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## Homework Statement

For part c ii) of this question, the new distance is now 1/4 of the previous distance as both acceleration are the same.

But why are both the acceleration the same?

## Homework Equations

v^2= u^2 +2as

## The Attempt at a Solution

I know that the method is by using the above equation, and taking v as 0, we get s = u^2/2a.

So when the initial speed is now halved, we get s' = (u/2)^2 /2a , and ∴ s' = u^2/8a , which is ¼ of the previous distance.

But why are the acceleration the same for both cases?