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Acceleration from Time & Distance.

  1. Oct 11, 2008 #1
    Hi Everyone,

    I haven't had any Physics education so I'll apologise in advance because this question is probably going to be stupid.

    If I had something starting from rest (u = 0) and I have a distance (s = 0.8m), and time it takes to travel that distance (t = 1.9s), which way would you go about working out the acceleration?

    I can see two ways, but one of them is obviously incorrect because you end up with different answers.


    [tex]s = ut + 1/2at^2[/tex]

    Since u = 0

    [tex]s = 1/2at^2[/tex]


    [tex]a = 2s/t^2[/tex]

    Which with my numbers would give, a = 1.6/3.61 = 0.443ms^-2


    Find the velocity using s/t, so v = 0.8/1.9 = 0.42ms^-1

    then a = (v - u) / t = (0.42 - 0) / 1.9 = 0.221ms^-2


    Is one of those methods correct? I thought they would both come out the same... but they don't so I seem to be doing something horribly wrong.

    Any help would be greatly appreciated, many thanks,

    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 11, 2008 #2


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    Your statement of the problem allows many accelerations to be correct.

    1. Your first answer is correct for constant acceleration beginning at t=0.

    2. If your second answer for the velocity v is correct, then your answer for the acceleration is wrong. In this case, the velocity jumps from 0 to v at t=0, ie. infinite acceleration at t=0. After that, the velocity is constant and there is zero acceleration for the rest of the trip.
  4. Oct 11, 2008 #3


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    That was, in fact, the problem that lead to the development of the calculus. If you were in a spaceship far above the plane of the solar system, you could take a photograph, measure the distance from the sun to the planet and, if gravity were just a function of distance, calculate the force and so acceleration. But acceleration, as defined then, could not be calculated at a given instant as could the distance between the sun and the planet.

    So: to calculate acceleration either 1) get the position as a function of time and differentiate twice or 2) get the position at 3 different times so you can calculate two different (average) speeds and from that an average acceleration.
  5. Oct 11, 2008 #4
    If you can assume uniformly accelerated rectilinear motion, your first method is correct and the only problem with your second method is that you are using the average velocity to calculate the acceleration. The average velocity of 0.42 m/s, that you calculated is midway between the start, 0 and the final velocity at 1.9 seconds. So the final velocity is 0.84 m/s. If you use that number to calculate the average acceleration you get 0.84 / 1.9 = .442 m/s^2 which agrees with your first method. Again, all of this is assuming uniformly accelerated straight line motion.
  6. Oct 11, 2008 #5
    Thanks for the replies everyone! That's a great help.
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