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. How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g?

Answer 1

0.381 m

Step-by-step explanation:

Distance traveled S is found by

`S = ut + (1)/(2)a{t^2}`

Where S is distance traveled, u is initial velocity, t is time, a is acceleration

Since we acceleration a is given as 60g, where g is gravitational constant of 9.81 then a=60*9.81=588.6

The initial velocity u is zero hence ut=0

Substituting a with 588.6, t with 36 ms

`\begin{array}{c}\\S = 0 + (1)/(2)\left( {588.6 {\rm{m/}}{{\rm{s}}^2}} \right){\left( {\left( {36 {\rm{ms}}} \right)\left( {\frac{{1 {\rm{s}}}}{{{{10}^3} {\rm{ms}}}}} \right)} \right)^2}\\\\ = 0.3814128 {\rm{m}}\\\end{array}`

S=0.381 m