Question

during an auto accident, the vehicles air - bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. according to safety standards, the bags produce a maximum acceleration of 60g, but lasting for only 38 ms (or less). part a how far (in meters) does a person travel in coming to a complete stop in 38 ms at a constant acceleration of 60g? express your answer in meters to two significant figures.

Explanation:

Step1: Convert time to seconds

\( t = 38 \, \text{ms} = 38 \times 10^{-3} \, \text{s} = 0.038 \, \text{s} \)

Step2: Calculate acceleration

\( a = 60g = 60 \times 9.8 \, \text{m/s}^2 = 588 \, \text{m/s}^2 \)

Step3: Find initial velocity

From \( v = u + at \), \( v = 0 \), so \( u = at = 588 \times 0.038 \, \text{m/s} \)

Step4: Compute distance

Using \( d = \frac{u}{2}t \), substitute \( u = at \), get \( d = \frac{1}{2}at^2 \times 2 = 30gt^2 \)
\( d = 30 \times 9.8 \times (0.038)^2 \approx 0.42 \, \text{m} \)

Answer:

0.42