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Wanted: Equation for Distance Traveled Given Time

  1. Sep 25, 2004 #1
    Hi --

    I know this is so easy but right now I am a box of rocks. I think I am doing too much calculus and physcis problems and just can not think of the simple things.

    What I am trying to do is solve this problem.


    A crate with mass 34.0 initially at rest on a warehouse floor is acted on by a net horizontal force of 140

    1) What acceleration is produced?

    Answer is 4.12 m/s^2

    2) How far does the crate travel in 14.0 s?

    I know the initial position is 0 since the crate is at rest, and the acceleration is constant here so I need to figure the final position after 14.0 seconds at this acceleration.

    Any help is appreciated. Thanks.
  2. jcsd
  3. Sep 25, 2004 #2
    Google saved the day for me.

    x = v_i(t) + .5a(t^2)
  4. Sep 25, 2004 #3
    well d^2(x)/dt^2 = a right?
    integrate once to get dx/dt = at + c . We know v(0) = 0 since it was initially at rest dx/dt = at

    integrate again to get position: x(t) = 1/2*a*t^2 + c. we can use the original starting point as x(0) = 0 so c = 0

    You are given a time and an acceleration so... plug it in. Don't worry it happens to the best of us (me not being the best but shrug)
  5. Sep 25, 2004 #4
    thanks for the calculus approach. i think now i can remember what to do using calculus :smile:
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