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?
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?