Question

During an auto accident, the vehicle's airbags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, airbags 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?

Answer 1

Answer:

0.38 m

Explanation:

As we know that the person due to the airbag action, comes to a complete stop, in a time of 36 msec or less, and during this interval, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:

vf = v₀ -a*t  

If vf = 0, we can solve for v₀:

v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s

With these values of v₀, a and t, we can find Δx, applying any kinematic equation that relates these parameters with the displacement.

Just for simplicity, we can use the following equation:

 $vf^{2} - vo^{2} =2*a*d$

where vf=0, v₀ =21.2 m/s and a= -588 m/s².

Solving for d:

 $d =\frac{21.2 m/s}{588 m/s2} = 0.38 m$

⇒ d = 0.38 m