A driver in a 1000 kg car traveling at 32 m/s down a slope, which is 10 degrees

Question

A driver in a 1000 kg car traveling at 32 m/s down a slope, which is 10 degrees below the horizontal, slams on the brakes and skids to a stop. If the coefficient of static friction between the tires and the road surface is 0.95 and the coefficient of kinetic friction is 0.80, how long will the skid marks be? [Notice that the answers are shown to 2 significant figures.] Working this kind of calculation backwards allows crash inspectors to decide how fast a car was travelling by measuring the skid marks.

Explanation / Answer

after breaks kinetic friction will work.

f = u.N = 0.80 x 1000g x cos10   

mgsin10 - f = ma

1000g x sin10 - 0.80 x 1000g x cos10 = 1000a

a = - 6.02 m/s2

v^2 - u^2 = 2ad

0^2 - 32^2 = 2 x (-6.02) x d

d =85.06 m

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