Problem:

Question

A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 "g's.

Part A:

Calculate the magnitude of the force on a 71-kg person accelerating at this rate.

Answer=> F= 2.09x10^4 N

Part B:

What distance is traveled if brought to rest at this rate from 95 km/h ?

Answer=> d=______m

Answer 1

A. F = M*g*30 = 71*9.8*30 = 20,874 N. = 2.09*10^4 N.

B. Vo = 95km/h = 95,000m/3600s = 26.4 m/s.

F = M*a = -20,874. a = -20,873/71 = -294 m/s^2.

V^2 = Vo^2 + 2a*d. V = 0, Vo = 26.4 m/s, a = -294 m/s^2, d = ?.