The final velocity of the passenger is zero as he is brought to rest by the inflated bag.
[tex]V_f = 0[/tex]
Apply the equation of motion
[tex]V_f^2 = V_i^2 +2as[/tex]
Replacing with our values,
[tex]0 = 28^2+2(a)(0.55)[/tex]
[tex]a = \frac{28^2}{2(0.55)}[/tex]
[tex]a = 712.72m/s^2[/tex]
Calculate the force using the force equation,
[tex]F = ma[/tex]
[tex]F = (49kg)(712.72m/s^2)[/tex]
[tex]F = 34.923kN[/tex]
Therefore the magnitude of force acts on the passenger's upper torso is 34.923kN
