91影视

Mass of the person$\left(m\right)=60kg$

Initial velocity of the car$\left(v_i\right)=15m/s$

Formula used :

The constant acceleration, the third equation of the kinematics is used. The initial and final velocities can be related with the following relation.

$v_f^2=v_i^2+2a鈭�x\left(I\right)$

The final velocity $\left(v_f\right)$is zero,

Therefore acceleration is given by $a=-\frac{v_i^2}{2a鈭�x}\left(II\right)$

The force acting on the passenger is given by

$F=ma=-\frac{m{v}_i^2}{2鈭�x}\left(III\right)$

Conclusion :

When the air bag is deployed, and the seat belt is hold the stopping distance will be

$鈭�x=1$

using the equation $\left(III\right)$

$F=ma=-\frac{m{v}_i^2}{2鈭�x}$

Force can be calculated by substituting the values of mass of the person, initial velocity of the car and the stopping distance.

$F=-\frac{\left(60kg\right){(15m/s)}^2}{\left(2\right)\left(1\right)}=-6750N$

Negative sign indicates that it is pushing the passenger back to the seat.

The net force on the person if he/she is wearing a seat belt and if air bag deploy is$-6750N$

When there are no restraints, then the stopping distance will be equal to

$鈭�x=5mmor鈭�x=0.005m$

Using the equation $\left(III\right)$

$F=ma=-\frac{m{v}_i^2}{2鈭�x}$

Force can be calculated by substituting the values of mass of the person, initial velocity of the car and the stopping distance.

$F=-\frac{\left(60kg\right){(15m/s)}^2}{\left(2\right)\left(0.005\right)}=-1.35脳{10}^{6}N$

This force is much larger than the restrained case.

Conclusion :

The net force that stops the person if he/she is not restrained is$-1.35脳{10}^{6}N$