This text is meant to accompany class discussions. It is not everything there is to know about uniform circular motion. It is meant as a prep for class. More detailed notes and examples are given in the class notes, presentations, and demonstrations (click here.)

The acceleration of the car on the the previous page is 1 m/s^{2}. The acceleration is constant. At the start, (0 seconds,) the acceleration is 1 m/s^{2}; at 1 s it is 1 m/s^{2}; at 2 seconds it is still 1 m/s^{2}; at 3 seconds it remains 1 m/s^{2}; at 4 seconds it is still 1 m/s^{2}, and at 5 seconds it is also 1 m/s^{2}. The graph will look like the one below.

Look at the slope on the previous graph of velocity vs time.

The slope on the velocity vs time graph equals the acceleration.

Let's change the situation again. Instead of the car traveling at a constant acceleration suppose its acceleration changes by 1 m/s^{2} every second. This is a constant jerk of 1 m/s^{3}. After 1 second, it would be accelerating at 2 m/s^{2}; after 2 seconds, it would be accelerating at 3 m/s^{2}; after 3 seconds it would be accelerating at 4 m/s^{2}; at 4 seconds its acceleration would be 5 m/s^{2}, and after 5 seconds its acceleration would be at 6 m/s^{2}. The graph of the car's acceleration would look like the one below.

This motion may be hard to imagine. It is like pressing harder and harder on the car's accelerator each second of the motion. Below is a table showing how the velocity and position would change each second.

Time (s)

Velocity (m/s)

Position (m)

0

1.0

1.0

1

2.5

2.67

2

5.0

6.33

3

8.5

13.00

4

13.0

23.67

5

18.5

39.33

Towards the end of the motion the displacement begins to jump larger and larger amounts.

Example

Question

Paper Solution

Video Solution

What is the acceleration from the v vs t graph below?