A car traveling at $\mathbf{53}\mathbf{}\mathbf{Km}\mathbf{/}\mathbf{h}$hits a bridge abutment. A passenger in the car moves forward a distance of$\mathbf{65}\mathbf{}\mathbf{cm}$(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$\mathbf{41}\mathbf{}\mathbf{kg}$?

• Speed of the car, ${\mathrm{v}}_{\mathrm{i}}=53\mathrm{km}/\mathrm{h}\mathrm{or}14.72\mathrm{m}/\mathrm{s}$.
• Distance traveled by the car,$\mathrm{x}=65\mathrm{cm}\mathrm{or}0.65\mathrm{m}$.
• Mass of passengers, $m=41\mathrm{kg}$.

As we know the initial and final speed of the passenger and the distance traveled by him, using kinematic equations, we can find the acceleration (Deceleration) of the person.

Using this acceleration and Newton’s 2^nd law we can find the force acting on it.

Formula:

The third kinematic equation of motion,${v_i}^2={v_f}^2+2ax$ (i)

The force according to Newton’s second law,$F=ma$ (ii)

Using the given values and equation (i), we get the acceleration of the car as:

$$\begin{gathered} {(14.7\mathrm{m}/\mathrm{s})}^2=0+2\times \mathrm{a}\times 0.65\mathrm{m} \\ 216.09=1.3\times \mathrm{a} \\ \mathrm{a}=-\frac{216.1}{1.3} \\ =-166.22\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Now, using equation (ii) and the given values, we get the magnitude of force as follows:

$$\begin{gathered} F=41\mathrm{kg}\times (-166.22\mathrm{m}/{\mathrm{s}}^2) \\ =-6815.52\mathrm{N} \\ =-6.8\times {10}^3\mathrm{N} \end{gathered}$$

Negative sign shows that the force is acting in the opposite direction of the motion and the magnitude value of force is$6.8\times {10}^3\mathrm{N}$ .