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Math Help - Calculate acceleration with time and distance?

  1. #1
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    Calculate acceleration with time and distance?

    In the 74s after lift-off, the shuttle Challenger travels 36km. Assuming constant acceleration, calculate the acceleration of the shuttle in m/s^2.

    G:
    t = 74s
    d = 36km

    Which equation would I use to solve this? None of the equations I have will work, and I'm not sure which kind of velocity to calculate (instantaneous, average, initial, final). Everything I tried so far failed. The correct answer (according to my sheet) is 13m/s^2.
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  2. #2
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    Quote Originally Posted by shadow6 View Post
    In the 74s after lift-off, the shuttle Challenger travels 36km. Assuming constant acceleration, calculate the acceleration of the shuttle in m/s^2.

    G:
    t = 74s
    d = 36km

    Which equation would I use to solve this? None of the equations I have will work, and I'm not sure which kind of velocity to calculate (instantaneous, average, initial, final). Everything I tried so far failed. The correct answer (according to my sheet) is 13m/s^2.
    assuming the shuttle started from rest at t = 0 ...

    d = \frac{1}{2}at^2

    solve for a

    remember that d must be in meters
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  3. #3
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    Thanks for the help, got it right finally.

    Though, the second part of the question asks to find the speed of the shuttle in km/h. I did the following.

    S = a(t)
    S = 13m/s^2(74s)
    S = 962m/s
    S = 962(.001km)(.0003h)
    S = 962(3.33km/h)
    S = 3203.46km/h.

    The answer on the page however, says 3500km/h. Don't know if this was due to rounding or not, though I doubt that based on the high inaccuracy.

    I also tried this.

    S = d / t
    S = 36km / 74s
    S = 0.49km/s
    S = 0.49km/(0.0003h)
    S = 0.000147km/h

    I thought for sure this one would work, but I guess not.

    What would be a good equation for calculating the shuttle's speed?
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  4. #4
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     <br />
a \approx 13.15 \, \, m/s^2<br />

     <br />
v = at \approx 973 \, \, m/s \, \, \approx 3500 \, \, km/hr<br />
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