A 1500kg car traveling at 28 m/s hits a bridge abutment coming to rest in 0.04s

Question

A 1500kg car traveling at 28 m/s hits a bridge abutment coming to rest in 0.04s (40 milliseconds). The driverof the car, who has a mass of 45 kg, moves forwards a distance of 75 cm while being  brought to rest by an inflated air bag. The NHTSA standard for sudden impact acceleration on a human that  would cause severe injury, or death, is 75 g's for a "50th percentile male", 65 g's for a "50th percentile female", and 50 g's for a "50th percentile child.

a) Calculate the accelerations of the car and of the passenger, and state them in units of both ms-2

and as a multiple of the gravitational acceleration 9.8 ms-2

.

b) Assuming that the forces that stop the car and stop the passenger are constant, what are the

magnitudes of forces acting on the car, and on the passenger?

c) How exactly does the airbag minimize physical bodily injury?

Explanation / Answer

a) car acceleration v= u +at

V=0 U=28m/s t=.04 s

0=28+a*.04

a=-700 m/s

car acceleration= -700 m/s or 71.42 g(gravitational unit)

man acceleration=?

man's initial speed= 28m/s final speed=0 distance moved=.75 m

V^2 = U^2 +2as

0= 28^2 + 2a* .75

a= -522.66 m/s   

man's acceleration= -522.66m/s or 53.33g

B) force on car= m*a = 1500*-700 = -105000 N

force on man= 45*- 522.66 = -23519.7 N

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