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, the bags produce a maximum acceleration of 60 g, but lasting 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 60g?

At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop

Explanation:

The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s

To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

we write out the equation of motion thus.

S = ut + 0.5at²

wgere

S = distance to come to complete stop

u = final velocoty = 0 m/s

a = acceleration = 60g = 60 * 9.81

t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0*0.036 + 0.5*60*9.81*0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop