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Displacement equation due to acceleration

  1. Nov 10, 2015 #1
    Using the formulas: s = [tex] \frac{1}{2} \alpha t^2 [/tex]
    v = [tex] \frac{d}{t} [/tex]
    a = [tex] \frac{v}{t} [/tex]​

    When we divide distance "s" by time we get velocity:
    v = [tex] \frac{\frac{1}{2} \alpha t^2}{t} [/tex] = [tex] \frac{1}{2} \alpha t [/tex]
    When we divide velocity "v" by time we get acceleration:
    a = [tex] \frac{\frac{1}{2} \alpha t}{t} [/tex] = [tex] \frac{1}{2} \alpha [/tex]​

    ½a ≠ a



     
  2. jcsd
  3. Nov 10, 2015 #2
    You need to use calculus here. The instantaneous velocity is not the same as the average velocity.
     
  4. Nov 10, 2015 #3
    So the velocity = the derivative with respect of time of acceleration?
     
  5. Nov 10, 2015 #4
    You got that backwards. Acceleration is derivative of velocity wrt time
     
  6. Nov 10, 2015 #5
    So does that make v = ∫a ?
     
  7. Nov 10, 2015 #6
    yeah, plus an integration constant
     
  8. Nov 10, 2015 #7
    Thank you.
     
  9. Nov 10, 2015 #8

    russ_watters

    User Avatar

    Staff: Mentor

    Note that under constant acceleration, the calculus is easy and you can probably even see without calculus that the average speed under a linear acceleration is half the final speed.
     
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