The front of a 1,500 kg car is designed as a \"crumple zone\"that collapses to a

Question

The front of a 1,500 kg car is designed as a "crumple zone"that collapses to absorb the shock of a collision. If a cartravelling 35.0 m/s stops uniformly in 1.65 m,
a) how long does the collision last? b) what is the magnitude of the average force
c) what is the acceleration of the car? Express your answeras a multiple of the accleration of gravity. (For example,"2g".)

The front of a 1,500 kg car is designed as a "crumple zone"that collapses to absorb the shock of a collision. If a cartravelling 35.0 m/s stops uniformly in 1.65 m,
a) how long does the collision last? b) what is the magnitude of the average force
c) what is the acceleration of the car? Express your answeras a multiple of the accleration of gravity. (For example,"2g".)

Explanation / Answer

The acceleration of the car is derived from using the relation           v2 = u2 + 2as Where v is final velocity of the car = 0 m/s            u isinitial velocity of the car = 35 m/s            a isacceleration of the car            s isdistance traveled by the car = 1.65 m So that we have            a = -u2 / 2s              = - 35 * 35    /   2 * 1.65              = - 371.21 m/s2 a)       The time taken by the car to travelthe distance 1.65 m is              s = u t + 0.5 a t2         1.65 = 35 t + 0.5(-371.21 ) t2         1.65 = 35 t - 185.6 t2       solving this we get          t = 0.0948 s b) The average force experienced by the car is         F = m a       Where m is mass of the car = 1500kg                 a is acceleration of the car = -371.21 m/s       So that we have         F = m a            =1500 * -371.21           = - 556815 N c) The acceleration of the car a = -371.21m/s2 = -37.81 g

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