Question

The driver of a 1.19 103 kg car traveling on the interstate at 35.0 m/s (nearly 80.0 mph) slams on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the brakes are applied, a constant kinetic friction force of magnitude 7.96 103 N acts on the car. Ignore air resistance. (a) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle? m (b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collision occur? m/s

Answer 1

a) Acceleration due to brakes a= -7.96e3/1.19e3 = -6.689 m/s^2

applying third equation of motion,

v^2 = u^2 +2as

0 = 35^2 -2*6.689 *s

s = 35^2/[2*6.689]

= 91.568 m

b) v^2 = 35^2 - 2*6.689*30

= 823.66

v = sqrt 823.66

= 28.7 m/s Answer