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MCAT Physical : Motion in Two Dimensions

Study concepts, example questions & explanations for mcat physical, all mcat physical resources, example questions, example question #1 : motion in two dimensions.

A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.

What is the horizontal acceleration of the ball during its flight?

Without an additional force acting in the horizontal direction during flight (we are told we can neglect air resistance), there is no acceleration. Remember that Newton's second law, F = ma, requires that a force act on an object to produce acceleration. Here, we have no additional force and thus no acceleration.

Example Question #2 : Motion In Two Dimensions

A ball of mass 3kg is thrown into the air at an angle of 45 o above the horizontal, with initial velocity of 15m/s. Instantaneously at the highest point in its motion, the ball comes to rest. Approximately what is the magnitude of acceleration at this point? Assume that air resistance is negligible.

After being thrown, the ball is only acted on by the force of gravity. Since this force is constant throughout the motion, acceleration must also remain constant, and be equal to the gravitational acceleration of 9.8 m/s 2 (approximately 10 m/s 2 )

Example Question #3 : Motion In Two Dimensions

kinematics practice problems mcat

We need to calculate the vertical component on the daredevil's velocity using the equation:

kinematics practice problems mcat

Given the angle of trajectory and the initial velocity, we can solve.

kinematics practice problems mcat

At the maximum height, the daredevil will have zero vertical velocity. Using this as the final vertical velocity, we can calculate the distance traveled using the appropriate kinematics equation.

kinematics practice problems mcat

Example Question #4 : Motion In Two Dimensions

Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top. 

They then decide to sled down the hill, but disagree about who will go first. 

Scenario 1:

Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.

Scenario 2:

Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.

Scenario 3:

Unable to agree, Sam and Sally tether themselves with a rope and go down together.

In Scenario 3, before they begin moving, Sam is hanging down the hill facing forward, and Sally has fixed her shoes in the snow to keep her and Sam from beginning down the hill together. If the hill is at a 30 o angle to the horizontal, and Sam is stationary but tied to Sally by a rope, how much force does Sam experience down the surface of the hill?

kinematics practice problems mcat

F = (60kg)(10m/s 2 ) * sin(30 o ) = 300J

Example Question #5 : Motion In Two Dimensions

In Scenario 3, before they begin moving, Sam is hanging down the hill face first, and Sally has fixed her shoes in the snow to keep her and Sam from beginning down the hill together. If the hill is at a 30 o angle to the horizontal, and Sam is stationary but tied to Sally by a rope, how much tension is in the rope?

kinematics practice problems mcat

F = 60kg * 10m/s 2 * sin(30) = 300J

Tension in the rope will be equal to this value, as these forces must cancel to result in a net force of zero.

Example Question #6 : Motion In Two Dimensions

Which of the following expressions shows the force that Sally and Sam would feel down the hill at a 30 o incline if they both rode on Sam’s sled together?

kinematics practice problems mcat

Example Question #7 : Motion In Two Dimensions

A 2kg box is at the top of a frictionless ramp at an angle of 60 o . The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

Which of these statements is true?

As the angle of the ramp is decreases, the acceleration on the box due to gravity decreases

The force on the box is greater at the bottom of the ramp than at the top of the ramp

The box has a greater acceleration at the bottom of the ramp

The normal force is always perpendicular to the angle of the ramp

Normal force will always be perpendicular to the surface in question.

kinematics practice problems mcat

Example Question #8 : Motion In Two Dimensions

If the angle of the ramp is decreased, which of the following statements is false?

The normal force will decrease

The net force on the box will decrease

The force due to gravity will decrease

The time to reach the ground will increase

When given a question about the angle of a ramp, compare it to the extreme angles: 0 o and 90 o .

1. When the ramp has an angle of 0 o , the net force 0. The force due to gravity must equal the normal force; thus the normal force is at a maximum value.

kinematics practice problems mcat

When we decrease the angle of the ramp, we get closer to scenario 1. As a result, we can conclude that the normal force on the box increases, rather than decreases.

Example Question #9 : Motion In Two Dimensions

Find the horizontal component of velocity once the ball has left the cannon.

This asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle (solved to be 10m/s in the previous problem), and that by knowing the hypotenuse value we can solve for the horizontal component by using cosine.

v x = (10m/s)(cos(45 o )) = 7.1m/s

Example Question #101 : Newtonian Mechanics And Motion

What is the initial vertical component of velocity of the ball?

This question also asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle, and that by knowing the hypotenuse value (solved as 10m/s in a previous problem) we can solve for the vertical component using sine.

v y = (10m/s)(sin(45 o )) = 7.1m/s

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Free MCAT Practice Question – Physics

  • by Allison Chae
  • May 10, 2013
  • MCAT Blog, MCAT Long Form, MCAT Prep
  • Reviewed By: Liz Flagge

Mastering the Physical Sciences portion of the MCAT first requires mastering the fundamental science concepts that the MCAT will test.

By far, the physics questions the MCAT most likes to ask relate to kinematics and forces. Try your hand at this sample problem (click to enlarge):

Blog-Discrete-P1a

Explanation

Like most questions on the MCAT, this one can be approached a couple of different ways.

First, if you have no idea what you’re doing, you’re going to have to guess. The MCAT covers a huge array of topics, and sometimes we’ll see a topic we just didn’t get time to study. If that happens, don’t panic. It’s okay to guess – in fact, you have to! Don’t leave anything blank on the test! If you’re going to have to guess, guess smart.

Try to avoid picking answers that are just the simple addition or subtraction of numbers listed in the question itself. For example, in this question you probably shouldn’t guess 9, since that’s just the subtraction of the 10 and the 1 listed in the question. That’s a “panic answer” and the MCAT writers put it in there as a trap. Just guess between A, B, C and move on quickly. There’s more questions to get right!

Next, let’s say you remember a bit of physics but don’t really know how to solve this question. It’s hugely important on the test to be familiar with units. In fact, I routinely tell my tutoring students that if they learn nothing else in physics, they need to LEARN UNITS! In this question, we’re asked for acceleration. Two of the answer choices have the units listed as Newtons. That’s wrong! Acceleration is in meters per seconds squared. If you know your units, you can quickly eliminate B and D. Just pick either A or C and move on.

Finally, to answer this question down you’re going to have to break down the applied 10N force. The child is sliding the block horizontally along the floor, so you need the horizontal component of the applied force. Remember that you use cos to find the horizontal components of force. The equation would thus be:

Fx = 10N x cos 60 = 10 x 0.5 = 5 N

We’re told that the frictional force is 1 N. The frictional force would oppose the applied force, so we would find the net force by subtracting friction:

F net = 5 N – 1 N = 4 N

Now that we know the net force, we can use Newton’s Second Law, F = ma, to find the answer. This is a biggie! F=ma is probably the most important equation in all of MCAT physics.

F = ma 4 N = 2 kg x a a = 2

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MCAT Physics: Practice tests and explanations

Home > MCAT > MCAT Physics >

  • MCAT Physics Practice Test 1: Kinematics and Dynamics
  • MCAT Physics Practice Test 2: Work and Energy
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  • MCAT Physics Practice Test 4: Fluids
  • MCAT Physics Practice Test 5: Electrostatics and Magnetism
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MCAT Kinematics Formula Study Guide Cheat Sheet

July 1, 2014 By Leah4sci 7 Comments

kinematics practice problems mcat

Kinematic equations are the foundation to mastering MCAT physics . Yes you have to ‘memorize' and know the standard formulas, but it goes deeper than that. You have to understand the units, when to use which equation, and be VERY comfortable applying these concepts to the more advanced Physics topics on the MCAT.

To help you break down the kinematic equations I've put together this free cheat sheet which goes along with my MCAT Translational Motion Tutorial Videos . Save a copy to your computer and print another to study on the go. But don't keep this to yourself.

I put a great deal of time/effort into this guide and will greatly appreciate if you share on Facebook/Twitter/Google+… using the links above.

Click the image below for the full version

Kinematics Cheat Sheet MCAT Physics Study Guide Preview

Watch the associated MCAT Physics Tutorial Videos

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March 11, 2018 at 12:44 am

Dear Leah4sci,

Thank you for making your content free! We really appreciate it! Do you have a link for donations?

' srcset=

October 10, 2017 at 2:32 pm

Can you please explain the angles in projectile motion the way it affects range,height and other parameters??

' srcset=

May 3, 2017 at 11:05 pm

Did you take the old MCAT or new one?

' srcset=

July 27, 2015 at 5:07 pm

OMG I just found your website by way of youtube. I love the way you explain stuff. I wish I had you for physics in high school and college

' srcset=

October 1, 2015 at 8:54 pm

Thanks 🙂 Glad to help When are you taking your MCAT?

' srcset=

June 29, 2015 at 4:20 pm

Your videos are super great!!! THANKS for putting in your time and effort and helping those who need it.

June 30, 2015 at 6:22 pm

Thank you Joanne. When are you testing?

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Organic Chemistry Reference Material and Cheat Sheets

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Alkene Reactions Overview Cheat Sheet – Organic Chemistry

The true key to successful mastery of alkene reactions lies in practice practice practice. However, … [Read More...]

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KET Keto enol tautomerization reaction and mechanism leah4sci

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Keto Enol Tautomerization or KET, is an organic chemistry reaction in which ketone and enol … [Read More...]

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Work Kinetic Energy Theorem

The work-energy theorem states that the work done by all forces acting on a particle equals the change in the particle’s kinetic energy.

A force does work on the block and sets it in motion. The kinetic energy of the block (the energy that it possesses due to its motion) increases as a result of the amount of work .

kinematics practice problems mcat

V i and V f is the velocity of the particle before and after the application of force, and m is the particle’s mass.

kinematics practice problems mcat

With the knowledge of this relationship, try the energy approach first before applying kinematics when solving a problem, as the energy approach is much easier.

Practice Questions

Khan Academy

MCAT Official Prep (AAMC)

Sample Test C/P Section Question 57

Practice Exam 1 C/P Section Passage 4 Question 21

Practice Exam 3 C/P Section Passage 2 Question 7

• The work W done by the net force on a particle equals the change in the particle’s kinetic energy KE

• The work-energy theorem can be derived from Newton’s second law.

• Prioritize energy approach to kinematics in problem-solving.

the kinetic energy of an object is the energy that it possesses due to its motion

work:  a measure of energy transfer that occurs when an object is moved over a distance by an external force

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MCAT Physics Practice Questions & Answers

Including expert analysis.

MCAT Physics Practice Passages

It is important to go over MCAT physics practice questions and answers to understand what is expected of you in the Chemical and Physical Foundations of Biological Systems (CPBS) section of the MCAT and what to include in your MCAT prep .

Disclaimer: MCAT is a registered trademark of AAMC. BeMo and AAMC do not endorse or affiliate with one another.

>> Want us to help you get accepted? Schedule a free strategy call here . <<

Article Contents 7 min read

Physics concepts covered on the mcat.

To study effectively for the Chemical and Physical Foundations of Living Systems section of the MCAT, one should thoroughly understand MCAT physics equations and topics. Some of the important physics concepts covered on the MCAT are Atomic and Nuclear Phenomena, Circuits, Electrostatics, Fluids, Kinematics, Light and Optics, Magnetism, Thermodynamics, Units and Dimensional Analysis, Waves and Sound and Work and Energy.

As per the AAMC, these foundational concepts are about the physical processes that allow complex organisms to transport materials, sense their environment, process signals, and respond to changes. This is further subdivided into five categories:

This section is designed to test introductory-level biology, organic and inorganic chemistry, physics concepts, and biochemistry concepts that are typically taught in medical school prerequisites offered at most colleges and universities in the US and Canada. In addition to testing your knowledge, this MCAT section also evaluates your expertise in basic research methods and statistics concepts. It also requires you to demonstrate your scientific inquiry and reasoning.

You may wonder how much physics you’ll see in this section of the MCAT exam and how many questions you’ll get about a particular foundational concept? There are 59 questions that include a combination of passages-based and discreet questions. This section of the test lasts for 95 minutes, and of all these questions, introductory physics makes up approximately 25%. While First-semester biochemistry, Introductory biology, General chemistry, and Organic chemistry make up around 25%, 5%, 30% and 15% of this section, respectively.

Now, let’s get to some practice passages to test your knowledge!

Check out the MCAT physics equations you must know!

The “two charged blocks” problem was posed to probe thinking about potential energy in electrostatics. This problem consists of two different scenarios (see Fig. 1): in one, two positively charged blocks are being pushed toward one another (the “like-charged blocks” scenario); in the other, two oppositely charged blocks are being moved apart (the “oppositely charged blocks” scenario).

The blocks begin and end at rest. In each case, students were asked to identify whether the electric potential energy of the two-block system increases, decreases, or stays the same. Students might intuitively recognize which sets of objects could gain more kinetic energy if released from rest in the positions shown and use that to draw conclusions about potential energy. More formally, they could reason that positive work is done on each system. Therefore, the potential energy must increase (kinetic energy does not change). Finally, students could reason mathematically by applying the formula for the electric potential energy of a system of two-point charges, q1 and q2, separated by a distance.

R∶ U e = kq1q2/R.

This approach requires careful attention to the sign of each of the charges and their overall product.

Adapted from: Lindsey BA. Student reasoning about electrostatic and gravitational potential energy: An exploratory study with interdisciplinary consequences. Physical Review Special Topics-Physics Education Research. 2014 Jan 17;10(1):013101.

Question 1: How would the potential energy change in case of situation a when two blocks of similar charges move towards each other, as described in the figure?

A.   The energy of the system would increase as the blocks are moved toward each other

B.   The system’s energy would decrease as the blocks are moved toward each other

C.   There would be no change in the energy of the system as the blocks are moved toward each other

D.   There is insufficient information to determine the response

A passage in the musical Man of la Mancha relates to Newton’s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, “Of course, I hit her back, Your Grace, but she’s a lot harder than me, and you know what they say, ‘Whether the stone hits the pitcher, or the pitcher hits the stone, it’s going to be bad for the pitcher.’”

This is precisely what happens whenever one body exerts a force on another—the first also experiences a force (equal in magnitude and opposite in direction). Numerous everyday experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in Newton’s third law of motion.

This law represents a certain symmetry in nature: Forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as “action-reaction,” where the force exerted is the action, and the force experienced as a consequence is the reaction. Newton’s third law has practical uses in analyzing the origin of forces and understanding which forces are external to a system.

Adapted from - Hamm K. 6.4 Newton’s Third Law of Motion: Symmetry in Forces. Biomechanics of Human Movement. 2020 Aug 1.

Question 1: Keeping what you leant from the passage, solve the following problem - An ice cream seller pushes a cart on a straight road. His mass is 70.0 kg, the cart’s weight is 10 kg, and the ice-cream’s weight along with dry ice is 20 kg. What will be the acceleration produced when the ice-cream exerts a backward force of 200 N on the floor? All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 20.0 N.

A.   -1.5m/s2

B.   1.5m/s2

C.   -1.8 m/s2

D.   1.8 m/s2

Question 2: A coin is tossed straight up into the air. After it is released, it moves upward, reaches its highest point and falls back down again. What force is acting on the coin at its highest point? Air resistance can be neglected.

A.   Force is zero

B.   Force is up and constant

C.  Force is down and constant

D.   Force is down and decreasing

Question 3: Two trucks, one of mass 1000kg and one of mass 2000kg, collide head on. The truck with more mass experiences a(n) __________ force and a(n) __________ acceleration with respect to the smaller truck.

A.   larger . . . smaller

B.   smaller . . . larger

C.   equal . . . smaller

D.   equal . . . larger

If it is understood that the electric circuit is a system consisting of the three elements, viz., the drive, the closed flow of matter, and the hindrance (obstacle), the transition to a quantitative treatment can be facilitated by explaining the process as a dynamic equilibrium. If the stationary flow of electricity is interpreted as the result of a dynamic balance between drive and hindrance, the following semiquantitative relations follow immediately:

— The stronger the hindrance, the smaller the flow (if the drive remains constant).

— The stronger the drive, the stronger the flow (if the hindrance remains constant).

When proceeding to a quantitative definition of terms, it is possible to discuss with students the question under what conditions two completely different resistors from different materials can be regarded as having the same resistance. One can also ask the question under what circumstances it can be concluded that the resistance of a resistor remains constant whilst the electric current is changing. If the students accept that the resistance is not a fixed-term but that it can be different at different temperatures, or for different velocities of the moving particles, etc., the condition that doubling the drive gives twice a current is a straightforward and understandable answer to this question. Ohm's law V/I = constant or R = constant can be presented as a simple and reasonable hypothesis that must be proven or refuted by experiment.

Adapted from - Härtel H. The electric circuit as a system: A new approach. European Journal of Science Education. 1982 Jan 1;4(1):45-55.

Question 1: What would happen if the resistance (hindrance) suddenly became less?

A.   Nothing would change, and the system would work as usual.

B.   Voltage and current would be equivalent to each other

C.   The circuit would break

D.   None of the above

Question 2: Suppose an electric current of 1 microamp (1 μA) was to go through a resistance of 3 mega-ohms (3 MΩ). How much voltage would be “dropped” across this resistance?

A.   3.0 volts

B.   3.45 volts

C.   30 volts

D.   34.5 volts

Question 3: How current and voltage are experienced in resistor 1 if two resistors are placed in parallel and resistor 1 has twice the resistance as resistor 2.

A.   Voltage remains the same, and current decreases

B.   Current remains the same, and voltage increases

C.   Both current and voltage remain the same

D.   Both current and voltage increases

Physics makes up about 25% of the Chemical and Physical Foundations of Biological Systems (CPBS) section of the test.

MCAT tests physics concepts you would cover in introductory science courses in college and university.

The best practice is to employ active learning and practice tests. Some of the tricks that you might use are –

A.   Strategize - Prioritize high yield MCAT topics , equations, and concepts that are outlined in the AAMC’s official content list

B.   Know your units- Get comfortable performing quick unit conversions because sometimes that’s all you need to do to find the correct answer among the MCAT answer options. Practice messing with one of the variables to see what happens to the rest of the equations.

C.   Flashcards and memorization- Use flashcard for writing down formulas, concepts and ideas. Be sure to also incorporate a lot of practice with sample physics questions and MCAT diagnostic tests.

D.  Practice - Understanding physics equations and concepts takes practice. Work to build your understanding through practice and internalize the concepts you cover. You can use different practice tests and platforms like UWorld, Examkrackers, Khan Academy, and more.

Using active study strategies will help you internalize physics concepts.

 E. Be an Instructor - Explaining concepts to others can also reveal your weaknesses and parts of the concepts you need to focus on.

F. Get a tutor for MCAT physics - If you're struggling with your physics prep, it might be wise to get an MCAT tutor who can help you go over the necessary MCAT physics equations. 

The best MCAT test prep involves content review, active study strategies, and full-length practice exams. 

You can consider signing up for an MCAT private tutoring service that will help you apply your physics knowledge on the test. 

Yes, they can. Not only can they help you find the right study resources and materials, but they can also help you learn how to apply your knowledge on the day of the test.

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Kinematics on the MCAT

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Chances are when you think about physics, this may be one of the first things that come  to mind! Forces, velocity, displacement, etc.: these are probably the words and terms that pop up in your head when you hear the word physics!

Similar to other physics topics, we can already have a good idea of what we’re trying to study based on the name itself: the prefix “-kine” refers to motion, which makes sense as kinematics is the branch of physics that studies the motion of objects. 

After finishing this guide, you’ll have a basic understanding of various topics in kinematics which you can delve more into in detail with our individual study articles. Let’s get started!

Kinematics on the MCAT: What You Need to Know

Topics on kinematics will be tested on the Chem/Phys section of the MCAT and can appear both as passage based and fundamental discrete questions. 

If you save this section for a little later in your prep, don’t feel worried! We calculated that you can probably expect around 2-3 questions that can possibly come up!

Introductory physics accounts for 25% of the content covered in the Chemical and Physical Foundations of Biological Systems.

Important Sub-Topics: Kinematics

This portion of MCAT physics content review is probably the one most math heavy in regards to trigonometry (ewww…). There will be just a few trigonometric concepts that will come up when covering kinematics. 

However, these concepts are fairly easy to comprehend and recall (if you’ve taken a little trigonometry before)! Additionally, we’ll always try to keep the content of these outside trigonometric concepts to the bare essentials in terms of what you need to know.

1. Mathematics of Vectors and Scalars

If you’ve watched the movie Despicable Me , you won’t have a hard time remembering what vectors are. Just like the main villain, vectors are quantities that have both direction   and magnitude which is usually represented by an arrow. Scalars are quantities that only have magnitude . Consider the example below!

Mathematics of Vectors and Scalars

The difference between vector and scalar quantities is even better pronounced when you apply them to examples. For instance, values like displacement and velocity are vector quantities while distance and speed are the scalar equivalents. 

Mathematics of Vectors and Scalars - A

Vectors and scalar quantities can both be added and subtracted. Usually, the MCAT will focus on vector mathematics because they want to test your ability to take into account the vector direction. 

Mathematics of Vectors and Scalars - B

As shown, we simply add the individual x and y components of the individual vectors in order to get the resultant vector. Conversely, to get the resultant vector from subtraction, we simply subtract the x and y components from the individual vectors. 

(Coming Soon!) Full Study Notes : Mathematics of Vectors and Scalars

For more in-depth content review on mathematics of vectors and scalars, check out these detailed lesson notes created by top MCAT scorers. 

2. Displacement, Distance, Velocity, and Speed

Now we can actually get into applying the vector and scalar quantities to more applicable real life examples. Let’s first start with displacement and distance as they both relate to the position. 

Displacement refers to the total amount of change in the position of an object from a starting point. Distance refers to the total distance that an object traveled, regardless of how far it is from its original position. As mentioned above, displacement and distance are vector and scalar quantities, respectively. 

Kinematics - Displacement vs. Distance

Similarly, velocity and speed are also related concepts to displacement and distance. Velocity is simply the average displacement of an object per unit of time, usually in seconds while speed refers to the average distance traveled per unit of time. 

Just like displacement and distance, velocity and speed are also vector and scalar quantities, respectively. The calculations below take the values from the example above!

Displacement, Distance, Velocity, and Speed - A

(Coming Soon!) Full Study Notes : Displacement, Velocity, and Speed

For more in-depth content review on displacement, velocity and speed, check out these detailed lesson notes created by top MCAT scorers. 

3. Acceleration and Motion

This term can sometimes be a little confusing at first sight, but is easier when broken down into an example! Acceleration refers to how fast an object’s velocity is changing. When an NBA player like Zach Lavine changes his velocity to drive down the lane for a slam dunk, he accelerates to get that dunk (and hopefully, a poster!).

Acceleration and Motion - A

Suppose you drop a ball down from a cliff. 1 second after the ball drops, the ball is traveling at a velocity of 9.8 m/s2. This is because it started at a velocity of 0 at 0 seconds and accelerated (change velocity) via the gravity acceleration constant. Following this, 2 seconds after the ball drops, the ball is now traveling at a velocity of 19.6 m/s.

A lot of the tested kinematic equations that deal with motion on the MCAT deal with constant acceleration, where velocity is changing at a constant rate. A couple of these equations are listed below!

The left equation can be used to solve for a final velocity of an object, while the right equation can be used to determine the displacement of an object!

Acceleration and Motion - B

The 2 most common types of motions with constant acceleration that the MCAT will test is linear and projectile motion. The only difference between the 2 is the dimensions that are used: linear motion deals with one dimensional motion (only x or y) while projectile motion deals with both dimensions (x and y). 

Acceleration and Motion - C

(Coming Soon!) Full Study Notes : Acceleration and Motion

For more in-depth content review on acceleration and motion, check out these detailed lesson notes created by top MCAT scorers. 

4. Types of Forces

Before getting into the different types of forces that you’ll encounter on the MCAT< let’s first give a basic definition of what a force is! Force is often described as a pushing or pulling interaction between objects given in the SI units, newtons. 

One type of force that will for sure come up on the MCAT is gravitational force, specifically in the context of weight. From here, we can differentiate between the values of mass and weight. 

Mass refers to the total amount of matter within an object while weight is the gravitational force acting on an object’s mass, as given by the equation below. Notice the similarity between this equation and Newton’s second law as we’ll discuss in the next section!

Types of Forces - A

Another type of force you’ll encounter on the MCAT is frictional force which is the force that opposes the motion of an object. These forces can be further divided into static friction and kinetic friction. 

As indicated by the name, static friction is the frictional force on a stationary object where it prevents it from moving. Conversely, kinetic friction is the frictional force that opposes a moving object! 

Types of Forces - B

(Coming Soon!) Full Study Notes : Types of Forces

For more in-depth content review on types of forces, check out these detailed lesson notes created by top MCAT scorers. 

5. Newton’s Laws

Just like with thermodynamics and other physics concepts, we can attribute the various observations in kinematics to 3 main laws, more specifically Newton’s Laws! The most common quote associated with Newton’s first law is that an object in motion (or at rest) will stay in motion (or at rest) until acted upon by an external force. 

However, a more correct wording would be that an object at a constant velocity (or at rest) will stay at that constant velocity (or at rest) until acted upon by an unbalanced, external force.

A good example of the above definition is static friction! Suppose the maximum static friction force applied on an object at rest is 10 N and that there’s an applied force at 5 N. 

Even though there’s an applied external force, there is still a balance which is why the object stays at rest. When a force of 11 N is applied, the net force is unbalanced and the object begins to move! 

Newton’s First Law

Likewise, a common iteration of Newton’s second law is that force is equal to mass times acceleration. More specifically, when a net force acts on an object, the object with mass will accelerate!

However, note that when the force is removed from the object, the object then moves at a constant velocity which accelerated too!

Newton’s Laws - Second Law

You’ve maybe heard Newton’s third law where people say, “for every reaction, there is an equal and opposite reaction.” To add to this great summation, we can say that object A exerts a force on object B, object B exerts an equal, opposing force on object A. 

A great example of this is when you push up with your feet! In this process, you apply a force, FAB, on the ground. Based on Newton’s third law, the ground provides an equal force, FBA, on you, which actually causes you to push up!

Newton’s Laws - Third Law

A common misconception that’s made is that because the forces are of equal magnitude, no motion can occur. However, it’s important to make the distinction that the forces act on different objects! It’s only when forces of equal magnitude and opposite direction act on the same system that no motion occurs! 

(Coming Soon!) Full Study Notes : Newton’s Law

For more in-depth content review on newton's 3 laws of motion, check out these detailed lesson notes created by top MCAT scorers. 

6. Mechanical Equilibrium

Though at first sight this term might seem scary, it’s actually fairly easy to comprehend when put into the context of examples. An important concept to understand with mechanical equilibrium is that it’s just an application of Newton’s first law. In this sense, mechanical equilibrium is when the net force acting on an object is 0. 

In translational equilibrium, the motion in question does not involve rotation, rather focusing on linear and projectile motion. While an object can still be moving at a constant velocity in translational equilibrium, the MCAT will most likely test translational equilibrium when an object is not moving. 

Mechanical Equilibrium - A

Conversely, rotational equilibrium occurs when the net force causing rotational motion on an object is 0. The force that causes rotational motion is specifically called torque (𝛕)! Similar to translational equilibrium, the MCAT will most likely test cases of rotational equilibrium where an object is not moving. 

The most common example you’ll probably come across is a problem involving a seesaw with two objects placed on opposite sides of the fulcrum. 

Mechanical Equilibrium - B

(Coming Soon!) Full Study Notes : Mechanical Equilibrium

For more in-depth content review on mechanical equilibrium, check out these detailed lesson notes created by top MCAT scorers. 

Important Definitions and Key Terms

Below are some high yield definitions and key terms to refer to when reviewing concepts and ideas about kinematics!

Additional FAQs - Kinematics on the MCAT

Are kinematics on the mcat, are kinematic equations given on the mcat, how do you memorize kinematic equations for the mcat, what are the 4 kinematic equations – mcat, additional reading links  (coming soon)   – study notes for kinematics on the mcat.

  • Mathematics of Vectors and Scalars
  • Displacement, Velocity, and Speed
  • Acceleration and Motion
  • Types of Forces 
  • Newton’s Law  
  • Mechanical Equilibrium  

Additional Reading: Physics Topics on the MCAT:

  • Circuits on the MCAT
  • Electrostatics on the MCAT
  • Fluids on the MCAT
  • Atomic and Nuclear Phenomena on the MCAT
  • Light and Optics on the MCAT
  • Magnetism on the MCAT
  • Thermodynamics on the MCAT
  • Units and Dimensional Analysis on the MCAT
  • Waves and Sound on the MCAT
  • Work and Energy on the MCAT

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Course: MCAT   >   Unit 9

Kinetics questions.

  • Introduction to reaction rates
  • Rate law and reaction order
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  • (Choice A)   R=k[A] 2 ‍   [B] 1 ‍   [C] 1 ‍   A R=k[A] 2 ‍   [B] 1 ‍   [C] 1 ‍  
  • (Choice B)   R=k[A] 4 ‍   [B] 2 ‍   [C] 1 ‍   B R=k[A] 4 ‍   [B] 2 ‍   [C] 1 ‍  
  • (Choice C)   R=k[A] 2 ‍   [B] 1 ‍   [C] 0 ‍   C R=k[A] 2 ‍   [B] 1 ‍   [C] 0 ‍  
  • (Choice D)   R=k[A] 1 ‍   [B] 2 ‍   [C] 0 ‍   D R=k[A] 1 ‍   [B] 2 ‍   [C] 0 ‍  

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kinematics practice problems mcat

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UsingKinEqns1ThN.png

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

IMAGES

  1. MCAT Kinematic Equations Study Guide Cheat Sheet

    kinematics practice problems mcat

  2. Help with EK1001 Physics Kinematics question

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  3. MCAT Physics Kinematic Equations and Knowing Which To Use

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  4. Everything MCAT: MCAT Learning Material: Physics Lesson 1 Kinematics

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  5. Kinematics Worksheet With Answers

    kinematics practice problems mcat

  6. Kinematic Equations Worksheet With Answers

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VIDEO

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