable for each specific situation.

Here the basic problem solving steps to use these equations:

Step one – Identify exactly what needs to be determined in the problem (identify the unknowns).

Step two – Find an equation or set of equations that can help you solve the problem.

Step three – Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units.

Step four – Check the answer to see if it is reasonable: Does it make sense?

Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in a physics class and for applying physics in everyday and professional life.

  • Curation and Revision. Provided by : Boundless.com. License : CC BY-SA: Attribution-ShareAlike
  • OpenStax College, College Physics. September 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42125/latest/?collection=col11406/1.7 . License : CC BY: Attribution
  • kinematics. Provided by : Wiktionary. Located at : http://en.wiktionary.org/wiki/kinematics . License : CC BY-SA: Attribution-ShareAlike
  • Motion diagram. Provided by : Wikipedia. Located at : http://en.wikipedia.org/wiki/Motion_diagram . License : CC BY-SA: Attribution-ShareAlike
  • motion. Provided by : Wiktionary. Located at : http://en.wiktionary.org/wiki/motion . License : CC BY-SA: Attribution-ShareAlike
  • diagram. Provided by : Wiktionary. Located at : http://en.wiktionary.org/wiki/diagram . License : CC BY-SA: Attribution-ShareAlike
  • stroboscopic. Provided by : Wikipedia. Located at : http://en.wikipedia.org/wiki/stroboscopic . License : CC BY-SA: Attribution-ShareAlike
  • Boundless. Provided by : Amazon Web Services. Located at : http://s3.amazonaws.com/figures.boundless.com/51129b2ee4b0c14bf464ce1b/1.jpg . License : CC BY: Attribution
  • Provided by : Wikimedia. Located at : http://upload.wikimedia.org/wikipedia/commons/3/3c/Bouncing_ball_strobe_edit.jpg . License : CC BY: Attribution

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1D Kinematics Problem Solving

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Recommended Course

Classical mechanics.

Hardcore training for the aspiring physicist.

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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1.6: Motion Equations for Constant Acceleration in One Dimension

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Learning Objectives

  • Calculate displacement of an object that is not accelerating, given initial position and velocity.
  • Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
  • Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.

fig-ch01_patchfile_01.jpg

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

Notation:  t ,  x ,  v ,  a

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 \(\Delta t=t_{\mathrm{f}}-t_{0}\), taking \(t_{0}=0\) means that \(\Delta t=t_{\mathrm{f}}\), 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—now, \(\Delta t=t\). It also simplifies the expression for 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,

\[\left.\begin{array}{l} \Delta t=t \\ \Delta x=x-x_{0} \\ \Delta v=v-v_{0} \end{array}\right\} \nonumber \]

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,

\[\bar{a}=a=\text { constant } \nonumber \]

so we use the symbol \(a\) for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor 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 accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration.

SOLVING FOR DISPLACEMENT (Δx) AND FINAL POSITION (\(x\)) FROM AVERAGE VELOCITY WHEN ACCELERATION (\(a\)) IS CONSTANT

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

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

Substituting the simplified notation for Δx and Δt yields

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

Solving for \(x\) yields

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

where the average velocity is

\[\bar{v}=\frac{v_{0}+v}{2}(\operatorname{constant} a). \nonumber \]

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. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation \(\bar{v}=\frac{v_{0}+v}{2}\) to check this, we see that

\[\bar{v}=\frac{v_{0}+v}{2}=\frac{30 \mathrm{~km} / \mathrm{h}+60 \mathrm{~km} / \mathrm{h}}{2}=45 \mathrm{~km} / \mathrm{h}, \nonumber\]

which seems reasonable.

Example \(\PageIndex{1}\): Calculating Displacement: How Far does the Jogger Run?

A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?

Draw a sketch.

fig-ch01_patchfile_01.jpg

Figure \(\PageIndex{2}\)

The final position \(x\) is given by the equation

\[x=x_{0}+\bar{v} t. \nonumber\]

To find \(x\), we identify the values of \(x_{0}\), \(\bar{v}\), and \(t\) from the statement of the problem and substitute them into the equation.

1. Identify the knowns. \(\bar{v}=4.00 \mathrm{~m} / \mathrm{s}, \ \Delta t=2.00 \mathrm{~min}\), and \(x_{0}=0 \mathrm{~m}\).

2. Enter the known values into the equation.

\[x=x_{0}+\bar{v} t=0+(4.00 \mathrm{~m} / \mathrm{s})(120 \mathrm{~s})=480 \mathrm{~m} \nonumber \]

Velocity and final displacement are both positive, which means they are in the same direction.

The equation \(x=x_{0}+\bar{v} t\) gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on \(\bar{v}\) rather than on \(\bar{v}\) raised to some other power, such as \(\bar{v}^{2}\). When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h.

fig-ch01_patchfile_01.jpg

SOLVING FOR FINAL VELOCITY

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

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

Substituting the simplified notation for Δv and Δt gives us

\[a=\frac{v-v_{0}}{t}(\text { constant } a). \nonumber \]

Solving for \(v\) yields

\[v=v_{0}+a t(\text { constant } a). \nonumber \]

Example \(\PageIndex{2}\): Calculating Final Velocity: An Airplane Slowing Down after Landing

An airplane lands with an initial velocity of 70.0 m/s and then slows down at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) for 40.0 s. What is its final velocity?

Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is slowing down.

fig-ch01_patchfile_01.jpg

Figure \(\PageIndex{4}\)

1. Identify the knowns. \(v_{0}=70.0 \mathrm{~m} / \mathrm{s}, \ a=-1.50 \mathrm{~m} / \mathrm{s}^{2}, \ t=40.0 \mathrm{~s}\).

2. Identify the unknown. In this case, it is final velocity, \(v_{\mathrm{f}}\).

3. Determine which equation to use. We can calculate the final velocity using the equation \(v=v_{0}+a t\).

4. Plug in the known values and solve.

\[v=v_{0}+a t=70.0 \mathrm{~m} / \mathrm{s}+\left(-1.50 \mathrm{~m} / \mathrm{s}^{2}\right)(40.0 \mathrm{~s})=10.0 \mathrm{~m} / \mathrm{s} \nonumber\]

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

fig-ch01_patchfile_01.jpg

In addition to being useful in problem solving, the equation \(v=v_{0}+a t\) gives us insight into the relationships among velocity, acceleration, and time. From it 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 \(\left(v=v_{0}\right)\), as expected (i.e., velocity is constant)
  • if \(a\) is negative, then the final velocity is less than the initial velocity

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

MAKING CONNECTIONS: REAL-WORLD CONNECTION

fig-ch01_patchfile_01.jpg

An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified—short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.

SOLVING FOR FINAL POSITION WHEN VELOCITY IS NOT CONSTANT (a≠0)

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

\[v=v_{0}+a t. \nonumber\]

Adding \(v_{0}\) to each side of this equation and dividing by 2 gives

\[\frac{v_{0}+v}{2}=v_{0}+\frac{1}{2} a t. \nonumber\]

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

\[\bar{v}=v_{0}+\frac{1}{2} a t. \nonumber \]

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

\[x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}(\text { constant } a). \nonumber \]

Example \(\PageIndex{3}\): Calculating Displacement of an Accelerating Object: Dragsters

Dragsters can achieve average accelerations of \(26.0 \mathrm{~m} / \mathrm{s}^{2}\). Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

fig-ch01_patchfile_01.jpg

Figure \(\PageIndex{8}\)

We are asked to find displacement, which is \(x\) if we take \(x_{0}\) to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation \(x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}\) once we identify \(v_{0}\), \(a\), and \(t\) from the statement of the problem.

1. Identify the knowns. Starting from rest means that \(v_{0}=0\), \(a\) is given as \(26.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(t\) is given as 5.56 s.

2. Plug the known values into the equation to solve for the unknown \(x\):

\[x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}. \nonumber\]

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

\[x=\frac{1}{2} a t^{2}. \nonumber\]

Substituting the identified values of \(a\) and \(t\) gives

\[x=\frac{1}{2}\left(26.0 \mathrm{~m} / \mathrm{s}^{2}\right)(5.56 \mathrm{~s})^{2}, \nonumber\]

\[x=402 \mathrm{~m}. \nonumber\]

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 the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.

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

  • displacement depends on the square of the elapsed time when acceleration is not zero. In Example \(\PageIndex{3}\), the dragster covers only one fourth of the total distance in the first half of the elapsed time
  • if acceleration is zero, then the initial velocity equals average velocity \(\left(v_{0}=\bar{v}\right)\) and \(x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}\) becomes \(x=x_{0}+v_{0} t\)

SOLVING FOR FINAL VELOCITY WHEN VELOCITY IS NOT CONSTANT (a≠0)

A fourth useful equation can be obtained from another algebraic manipulation of previous equations.

If we solve \(v=v_{0}+a t\) for \(t\), we get

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

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

\[v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)(\text { constant } a). \nonumber \]

Example \(\PageIndex{4}\): Calculating Final Velocity: Dragsters

Calculate the final velocity of the dragster in Example \(\PageIndex{3}\) without using information about time.

fig-ch01_patchfile_01.jpg

Figure \(\PageIndex{9}\)

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

1. Identify the known values. We know that \(v_{0}=0\), since the dragster starts from rest. Then we note that \(x-x_{0}=402 \mathrm{~m}\) (this was the answer in Example \(\PageIndex{3}\)). Finally, the average acceleration was given to be \(a=26.0 \mathrm{~m} / \mathrm{s}^{2}\).

2. Plug the knowns into the equation \(v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)\) and solve for \(v\).

\[v^{2}=0+2\left(26.0 \mathrm{~m} / \mathrm{s}^{2}\right)(402 \mathrm{~m}). \nonumber\]

\[v^{2}=2.09 \times 10^{4} \mathrm{~m}^{2} / \mathrm{s}^{2}. \nonumber\]

To get \(v\), we take the square root:

\[v=\sqrt{2.09 \times 10^{4} \mathrm{~m}^{2} / \mathrm{s}^{2}}=145 \mathrm{~m} / \mathrm{s} . \nonumber\]

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 \(v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)\) can produce further 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 magnitude of acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)

Putting Equations Together

In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.

SUMMARY OF KINEMATIC EQUATIONS (CONSTANT \(a\))

\[\begin{gathered} x=x_{0}+\bar{v} t \\ \bar{v}=\frac{v_{0}+v}{2} \\ v=v_{0}+a t \\ x=x_{0}+v_{0} t+\frac{1}{2} a t^{2} \\ v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right) \end{gathered} \nonumber\]

Example \(\PageIndex{5}\): Calculating Displacement: How Far Does a Car Go When Coming to a Halt?

On dry concrete, suppose a car can slow down at a rate of \(7.00 \mathrm{~m} / \mathrm{s}^{2}\), whereas on wet concrete the maximum magnitude of acceleration is only \(5.00 \mathrm{~m} / \mathrm{s}^{2}\). Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

fig-ch01_patchfile_01.jpg

Figure \(\PageIndex{10}\)

In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.

Solution for (a)

1. Identify the knowns and what we want to solve for. We know that \(v_{0}=30.0 \mathrm{~m} / \mathrm{s} ; \ v=0 ; \ a=-7.00 \mathrm{~m} / \mathrm{s}^{2}\) (\(a\) is negative because it is in a direction opposite to velocity). We take \(x_{0}\) to be 0. We are looking for displacement \(\Delta x\), or \(x-x_{0}\).

2. Identify the equation that will help up solve the problem. The best equation to use is

\[v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right). \nonumber\]

This equation is best because it includes only one unknown, \(x\). We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for \(x\), but they require us to know the stopping time, \(t\), which we do not know. We could use them but it would entail additional calculations.)

3. Rearrange the equation to solve for \(x\).

\[x-x_{0}=\frac{v^{2}-v_{0}^{2}}{2 a} \nonumber\]

4. Enter known values.

\[x-0=\frac{0^{2}-(30.0 \mathrm{~m} / \mathrm{s})^{2}}{2\left(-7.00 \mathrm{~m} / \mathrm{s}^{2}\right)} \nonumber\]

\[x=64.3 \mathrm{~m} \text { on dry concrete. } \nonumber\]

Solution for (b)

This part can be solved in exactly the same manner as Part A. The only difference is that the acceleration is \(-5.00 \mathrm{~m} / \mathrm{s}^{2}\). The result is

\[x_{\text {wet }}=90.0 \mathrm{~m} \text { on wet concrete. }. \nonumber\]

Solution for (c)

Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver’s reaction time.

1. Identify the knowns and what we want to solve for. We know that \(\bar{v}=30.0 \mathrm{~m} / \mathrm{s} ; \ t_{\text {reaction }}=0.500 \mathrm{~s} ; \ a_{\text {reaction }}=0\). We take \(x_{0 \text {-reaction }}\) to be 0. We are looking for \(x_{reaction}\).

2. Identify the best equation to use.

\(x=x_{0}+\bar{v} t\) works well because the only unknown value is \(x\), which is what we want to solve for.

3. Plug in the knowns to solve the equation.

\[x=0+(30.0 \mathrm{~m} / \mathrm{s})(0.500 \mathrm{~s})=15.0 \mathrm{~m}. \nonumber\]

This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.

4. Add the displacement during the reaction time to the displacement when braking.

\[x_{\text {braking }}+x_{\text {reaction }}=x_{\text {total }} \nonumber\]

  • 64.3 m + 15.0 m = 79.3 m when dry
  • 90.0 m + 15.0 m = 105 m when wet

fig-ch01_patchfile_01.jpg

The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.

With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships.

MAKING CONNECTIONS: TAKE-HOME EXPERIMENT—BREAKING NEWS

We have been using SI units of meters per second squared to describe some examples of acceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking acceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, \(\bar{a}=\Delta v / \Delta t\). While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the acceleration in miles per hour per second. Convert this to meters per second squared and compare with other accelerations mentioned in this chapter. Calculate the distance traveled in braking.

Exercise \(\PageIndex{1}\)

A manned rocket accelerates at a rate of \(20 \mathrm{~m} / \mathrm{s}^{2}\) during launch. How long does it take the rocket reach a velocity of 400 m/s?

To answer this, choose an equation that allows you to solve for time \(t\), given only \(a\), \(v_{0}\), and \(v\).

\[v=v_{0}+a t \nonumber\]

Rearrange to solve for \(t\) .

\[t=\frac{v-v_{0}}{a}=\frac{400 \mathrm{~m} / \mathrm{s}-0 \mathrm{~m} / \mathrm{s}}{20 \mathrm{~m} / \mathrm{s}^{2}}=20 \mathrm{~s} \nonumber\]

Section Summary

  • To simplify calculations we take acceleration to be constant, so that \(\bar{a}=a\) at all times.
  • We also take initial time to be zero.

\[\left.\begin{array}{l} \Delta t=t \\ \Delta x=x-x_{0} \\ \Delta v=v-v_{0} \end{array}\right\} \nonumber\]

  • In vertical motion, \(y\) is substituted for \(x\).

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Mechanics: 1-Dimensional Kinematics

1-d kinematics: problem set overview.

There are 23 ready-to-use problem sets on the topic of 1-Dimensional Kinematics. The problems target your ability to use the average velocity and average acceleration equations, to interpret position-time and velocity-time graphs, and to use the kinematic equations to determine the answer to problems, including those which involve a free fall acceleration. Problems range in difficulty from the very easy and straight-forward to the very difficult and complex.  

Average Speed, Average Velocity, and Average Acceleration Equations

The equations for average velocity ( v ave ) and average acceleration ( a ave ) are stated below.

problem solving with kinematics equation 1

The ∆time is the change in time over which the position change or velocity change is measured. When using a stopwatch to measure this time, the ∆time is simply the time reading on the stopwatch. We refer to this as t in the equations. An inspection of these equations shows that there are clear mathematical relationships between the following sets of quantities:

From the top equation:

average velocity (v ave ), initial velocity (v i ), final velocity (v f ), displacement and time

From the bottom equation:

average acceleration (a ave ), initial velocity (v i ), final velocity (v f ) and time

Position-time graphs represent variations in an object's position over the course of time. Thus, an inspection of the values being plotted along the vertical axis allows one to determine position values and changes in position values. These axis values are related to the distance and displacement for an object's motion. The displacement of an object is simply the overall change in position; it considers only the initial and the final positions and ignores all in-between positions. Being a vector quantity, it includes a direction that is often expressed as a positive or a negative sign (e.g., + for rightward and - for leftward). The distance traveled by an object is the accumulation of all the ground covered. In effect, it is the sum of all the changes in position for each leg of a trip; such sums must ignore the + and - signs since distance is a scalar quantity. Being a scalar quantity, it is ignorant of direction and does not distinguish between a change in position to the left or to the right. If a person walks 2 meters, left and then 6 meters right, then the displacement is 4 meters to the right and the distance is 8 meters.

Slopes of such position-time graphs yield a ratio of the position change to time change for any specified interval of time. As such, the slopes are equivalent to the velocity of an object. Velocity (slope in this case) is a vector quantity that has a direction associated with it. Mathematically, the velocity direction is often represented by a positive or a negative value. Upward-sloping lines are associated with positive velocities and downward-sloping lines are associated with negative velocities. For a complex motion with multiple slopes, the average velocity can be determined by connecting the initial and final position and calculating the slope of the connecting line. Alternatively, the average velocity could be determined by reading the overall position change (displacement) off the axis and dividing by the change in time.

The average speed for any motion is simply the distance to time ratio. For a simple motion involving a constant velocity, the average speed is the absolute value of the slope. For a complex motion with several slopes, the average speed is the overall distance traveled divided by the time of travel. As mentioned above, the overall distance traveled can be determined by summing the absolute value of the individual position changes for each leg of the motion.  

Velocity-time plots represent variations in an object's velocity over the course of time. As such, the slope of such graphs provides a velocity change to time change ratio; this ratio represents the acceleration of the object. Some of the questions in these problem sets will test your ability to use the coordinates for two points on a line to determine the slope. The slope is simply the change in the y coordinates divided by the change in the x-coordinate. You can use our video to learn more about the process.

Velocity-time plots can also be used to determine the distance traveled by an object and the displacement of an object. The area between a plotted line and the time axis (which could be a positive or a negative value) represents the displacement of the object. The area spoken of here is formed by the plotted line, the time axis and two imaginary vertical lines drawn at two specified points in time. The resulting area could be represented by either a rectangle, a triangle, a trapezoid or a collection of such shapes. The graphic below depicts a variety of resulting shapes and the appropriate formulas for calculating areas from each of them.

problem solving with kinematics equation 1

For a complex motion like that represented below, the total area can be broken into a collection of smaller shapes. The area of the individual sections can then be added together to determine the total area.

problem solving with kinematics equation 1

The Big Four - Kinematic Equations

Kinematics (the topic of the current unit) is the science of describing the motion of objects. An object's motion can be described using words, diagrams, numbers, graphs and equations. The most commonly used of all equations are the four kinematic equations - affectionately known as the big four. These four equations allow a student to make a prediction of how fast (velocity and speed), how far (displacement and distance), or how much time is required of an object during a motion. The four equations are listed below.

  • d = v o • t + 0.5 • a • t 2

problem solving with kinematics equation 1

  • v f 2 = v o 2 + 2 • a • d
  • d = (v o + v f )/ 2 • t
  • d = displacement
  • a = acceleration
  • v o = original or initial velocity
  • v f = final velocity

Each of the above kinematic equations have four variables. The usefulness of the equations is that they allow a person to make a prediction about the value of one of the variables if given the value of three other variables. By knowing three, one can calculate a fourth. The problem-solving strategy used in this collection of problems will center around this idea. Each problem consists of a word-story problem in which information about an object's motion is given. The goal is to carefully read through each story problem to identify at least three pieces of known information in order to calculate a fourth requested piece of information. Often the known information is explicitly stated - "A car moving with an initial velocity of 23.4 m/s...". At other times there are statements included within the word problem such as "Starting from rest, ...? or "...comes to a stop." Such statements imply that the initial velocity is 0 m/s and the final velocity is 0 m/s (respectively).

While the equations are extremely useful, there is one condition which must be met in order for the equations to be used. The object under study must have a constant and uniform acceleration. If an object changes its acceleration at a given point during the motion such that it accelerates at one rate and then later accelerates at a second rate, then the motion must be divided into two phases and each phase must be analyzed separately.  

Free Fall and Acceleration Caused by Gravity

Some of the problems in these problem sets involve free fall type motion. Free fall is a type of motion in which the only force acting upon the moving object is the force of gravity. When an object is in free fall, its motion is influenced solely by the force of gravity and the object accelerates at 9.8 m/s/s. The 9.8 m/s/s value is the acceleration of any free falling object irregardless of the characteristics of the object. Such an acceleration is dependent solely upon the gravitational characteristics of the planet and is thus called the acceleration of gravity  or acceleration caused by gravity and represented by the variable g . While the value of g is 9.8 m/s/s on Earth, it is a different value on other planets where the gravitational characteristics are different. NOTE: Some classes use a rounded value of 10 m/s/s to simplify calculations.)

There are some unique aspects about the trajectory of a free falling object which are worth noting and serve very useful in one's approach to solving problems. Any object that is originally projected vertically upward and under the sole influenc of gravity will eventually reach a peak height before turning around and falling back downward. At the peak of the trajectory, the velocity of the object is momentarily 0 m/s. The time required to reach the peak is equal to the time required to fall from the peak back to its original position. The total time of flight is thus twice the time required to reach the peak. The velocity of the object one second prior to the peak is of the same magnitude as the velocity of the object one second after reaching the peak. And similarly, the velocity of the object two seconds prior to the peak is of the same magnitude as the velocity of the object two seconds after reaching the peak. You may find our video on Solving Free Fall Problems to be useful.  

Habits of an Effective Problem-Solver

An effective problem solver by habit approaches a physics problem in a manner that reflects a collection of disciplined habits. While not every effective problem solver employs the same approach, they all have habits which they share in common. These habits are described briefly here. An effective problem-solver...

  • ...reads the problem carefully and develops a mental picture of the physical situation. If needed, they sketch a simple diagram of the physical situation to help visualize it.
  • ...identifies the known and unknown quantities in an organized manner, often times recording them on the diagram itself. They equate given values to the symbols used to represent the corresponding quantity (e.g., v o = 0 m/s, a = 2.67 m/s/s, v f = ???).
  • ...plots a strategy for solving for the unknown quantity; the strategy will typically centers around the use of physics equations is heavily dependent upon an understanding of physics principles.
  • ...identifies the appropriate formula(s) to use, often times writing them down. Where needed, they perform the needed conversion of quantities into the proper unit.
  • ...performs substitutions and algebraic manipulations in order to solve for the unknown quantity.

Read more...  

Additional Readings/Study Aids:

The following pages from The Physics Classroom tutorial may serve to be useful in assisting you in the understanding of the concepts and mathematics associated with these problems.

  • Distance and Displacement
  • Speed and Velocity
  • Meaning of Shape on a Position-Time Graph
  • Meaning of Slope on a Position-Time Graph
  • Calculating the Slope of a Position-Time Graph
  • Meaning of Shape for a v-t Graph
  • Meaning of Slope for a v-t Graph
  • Relating the Shape to the Motion
  • Determining the Slope on a v-t Graph
  • Determining the Area on a v-t Graph
  • The Kinematic Equations
  • Problem-Solving
  • Sample Problems and Solutions
  • The Acceleration of Gravity
  • Representing Free Fall by Graphs
  • How Fast? and How Far?

Watch a Video

We have developed and continue to develop Video Tutorials on introductory physics topics. You can find these videos on our YouTube channel . We have an entire Playlist on the topic of Kinematics .  

IMAGES

  1. Solving problems with kinematic equations 1

    problem solving with kinematics equation 1

  2. Problem Solving in Kinematics Example

    problem solving with kinematics equation 1

  3. Solving Problems Using Kinematic Equations

    problem solving with kinematics equation 1

  4. PPT

    problem solving with kinematics equation 1

  5. Kinematics Formula

    problem solving with kinematics equation 1

  6. Kinematics

    problem solving with kinematics equation 1

VIDEO

  1. Physics 11 Rearranging Kinematics Equations Part 2

  2. Kinematics Problem Solving

  3. KINEMATICS CLASS 11|PROJECTILE MOTION |IITJEE|NEET|LECTURE 5

  4. 1D Kinematics

  5. 2.3 Types of fluid flow

  6. How to solve Kinematics 1D Problems from Basics for Class11, JEE and NEET

COMMENTS

  1. Kinematic Equations: Sample Problems and Solutions

    See solution below. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s 2. Determine the time for the feather to fall to the surface of the moon. See Answer See solution below. Rocket-powered sleds are used to test the human response to acceleration.

  2. Kinematic Equations: Explanation, Review, and Examples

    v=v_ {0}+at v = v0 +at This physics equation would be read as "the final velocity is equal to the initial velocity plus acceleration times time". All it means is that if you have constant acceleration for some amount of time, you can find the final velocity.

  3. 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.

  4. Choosing kinematic equations (video)

    Kinematic equations help solve for an unknown in a problem when an object has either a constant velocity or constant acceleration. This video will help you choose which kinematic equations you should use, given the type of problem you're working through. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted Kajsa Fröberg

  5. 2.4: Problem-Solving for Basic Kinematics

    Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. There are four kinematic equations when the initial starting position is the origin, and the acceleration is constant: v = v0 + at v = v 0 + a t.

  6. 2.6: Problem-Solving Basics for One-Dimensional Kinematics

    This page titled 2.6: Problem-Solving Basics for One-Dimensional Kinematics is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. The ability to apply broad physical ...

  7. Kinematics In One Dimension

    It explains how to solve one-dimensional motion problems using kinematic equations and formulas with objects moving at constant speed and with constant acceleration. This video contains...

  8. Choosing kinematic equations

    Choosing kinematic equations | One-dimensional motion | AP Physics 1 | Khan Academy - YouTube AP Physics 1: Algebra-Based Courses on Khan Academy are always 100% free. Start...

  9. Kinematic formulas in one-dimension (practice)

    Kinematic formulas in one-dimension. A child blows a leaf straight up in the air. The leaf reaches 1.0 m higher than its original height with a constant acceleration of 1.0 m s 2 upward. How much time did it take the leaf to get displaced by 1.0 m ?

  10. Using the Kinematic Equations to Solve Problems

    0:00 / 10:28 • Introduction Using the Kinematic Equations to Solve Problems - Part 1 The Physics Classroom 34.6K subscribers Subscribe Subscribed 1.4K 77K views 4 years ago Kinematics...

  11. PDF Using the Kinematic Equations to Solve Problems

    Find the kinematic equation that contains these four variables. Write the equation down. 4. Substitute known values into this equation. 5. Perform algebra and calculations to solve for the unknown variable. Example 1 Starting from rest, a car accelerates at 6.52 m/s2 for 3.80 s. Determine the distance traveled by the car during this time. Known ...

  12. Kinematics (Description of Motion) Problems

    There are three key kinematic equations. If you carefully select the equation which most directly describes the situation in your problem, you will not only solve the problem in fewer steps but also understand it better. The three equations, written for motion in the x-direction, are: x = x 0 + v 0 Δt + ½ a (Δt) 2 (relates position and time)

  13. Module 1 Problem-Solving for Basic Kinematics

    Key Terms kinematics: The branch of physics concerned with objects in motion. Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion.

  14. 1.1: Kinematics

    Conversely, a positive acceleration means that the change in the velocity points in the positive direction. Kinematics is the correct use of the parameters position, velocity, and acceleration to describe motion. Learning to use these three terms correctly can be made much easier by learning a few tricks of the trade.

  15. 1D Kinematics Problem Solving

    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}. vf 2 = 2(9.8 m/s2)(100 m) vf ≈ 44.3 m/s. The first kinematical equation gives the time to accelerate up to this speed:

  16. 2.4: Problem-Solving for Basic Kinematics

    Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. There are four kinematic equations when the initial starting position is the origin, and the acceleration is constant: v = v0 + at v = v 0 + a t.

  17. One-dimensional motion

    Choosing kinematic equations (Opens a modal) Practice. Setting up problems with constant acceleration Get 5 of 7 questions to level up! Kinematic formulas in one-dimension Get 5 of 7 questions to level up! Quiz 2. Level up on the above skills and collect up to 160 Mastery points Start quiz.

  18. 1.6: Motion Equations for Constant Acceleration in One Dimension

    Figure \(\PageIndex{1}\): Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr) ... In addition to being useful in problem solving, the equation \(v=v_{0}+a t\) gives us insight into the relationships among velocity, acceleration, and ...

  19. Equation Overview for 1-Dimensional Kinematics Problem Sets

    1-D Kinematics: Problem Set Overview There are 23 ready-to-use problem sets on the topic of 1-Dimensional Kinematics. The problems target your ability to use the average velocity and average acceleration equations, to interpret position-time and velocity-time graphs, and to use the kinematic equations to determine the answer to problems, including those which involve a free fall acceleration.