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Vt=d doesn't seem to hold true w/ constant acceleration

  1. Jun 17, 2015 #1

    Lets say I start off with constant acceleration of 1 m/s^2, and I want to evaluate it at 3 seconds.

    a = 1.
    a(3) = 1

    then, since at=v,

    v(t) = at = 1(3) = 3m/s then I can work backward v/t = a = 3/3 = 1

    then, since vt = d

    d(t) = vt = 3(3) = 9 then I can work backward d/t= v = 9m/3s = 3m/s

    This way, the units make sense acceleration is 1 m/s/s

    Take 9 meters. Divide it by 3 seconds. Then divide it again by 3 seconds. And voila, you arrive back at 1m/s/s


    However, if I integrate acceleration to get the kinematic equations, it all seems to fall apart and I don't know why.

    [itex]\int a (dt) = at +v_0 = v_f[/itex] ,

    so evaluating time = 3 with acceleration 1m/s^2,

    v(3) = 1(3)+0 = 3m/s then I can work backward v/t = a = 3/3 = 1 => all good still


    [itex]\int v (dt) = \frac{1}{2}at^2 +v_0(t)+r_0= r_f[/itex],

    [itex]r(3) = \frac{1}{2}(1)3^2+0+0=4.5[/itex] so if I work backward this time d/t=v = 4.5/3= 1.5 m/s

    But my velocity is 3m/s, not 1.5m/s

    So my conclusion is, Distance/Time = Velocity will only hold true if there is 0 acceleration. Is this correct?

  2. jcsd
  3. Jun 17, 2015 #2


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    Or if v is average velocity.

    Edit: which is what you get from integrating.
    Last edited: Jun 17, 2015
  4. Jun 17, 2015 #3


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    Nope. This is wrong. This is only true if v is a constant. Since your acceleration is not zero, v is NOT a constant, so this is incorrect.

    Think about it, If you have a constant v, you use this equation. You have a non-constant v, you ALSO used this equation. Something's got to give because it can't both be correct.

  5. Jun 17, 2015 #4
    Hi ZapperZ,

    Thank you for responding.

    May I ask, which part is wrong exactly? I am betting that just the first part is wrong and that the second part (which contradicts the first part) is correct, since my final conclusion is:

    I feel that by the end of the work, you are in agreement with my final conclusion. But maybe that you are correcting another point I have missed.

    When solving for velocity using each formula, at = v = 3m/s and at + v_0 = 3m/s,

    both arrive at the correct answer, but this is only when assuming initial velocity is 0.

    But it breaks down when solving for distance, regardless of whether I have initial velocity or initial position.

    Is saying "constant velocity" the same thing as saying "0 acceleration."?
  6. Jun 17, 2015 #5


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    Look at the last line I quoted in your post. The use of d=vt is invalid for varying v.

  7. Jun 19, 2015 #6
    Thank you. Much appreciated.
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