Airbag stopping distance with constant acceleration

[Mdhiggenz]

Homework Statement

During an auto accident, the vehicle's air bags deploy and slowdown the passengers more gently than if they had hit the windshieldor steering wheel. According to safety standards, the bags producea maximum acceleration of 60, but lasting for only 36 (or less).

How far (in meters) does a person travel in coming to a completestop in 36 at a constant accelerationof 60?

Homework Equations

s=1/2a(t)^2

The Attempt at a Solution

What I first did was analyze what I have. I know I have an acceleration of -588m/s^2 ( due to the a_x=-g) also I have the time which is 0.036s.

I initially wanted to use the forumula vx²=v₀x² +2a_x(x-x₀)

Here is where I constantly get confused in these types of problems I know one of the velocities is going to be zero but how do I know which one to make zero.

So what I wanted to do was acquire one of the velocities using v_x=v₀x+at. Then plug it into the above equation however the solution manual chose to use the following equation s=1/2a(t)^2.

Can someone explain to me why my thought process is wrong and why, and how they acquired the above equation.

Thank you

 

[Andrew Mason]

36 what? 60 what? What are the units? Where did you get your formula? The units are not correct. vx^2 has units of distance^3/time and ax has dimensions of distance^2/time^2

Assume the maximum acceleration (deceleration) is constant over the maximum stopping time. How would you determine initial speed from that? What is the final speed?

You may find it is simpler to use s = .5at^2

AM