Worked Example

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.)

Solution

- 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}.

### Like this:

Like Loading...

*Related*