Position versus time or velocity versus time graphs are a core topic taught in just about all classical mechanics physics courses. In addition, it’s a topic likely to be seen on the MCAT.

Questions on position or velocity versus time graphs often trip students up. In this post, I’ve broken down a consistent approach you can take when tackling these problems so you can turn them into free points.

## How to Approach Position and Velocity vs. Time MCAT Questions:

**1. What’s the Nature of the Question?**

Position or velocity versus time graphs generally show some figure with X and Y axes and connected line segments representing either the position or velocity of some moving object.

As with any figure, the first important thing to do is to check the axes. Determine whether the graph is showing POSITION versus time or VELOCITY versus time. The only way to get the question right is to know what the graph is representing (or guessing right, but 100% odds are better than 20%). Once you have identified what the graph is showing, you can now begin to solve the problem.

## Position vs. Time Example

*Image From Algebralab.org*

So you’re shown the above graph and now need to interpret it. As the axes show, this is representing position versus time.

With a position versus time graph, there are five pieces of information you can obtain:

**1. Position:**

Your actual position at any given time. To do this, simply look at the Y-value (representing the position) at any given X-value (time). You’re unlikely to get a question on this but it’s important to understand this point nonetheless.

**2. Displacement:**

This is found by choosing any two points on the graphed line and simply looking at the change in the Y-value. Remember that displacement is a vector: direction matters.

**3. Distance:**

Distance is a scalar. Direction doesn’t matter here. So to calculate the distance between two points, you need to calculate the individual displacements of the individual line segments (the change in the Y-value) between any two points, and then add up the absolute value of all those individual segments. In other words, if the graph shows a positive displacement of 5 meters, followed by a negative displacement of 5 meters, the total distance traveled is 10 meters (whereas the total displacement would have been zero meters).

**4. Instantaneous Velocity:**

You don’t need to know calculus for the MCAT, but it is helpful to realize that the tangent line of a position versus time graph (the slope at any specific point) gives you the instantaneous velocity.

**5. Average Velocity:**

By looking at the slope between any two points (versus the tangent at a specific point as above), you can obtain the average velocity between those two points:

## Velocity vs. Time Exam

**Image from Algebralab.org
*

With a velocity versus time graph, the important things to interpret are the following:

**1. Velocity:** At any given time, the velocity is found by simply looking at the Y-axis value at that specific X-value (time).

**2. Acceleration:** The slope of velocity versus time graphs gives you acceleration. For the MCAT you usually won’t need to differentiate between average or instantaneous acceleration. Two things will be important for acceleration problems:

**a.** If velocity is increasing over time (i.e. the slope is positive), acceleration is positive. This means that even if the velocity is negative, if it’s becoming less negative (i.e. going from say -25 m/s to -5 m/s), the acceleration is positive.

**b.** If velocity is decreasing over time (i.e. the slope is negative), acceleration is negative. This means that even if the velocity is positive, if it’s becoming less positive (i.e. going from say +25 m/s to +5 m/s), acceleration is negative.

**3. Displacement:** This is found by looking at the area under the curve of velocity versus time graphs. Since displacement is a VECTOR (direction matters), areas that are above the X-axis that are on the positive velocity side will be given positive values, while areas that are below the X-axis on the negative velocity side will be given negative values. (Note: The area “under the curve” is the area between the function and the X-axis).

**4. Distance:** The approach to distance is similar to displacement—again you need to calculate the different areas between the velocity curve and the X-axis. However, distance is a scalar. Therefore you add the absolute values of the areas (as opposed to displacement where areas beneath the X-axis were assigned negative values).

## Units-Based Approach

Here’s one more strategy that should solidify your understanding of this topic as well as give you some insurance on test day: **Always use units in graphs to your advantage.**

What do I mean by this? Well, ultimately when looking at the slopes or areas of the curves I’ve presented, we’re performing the following mathematical operations:

**Slopes:** Here, we’re effectively dividing Y-axis values by X-axis values, after all, slope = “rise over run” or m = ∆y/∆x

**Areas:** Here, we’re effectively multiplying Y-axis values by X-axis values.

Knowing the units (either position or velocity on the Y-axis and time on the X-axis) as well as the mathematical operation being performed (none in the case of simply reporting the value at a given time, division of Y by X in the case of a slope, multiplication of Y by X in the case of an area) will therefore tell you the units of the answer.

**Slope of position versus time graph:** This is dividing position (SI units of meters) by time (SI units of seconds). M/S gives you the units of speed or velocity, which is exactly what taking the slope of position versus time graphs is giving you.

**Slope of velocity versus time graph:** This is dividing velocity (SI units of meters/second) by time (SI units of seconds). M/S^2 gives you the units of—you guessed it—acceleration. And sure enough, taking the slope of velocity versus time gives you acceleration.

**Area of velocity versus time graph:** This is multiplying velocity (SI units of meters/second) by time (SI units of seconds), giving you a final unit of meters, which is the unit of distance or displacement. And that is precisely what this operation gives you.

**Area of position versus time graph:** This multiplies position (meters) by time (seconds) meaning the area here has units of meters*seconds. This probably doesn’t remind you of the units of anything you’re familiar with. And that is good because this isn’t a meaningful metric and isn’t something you will ever need to or should calculate for the MCAT.

## Conclusion

Ultimately, if you’re comfortable with the above rules and understand these topics, you will be ready to tackle any position or velocity versus time graph. As with most difficult concepts in physics, if you have a consistent approach these problems are very solvable even if they initially seem confusing or difficult.

If there are other topics you’d like us to break down for you, or if you have questions about this breakdown, let us know below. Good luck!