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?

1 Answer
David G.
Feb 13, 2017

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.

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