Kinematics Equations for the MCAT: Everything You Need to Know

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Learn key MCAT concepts about kinematics, plus practice questions and answers

Kinematics Equations for the MCAT banner

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Part 1: Introduction to kinematics

Kinematics is the science of how objects move, and it’s often the first module in an introductory physics class because it’s such an everyday topic. Any student who’s ever thrown something in the air will have at least some intuition about kinematics! 

On the MCAT, kinematics is a low-yield topic, and you’ll only need to be familiar with it in a biological context. That said, kinematics is the physics topic most used in everyday life. Understanding it can be an important first step in understanding more advanced topics in physics.

In the guide below, the most important terms are in bold font. When you see one, try to define it in your own words and use it to create your own examples. This is a great way to check your understanding, and phrasing terms in a way that makes the most sense to you and will make studying much easier (and much more effective!) in the long run.

At the end of this guide, there’s an MCAT-style practice passage and standalone questions that will not only test your knowledge on forces, energy, and work but also show you how the AAMC likes to ask questions.

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Part 2: Displacement, velocity, and acceleration 

Displacement, velocity, acceleration, and time are the main variables in kinematics. Most questions will give you information on some combination of these variables and ask you to solve for another. To do this, you’ll use the equations and definitions that relate each of them.

a) Defining the terms

First let’s define each term, starting with displacement. Displacement is a change in position, given as:

$$\Delta x=x-x_0$$ $$\mbox{where } \Delta x \mbox{ = displacement,}$$ $$x \mbox{ = final position,}$$ $$x_0 \mbox{ = the initial position}$$

Displacement measures how far something moves and in what direction. It is usually given in a standard unit of meters (m). Displacement is often confused for distance, but they are not the same. Displacement needs to have both distance and direction. For example, “5 meters east” is a displacement, but “5 meters” is not. 

Another way to think about this is that displacement only concerns the difference between the starting point and the final point. So, if you’re returning home after a cross-country road trip, your distance traveled might be thousands of miles, but your displacement will be zero because you ended in the same place that you started. 

Velocity is a change in position over a change in time, expressed as:

$$v =\frac{\Delta x}{\Delta t}$$ $$\mbox{where } v \mbox{ = velocity,}$$ $$\Delta x=x_2-x_1\mbox{ = displacement,}$$ $$\Delta t=t_2-t_1\mbox{ = change in time}$$

Velocity measures how fast and in what direction a position is changing. Its standard unit is meters per second (m/s). Similar to displacement and distance, velocity is often conflated with speed. Again, the difference is that velocity needs to specify a direction.  

Velocity calculated using the above equation is usually the average velocity, or a calculated average of all velocities over a period of time. The period of time may be as short as seconds or milliseconds. 

In contrast, instantaneous velocity refers to velocity that is measured at a single moment in time. Instantaneous velocity cannot be measured and must be calculated using derivatives and integrals. Since calculus is not tested on the MCAT, instantaneous velocity is out of the scope of the exam. However, average velocity is still fair game—and you may be asked to calculate velocity using a table of values.

Finally, acceleration is a change in velocity over a change in time, written as:

$$a =\frac{\Delta v}{\Delta t}$$ $$\mbox{where } a \mbox{ = acceleration,}$$ $$\Delta v=v_2-v_1\mbox{ = change in velocity,}$$ $$\Delta t=t_2-t_1\mbox{ = change in time}$$

Just like the previous two terms, acceleration measures the magnitude of the change as well as its direction. Its standard unit is meters per second squared (m/s²). Note: even if a velocity remains at the same magnitude for an object, there can be an acceleration if the object is changing direction.

You might have noticed how much we emphasized direction in each of those definitions. We included direction because displacement, velocity, acceleration are all vectors, meaning they have magnitude and direction. Each must have both! A good way to think about vectors is by imagining them as arrows. An arrow has to have some length, and it has to point in some direction (otherwise, it’s not an arrow!)

Figure 1. Vector addition

Figure 1. Vector addition

Vector-addition-MCAT-2.png

Vectors can also be combined. Consecutive displacements in the same direction, for example, will make a bigger displacement, and displacements in opposite directions will cancel out (as shown on the right side of Figure 1).

b) Defining the sign conventions

Most kinematics problems exist in one or two dimensions. It is usually most convenient to use a vertical y-axis and a horizontal x-axis to define direction. Convention defines up and right as positive, while down and left are negative.

So, if someone walks 10 meters to the right, we say Δx=+10 m. Similarly, if someone walks 10 meters to the left in 10 seconds, v= -1 m/s.

c) Graphical interpretation

The ability to interpret graphs of displacement, velocity, and acceleration versus time is an important skill in introductory physics. In addition to basic graph interpretation skills, there are a couple of relationships specific to kinematics graphs.

First, consider the slope of a displacement-vs-time graph. Slope is defined as rise over run, or the change in the vertical axis over the change in the horizontal axis.

Figure 2. The slope of a displacement vs. time graph

Figure 2. The slope of a displacement vs. time graph

In this case, then, the slope equals Δx/Δt. You might recognize that equation, too—it’s the definition of velocity! It turns out that the slope of a displacement-vs-time graph is velocity, and the slope of a velocity-vs-time graph is acceleration.

Next, consider a velocity-vs-time graph with just a constant velocity. What is the area under that line?

Figure 3. The area under the curve of a velocity vs. time graph

Figure 3. The area under the curve of a velocity vs. time graph

The shape is a rectangle, so its area is defined as width times height. In this case, width is the change in time and height is the constant velocity. Rearranging the velocity equation above tells us that this area must equal the displacement! In other words, the area under the curve on a velocity-vs-time graph is the displacement, and the area under the curve on an acceleration-vs-time graph is the change in velocity.

These relationships let you say a lot about the motion of an object even if you have one graph. Your tools in deciphering kinematics graphs are reiterated in the chart below.

Figure 4. Relating displacement, velocity, and acceleration

Figure 4. Relating displacement, velocity, and acceleration

If you are familiar with calculus, it might help to know that velocity is the time derivative of displacement and that acceleration is the time derivative of velocity. 

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Dr. Shemmassian

Dr. Shirag Shemmassian is the Founder of Shemmassian Academic Consulting and well-known expert on college admissions, medical school admissions, and graduate school admissions. For nearly 20 years, he and his team have helped thousands of students get into elite institutions.