Question

If an object with uniform acceleration (or deceleration) has a speed of `8` `ms^-1` at `t=0` and moves a total of `16` `m` by `t=7`, what was the object's rate of acceleration?

Answer 1

We can use the formula `s= ut+1/2 at^2`, and rearrange to make `a` the subject:

`a = (2(s-ut))/t^2 = (2(16-7 times 8))/7^2 = (2(-40))/49 = -1.63 ms^-2`

Explanation:

The total time taken is `t=7` `s`.

(Note that it's sometimes a little confusing: 't = 7' might describe a moment in time, 7 seconds after the start, or it might describe a period of time. In this case it's being used one way in the question and a different way in the answer.)

If the object had continued at the same velocity, `8` `ms^-1`, for `7` `s`, it would have covered `56` `m`. As it was, it only covered `16` `m`. From a common sense perspective, this means it must have slowed down (decelerated), so the negative number in our answer when we calculated the acceleration of the object makes perfect sense.