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Determining Acceleration Using the Slope of a Velocity vs. Time Graph

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  • 0:03 Kinematics with the X,Y Graph
  • 0:29 Acceleration = Slope of Graph
  • 1:59 Acceleration Example Problem
  • 4:15 Lesson Summary
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Lesson Transcript
Instructor: Angela Hartsock

Angela has taught college Microbiology and has a doctoral degree in Microbiology.

In this lesson, we will learn how to use the slope of the line on a velocity vs. time graph to calculate the acceleration of an object in straight line motion.

Kinematics with the X,Y Graph

The basic x, y graph has many interesting applications in physics and kinematics. In addition to succinctly describing the straight line motion of an object, you can use these graphs to calculate information like displacement, velocity, and acceleration. In this lesson, we'll continue our examination of the velocity vs. time graph and how it can be used to calculate the acceleration of an object in straight line motion.

Acceleration = the Slope of the Graph

You should be familiar with the velocity vs. time graph. This example graph illustrates how the velocity of a car changes as it drives along a straight track.

Graph of car velocity
graph of car movement

The equation for acceleration is a = Δv / Δ t. Remember, when using an equation with a delta (Δ), you need to calculate the change: Δ = final value - initial value

So,

Δv = final v - initial v

Δt = final t - initial t

But, velocity (v) is on the y axis and time (t) is on the x axis. So, we could also write this equation as:

a = Δy / Δ x

Does this look familiar? It should. This is the equation for the slope of a line on an x, y graph. So, the slope of a velocity vs. time graph gives the acceleration over that section of the graph. Confused? Let's look at an example.

Acceleration Example Problem

This velocity vs. time graph shows the motion of a rat running in a long, straight tube.

The acceleration and deceleration of a rat in a straight tube
graph of rat acceleration

First, the equation for acceleration is:

a = Δv / Δ t

Let's fill in what we know:

Δv = 20 m/s - 0 m/s = 20 m/s. His velocity started at 0 m/s and ended at 20 m/s so the change in velocity (Δv) was 20 m/s.

Δ t = 4 s - 0 s = 4 s. The time started when he started moving (0 seconds) and we only care about the first 4 seconds, so his change in time (Δ t) was 4 s.

Filling in the equation, we get:

a = (20 m/s) / (4 s) = 5 m/s^2

His acceleration over the first 4 seconds was 5 m/s^2. Remember, acceleration is a vector quantity and needs a directional component. In the case of straight line motion, the vector direction is equal to the sign of the magnitude. Since the 5 m/s^2 acceleration is positive, the vector direction is forward.

If we look at the last four seconds of the motion (t = 8s - 12s), the math is going to be nearly identical, but the change in velocity will be slightly different.

Δv = 0 m/s - 20 m/s = - 20 m/s

If we substitute this into the equation, we get:

a = (-20 m/s) / (4 s) = -5 m/s^2

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