The a-t, v-t and s-t graphs are inter-related:
- The gradient of the s–t graph is v.
- The gradient of the v–t graph is a.
- The area under the a–t graph is Δv.
- The area under the v–t graph is Δs.
Given the v-t graph for the motion of a ball over 5 seconds, derive the corresponding s-t and a-t graphs. (Assume “rightward is positive” sign convention.)
- The areas of +1.0, +0.5, -0.5 and -1.0 in the v-t graph correspond to the s-t graph increasing by 1.0 m and 0.5 m, before decreasing by 0.5 m and 1.0 m.
- As the ball continues to move rightward until t = 2 s, the maximum displacement occurs at t = 2 s.
- The ball returns to the starting point after 5 s because the positive area happens to match the negative area exactly.
- As the velocity changes, so does the gradient of the v-t
- Constant speed segments in v-t graphs translate into straight lines in s-t graphs.
- Increasing speed segments translate into steepening curves, whereas decreasing speed segments translate into flattening curves.
- The a-t graph is readily obtained from the gradient of the v-t
- The areas of 0, -2.0 and +1.0 in the a-t graph correspond to velocity remaining constant, before decreasing by 2.0 m s-1 and then increasing by 1.0 m s-1.