(a) 328.6 kg m/s
The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:
[tex]I=\Delta p = m\Delta v[/tex]
where
m = 62.0 kg is the mass of the passenger
[tex]\Delta v[/tex] is the change in velocity of the car (and the passenger), which is
[tex]\Delta v = 5.30 m/s - 0 = 5.30 m/s[/tex]
So, the linear impulse experienced by the passenger is
[tex]I=(62.0 kg)(5.30 m/s)=328.6 kg m/s[/tex]
(b) 404.7 N
The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:
[tex]I=F \Delta t[/tex]
where in this case
[tex]I=328.6 kg m/s[/tex] is the linear impulse
[tex]\Delta t = 0.812 s[/tex] is the time during which the force is applied
Solving the equation for F, we find the magnitude of the average force experienced by the passenger:
[tex]F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N[/tex]
