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Acceleration in function of time and distance

  1. May 1, 2012 #1
    Here is something that has got me confused a lot of times.

    Suppose I have a distance of 11m and a time of 5 seconds and I want to know acceleration.

    I would say that [itex] a = v / t[/itex] and [itex] v = d / t [/itex] so I could plug the second equation to the first equation having: [itex]a = d / t^2[/itex]

    Therefore I have [itex] a = 11 / 5^2m/s^2[/itex] but that is the wrong answer.

    We also have [itex]x=x_0+v_0 t+1/2 at^2[/itex] so: [itex] a = 2d / t^2 [/itex] which in that case is [itex] a = 2(11) / 5^2 [/itex] and this is right.

    But why is the reason that distance should be the double of it?? Can someone explain me please??

    Thank you
     
  2. jcsd
  3. May 1, 2012 #2
    acceleration is not v/t and velocity is not d/t you got these wrong. Go back to the definition.
     
  4. May 1, 2012 #3

    haruspex

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    In case it's not clear, Curl is telling you a = v/t is only true for constant acceleration, and where v is a change in speed and t the time over which it happened. Similarly v = d/t is for constant speed etc. Since you have acceleration, the second is definitely not valid.
     
  5. May 1, 2012 #4
    We also have x=x0 +v0 t+1/2at2

    IS a correct equation....in this case, however, the nomenclature looks a bit different from those explained in the previous post. You always need to consider the applicability of an equation you wish to use against the conditions in the problem you are solving...in this equation
    x0 is a fixed initial distance; v0 is a constant velocity and a is a constant acceleration....[it's potentially confusing when two of the constants have a subscript yet acceleration doesn't even though it is also a [fixed] constant.]
     
  6. May 3, 2012 #5
    Thank you guys
     
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