PRACTICE IT Use the worked example above to help you solve this problem. The dri

Question

PRACTICE IT Use the worked example above to help you solve this problem. The driver of a 1.19 x 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 7.89 × 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? 6.63 X The response you submitted has the wrong sign.m (b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collision occur? m/s EXERCISE HINTS: GETTING STARTED I IM STUCK! Use the values from PRACTICE IT to help you work this exercise. A police investigator measures straight skid marks 28.0 m long in an accident investigation. Assuming a friction force and car mass the same as in the problem above, what was the minimum speed of the car when the brakes locked? m/s Need Help? Read It

Explanation / Answer

(a)

Acceleration due to brakes

a = -7..89*103/1.19*103 = -6.63 m/s^2

applying third equation of motion,

v^2 = u^2 +2as

0 = 35^2 -2*6.63 *s

s = 35^2/[2*6.63]

s = 92.38 m

b)

applying third equation of motion,

v^2 = u^2 +2as

v^2 = 35^2 - 2*6.63*30

v = 30.44 m/s

problem2

3rd equation : v2 = u2 + 2as

a = -7..89*103/1.19*103 = -6.63 m/s^2

v = 0

s = 28 m

0 = u2 + 2*(-6.63)*28

u2 = 185.64

u = 13.62 m/s

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