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, the bags produce a maximum acceleration of 60g, but lasting for only 32 ms (or less). How far does a person travel in coming to a complete stop in 32 ms at a constant acceleration of 60g

Answer 1

Answer:

s = 0.3 m

Explanation:

In order to calculate the distance traveled by the person, we will use the second equation of motion:

$s = v_it + \frac{1}{2}at^2$

where,

s = distance traveled = ?

vi = initial speed = 0 m/s

t = time = 32 ms = 0.032 s

a = acceleration = 60g = (60)(9.81 m/s²) = 588.6 m/s²

Therefore,

$s = (0\ m/s)(0.032\ s)+\frac{1}{2}(588.6\ m/s^2)(0.032\ s)^2$

s = 0.3 m