Takes 5.6 times as long to stop from 59 mph as it takes for 25 mph
For decelerating from 25 mph it takes 25/A seconds.
For decelerating from 59 mph it takes 59/A seconds.
Using the formula d = 1/2AT^2 for the distance traveled under constant acceleration.
Distance it takes to stop from 59 mph is 1/2A(59/A)^2
Distance it takes to stop from 25 mph is 1/2A(25/A)^2
So the ratio becomes (1/2A(59/A)^2) / (1/2A(25/A)^2)
Looks pretty complicated, but we can start by cancelling the 1/2A on the top and bottom, giving
((59/A)^2) / ((25/A)^2)
Now handle the squaring
(59^2/A^2)/(25^2/A^2)
And cancel the 1/A^2 on the top and bottom
(59^2)/(25^2)
And finally, do the actual math
(59^2)/(25^2) = 3481/625 = 5.5696
So it will take 5.5696 times as far to stop from 59 mph as it takes from 25 mph.
Since we only have 2 significant figures, the rounded result is 5.6
