A car that weighs 1.0 × 10^4 N is initially moving at a speed of 37 km/h when th

Question

A car that weighs 1.0 × 10^4 N is initially moving at a speed of 37 km/h when the brakes are applied and the car is brought to a stop in 17 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

Last guy who tried this was wrong.

Explanation / Answer

mass = 1.0*10^4/9.8 = 1020.40 kg initial speed vi = 37 km/h = 10.27 m/s final speed vf = 0 stopping distance S = 17 m (a) According to equation of motion, vf^2 -vi^2 = 2aS ==> a = - (10.27)^2/(2*17) = - 3.106 m/s^2 the force required, F = ma = 1020.40*3.106 = 3170.22 N (b) From, vf = vi + at the time taken to come to stop, t = -vi/a = 3.30 seconds (c) if initial speed of the car is doubled, Breaking force , F' = ma = m(-vi^2/2S) ==> S = -mvi^2/2F' here the stopping distance is directly proportional to the vi^2, hence if initial speed doubled,same breaking force then stopping distance is increased by 4 times. (d) the stopping time is increased by 2 times.

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