Q25E - University Physics with Modern Physics

Question

During an auto accident, the vehicle’s air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, air bags produce a maximum acceleration of 60g that lasts for only 36 ms (or less). How far (in meters) does a person travel in coming to a complete stop in 36 ms  at a constant acceleration of 60 g ?

Step-by-Step Solution

Answer

The person travels in coming to a complete stop is, 0.38 m.

Step 1: Identification of the given data

The given data can be listed below as:

• The airbags produce a maximum acceleration of, 

$60\hspace{0.33em}g=60\times 9.8\hspace{0.33em}m/s^2=588\hspace{0.33em}m/s^2$.

• The airbags last for $36\hspace{0.33em}ms=36\times {10}^{-3}\hspace{0.33em}s=0.036\hspace{0.33em}s$.
• The person has to completely stop in 36 ms.
• The person has a constant acceleration, 60 g.

Step 2: Significance of Newton’s first law for the person

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.

Step 3: Determination of the displacement a person travels in coming 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 $588\hspace{0.33em}m/s^2$.

 

Substituting the values in the above expression, we get-

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

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