91Ó°ÊÓ

The given data can be listed below as:

$60 g=60×9.8 m/s^2=588 m/s^2$.

• The airbags last for $36 ms=36×{10}^{-3} s=0.036 s$.
• The person has to completely stop in 36 ms.
• The person has a constant acceleration, 60 g.

This law describes that an object will continue to move at a constant or uniform speed unless the object is acted by an external force.

The equation of the motion along a straight line gives the displacement a person needs to travel to a complete stop.

From Newton’s first law, the displacement of the person is expressed as:

$s=V_{i}t+\frac{1}{2}a{t}^2$

Here, s is the distance traveled,$v_i$ is the initial speed which is 0m/s as the person was a rest, t is time-traveled which is 0.036 s and a is the acceleration which is $588 m/s^2$.

Substituting the values in the above expression, we get-

$$\begin{gathered} s=(0\mathrm{m}/\mathrm{s}×0.036\mathrm{s})+\frac{1}{2}\left(588\mathrm{m}/{\mathrm{s}}^2\right)×(0.036\mathrm{s}{)}^2 \\ =0.38\mathrm{m} \end{gathered}$$

Thus, the person travels in coming to a complete stop 0.38 m.