Acar traveling at 43 km/h hits a bridge abutment. a passenger in the car moves forward a distance of 54 cm (with respect to the road) while being brought to rest by an inflated air bag. what magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 49 kg?

F = -6472.9 N

F= -6.47 kN

Explanation:

First of all you have to convert the data to SI  units

so for the velocity you have :

Vi = 43km/h *(1000m/1km)*(1h/3600s)  ---> using conversion factors

Vi= 11.9444 m/s

dX : distance the passanger moves

dX = 54cm*(1m/100cm)   -->  using conversion factors

dX = 0.54 m

Now to calculate the force we are going to use the sum of focers equals to mass for acceleration:

Sum F = m*a

We have to find a so we are going to use the velocity's formula as follows to solve a:

Vf ^2 = Vi^2 +2*a*dX

Vf=0  --> the passenger does not move after the airbag inflates.

a= -(Vi^2)/(2*dX)

you solve de acceleration with the data you hae and you will find

a = -132.1 m/ s^2

Now you can solve the Sum F equation

Sum F = 49 Kg * (-132.1 m/s^2)

F = -6472.9 N

F= -6.47 kN

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the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank