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I am working on a constant acceleration problem, and cannot figure out what I am doing wrong as I keep getting the wrong answer.

The situation is:

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 36 ms (or less).

And the question is:

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

So far to solve this I have been using the equation: x = init x + vt - 1/2 (a)(t^2)

I than do:

X = 0 + 0*(0.036)-1/2*(60)*(0.036^2)

X = 1/2(60)(0.036^2)

X = 0.04

However, this is not correct what am I missing/doing wrong here?

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closed as off-topic by ACuriousMind, Gert, user108787, user36790, John Rennie Sep 17 '16 at 5:39

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  • $\begingroup$ What is the numeric value of "60g"? $\endgroup$ – DJohnM Sep 17 '16 at 1:02
  • $\begingroup$ @DJohnM originally had no clue, but after reading some comments it may be gravity? $\endgroup$ – Jack Sep 17 '16 at 2:03
  • $\begingroup$ Hi Jack and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Sep 17 '16 at 5:39
  • $\begingroup$ @JohnRennie thanks for the explanation, I'll try to make my questions more open in the future. $\endgroup$ – Jack Sep 17 '16 at 18:00
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You left out g-- the acceleration is 60g, not 60. Use g=9.8 m/s^2 and make sure your units make sense. Always check units-- quantities are meaningless without the correct units attached to them.

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  • $\begingroup$ Huge thanks! I was unaware of what that g stood for and skipped over it accidentally when trying to figure out the question. Adding gravity fixed it. $\endgroup$ – Jack Sep 17 '16 at 2:06
  • $\begingroup$ You're welcome. By the way, some commenters thought the problem was you didn't include the initial velocity, but notice that what you really did was essentially run the problem backward in time. You started where the particle stopped, and accelerated it away from that point for a given time, ending up at the correct starting point, so that is a valid way to get the distance-- in fact, it's the clever way. $\endgroup$ – Ken G Sep 17 '16 at 16:05
  • $\begingroup$ Thanks for clarifying that, the explanation helped understand what the formula I was using was doing, I had read from the text book you could use final velocity there instead of initial but I wasn't completely sure why either, that explanation helped. $\endgroup$ – Jack Sep 17 '16 at 17:57
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The equation you are using is basically correct. The more general form, $x(t)= x_0 + v_0 t + \frac{1}{2} a t^2$, gives you the position as a function of time. Now, the acceleration you have to use is $a=-60g$. Since you care about distance travelled, you can take $x_0=0$. But, your initial velocity $v_0$ is not $0$. You have to calculate it knowing that in 36 ms you come to a complete stop if your accelerations is $-60g$.

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You need to find the initial velocity. In your answer, you put that initial velocity was $0$ but that is not correct. Since in auto accident, auto has a initial velocity, right? So, $0$ will be your final velocity. Hope you get it..

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