Assuming equal rates of acceleration in both cases, how much farther would you travel if braking from 59 mi/h to rest than from 25 mi/h ?

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

Ibelieve the answer is b

490 n  f=125n a=2.5m/s2