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Displacement in terms of acceleration and time

  1. Sep 6, 2009 #1
    So by manipulating two of the kinematic equations you can find this relationship:

    a= acceleration
    t = time
    v = velocity
    v0 = initial velocity
    x = displacement

    eq (1) = v = v0 + at --> at = v - v0

    eq (2) x = 1/2*(v - v0)*t --> 2x/t = v - v0

    substituting 2x/t for v - v0 : at = 2x/t --> x = (at^2)/2

    however when using a different method to derive this relationship i got a different answer...

    Acceleration in classical mechanics is usually represented as m/s/s (meters over seconds squared) and is always constant. Acceleration can also be thought of as the rate of change of velocity, which is how we will think of it from now on.

    suppose after 1 second the velocity changes by a. The velocity at 1 second must be a because a-0 = a

    because acceleration is constant at 2 seconds the velocity must also change by a
    and therefore the velocity at 2 seconds must be 2a because 2a - a = a

    Similarly the velocity at 3 seconds must be 3a because 3a-2a = a, and the velocity at n seconds is therefore n*a

    to find the total displacement we must add the velocities at each second.

    for example the total displacement after 3 seconds = the velocities at 3, 2 and 1 seconds added together = 3a+2a+a = 6a

    to find the equation for this we can find the first few terms of this sequence

    (total displacement (x)) 1a 3a 6a 10a 15a
    (seconds (t)) 1 2 3 4 5 ...

    This is actually a quadratic sequence and we can find the equation for the sequence using the formula for the partial sum of a quadratic sequence. The answer we get is:

    x = ((t^2 + t)*a)/2

    this is different from x = (at^2)/2 and i'm wondering why...which one is correct?

    The second way was probably hard to follow sorry...
  2. jcsd
  3. Sep 6, 2009 #2


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    Science Advisor
    Homework Helper

    I didn't really follow what you did, but x = ((t^2 + t)*a)/2 is nonsense. You cannot add something of units [s^2] and something of units .
  4. Sep 7, 2009 #3

    Doc Al

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    Staff: Mentor

    Your final relationship is only valid when v0 = 0 (and for constant acceleration).

    Your equation (2) is incorrect. It should be x = 1/2*(v + v0)t.

    OK. You're just saying here that v = v0 + at, or v = at when v0 = 0.

    Why is that? To find the total displacement, you must add the individual displacements that occured in each second.

    Of course it's much easier just to find the average velocity (at/2) and multiply by the time to get x = 1/2at^2, as usual.

    That makes no sense.
  5. Sep 7, 2009 #4


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    Possibly your confusion is that you are treating the kinematic equations as somehow discrete. The method you set forward would be appropriate if the speed was exactly a for the whole of the first second, then jumped to being exactly 2a for the whole of the second second, and so forth. But this is not the case: the kinematic equations are continuous, if you have a constant acceleration a and an initial velocity of zero, then the velocity gradually rises to a in the first second, and then continues to increase until it hits 2a at the end of the second second and so forth. The velocity at t=2.376 seconds is 2.376 times a, not 3a (as your scheme would have it where the velocity is constant for the whole of the third second).

    Put another way: your approach is implictly using the relationship x = vt to sum each of the displacements in each second, but this relation only holds if the velocity is constant: which it isn't, it is constantly increasing due to the acceleration.
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