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#

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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Answer 2

Nathan Ashcroft

The object's acceleration is 0.571 m/s^2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.