Question
6. A car is moving along a straight road at 60 mph [26.7 meter/sec ]. The driver makes slams on his breaks locking them so that the car skid s to a stop. Assume a constant braking force due to friction , and a kinetic friction coefficient of 0.70 between the tires and the road: (a) calculate the braking distance from the instant that the brakes were applied, (b0 how much work did the friction do in stopping the car?
Answers
Henry
Vo = 26.7 m/s
V = 0
u = 0.70
a. u = a/g
a = g*u = (-9.8) * 0.70 = -6.86 m/s^2.
V^2 = Vo^2 + 2a*d = 0 @ max distance.
d = -(Vo^2)/2a = -(26.7^2)/-13.72 = 52 m
V = 0
u = 0.70
a. u = a/g
a = g*u = (-9.8) * 0.70 = -6.86 m/s^2.
V^2 = Vo^2 + 2a*d = 0 @ max distance.
d = -(Vo^2)/2a = -(26.7^2)/-13.72 = 52 m
Owen
How does this work