Question #963

A car and its driver travelling at a velocity of 18 meters per second are stopped abruptly by a collision with the rear end of a packed van.The cars front end is pushed by 0.55 meters by rhe collision.The 60 kg driver who was restrained by the seat belt and shoulder strap is also stopped at 0.55 meters.Calculate
a. The time the car has stopped.
b. The average force acting on the driver.
c. The deceleration of the drive.

Expert's answer

The equations of the motion:

v=v_{0}+ at.

0 = 18 m/s - at

s=s_{0} +v_{0}t - at^{2}/2

0.55 = 18 t - at^{2}/2

We have to solve this system of equations:

at = 18; a = 18/t

0.55 = 18t - 18t/2 = 9tt = 0.55/9 = 0.06 s.

a = 18/0.06 = 300 m/s2

F = ma = 60*300 = 18,000 Na. 0.06s b. 18,000 N c. 300 m/s2

v=v

0 = 18 m/s - at

s=s

0.55 = 18 t - at

We have to solve this system of equations:

at = 18; a = 18/t

0.55 = 18t - 18t/2 = 9tt = 0.55/9 = 0.06 s.

a = 18/0.06 = 300 m/s2

F = ma = 60*300 = 18,000 Na. 0.06s b. 18,000 N c. 300 m/s2

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