It takes a minimum distance of 98.26 m to stop a car moving at 17.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

x_f = 212.5m

Explanation:

t = (x_f-x_0)/(.5*(v_f-v_0))

t = (98.26m-0m)/(.5(0m/s-17m/s))

t = 11.56s

a = (v_f-v_0)/t

a = (0m/s-17m/s)/11.56s

a = -1.47m/s²

t = (v_f-v_0)/a

t = (0m/s-25m/s)/-1.47m/s²

t = 17s

x_f = x_0+(.5*(v_f-v_0))*t

x_f = 0m+(.5*(0m/s-25m/s))*17s

x_f = 212.5m

option number c is the right answer.