Question

A car traveling at 28.0 m/s hits a bridge abutment. A passenger in the car, who has a mass of 49.0 kg, moves forward a distance of 55.0 cm while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso?

Answer 1

The final velocity of the passenger is zero as he is brought to rest by the inflated bag.

$V_f = 0$

Apply the equation of motion

$V_f^2 = V_i^2 +2as$

Replacing with our values,

$0 = 28^2+2(a)(0.55)$

$a = \frac{28^2}{2(0.55)}$

$a = 712.72m/s^2$

Calculate the force using the force equation,

$F = ma$

$F = (49kg)(712.72m/s^2)$

$F = 34.923kN$

Therefore the magnitude of force acts on the passenger's upper torso is 34.923kN