forces problem - Please Help ASAP

Question

An automobile is going up a grade of 15° at a speed of 25.5 m/s. The coefficient of static friction between the tires and the road is 0.70.
(a) What minimum distance does it take to stop the car?

(b) What minimum distance would it take if the car were going down the grade?

Answer 1

(a) going up the grade, two force act to slow car.  Friction and gravity.  X axis is along slope

Fn = mg cos 15.   Ffriction = 0.7 mg cos 15,  Fgx = mg sin 15

Fnet = 0.7mg cos 15 + mgsin 15.

a = Fnet/m  0.7 gcos 15 + g sin 15.   Once we have acceleration use  Vf^2 - Vi^2 = 2*a*d to get stopping distance.  Vf = 0.

(b) going downhill,  gravity works in the opposite direction of friction

a = 0.7 gcos 15 - gsin 15.   Use this new, smaller value of a to calculate the new, longer stopping distance.