Automobile Air Bags. The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than $\mathbf{250}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. If you are in an automobile accident with an initial speed of $\mathbf{105}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}\mathbf{}\mathbf{(}\mathbf{65}\mathbf{\hspace{0.33em}}\mathbf{mi}\mathbf{/}\mathbf{h}\mathbf{)}$ and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

The given data can be listed below as:

• The human body can survive an incident if the acceleration becomes less than $250\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$.
• The initial speed of the automobile is $105\mathrm{km}/\mathrm{h}=105\mathrm{km}/\mathrm{h}\times \left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\times \left(\frac{1\mathrm{hr}}{3600\mathrm{s}}\right)=29.17\mathrm{m}/\mathrm{s}.$

This law states that a particular body will continue to be in uniform motion unless it is resisted by an external force.

The difference between the square of the final velocity and the initial velocity divided by the acceleration gives the distance required to stop the airbag.

From Newton’s first law, the velocity required for the airbag is expressed as:

$v^2=v_0^2+2ad$

The final velocity v is zero as the airbag has come to rest,$v_0$ is the initial velocity 29.17 m/s and a is the acceleration of the airbag which is negative.

Substituting the values in the above expression, we get-

$$\begin{gathered} 0\mathrm{m}/\mathrm{s}=29.17\mathrm{m}/\mathrm{s}+2\times \left(-250\mathrm{m}/{\mathrm{s}}^2\right)\times d \\ d=0.065\mathrm{m} \end{gathered}$$

Thus, the airbag must stop at 0.065 m in order to survive the crash.