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Acceleration Question

  1. Sep 6, 2004 #1
    An Automobile starting from a rest at t=0 undergoes constant acceleration on a staight line. It is observed to pass two marks separated by 64m, first at t=8 and the second at t=12s. What is the value of the acceleration?

    I know how to find the Vel. from this (64m/4s) but i dont understand how im able to find the acceleration?

    The answer in the book is 1.6 m/s^2 but i what to understand how to do it. Thank You

  2. jcsd
  3. Sep 6, 2004 #2


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    Dearly Missed

    1.Your expression for the (instantaneous) velocity is totally wrong; you're mixing up with the average velocity.

    2. Ask yourself:
    How is "distance travelled in time t" related to constant acceleration?

    3. Welcome to PF!
  4. Sep 6, 2004 #3
    Man im so lost. Ok i relized i used the average velocity, but what i need to do is use the instantaneous velocity? I never did understand inst. vel?

    Thanks for welcoming me, i loved physics in high school but so far this college physics is :confused: :mad: :cry:
  5. Sep 6, 2004 #4


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    OK, let's take it slow&easy:
    What formulas do you know which is about constant acceleration?
  6. Sep 6, 2004 #5
    constant acceleration =(v(t2)-v(t1))/(t2-t1) = chance in v over the change in time.
  7. Sep 6, 2004 #6


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    Do you know d= (1/2)a t2 for constant acceleration starting from rest?
  8. Sep 6, 2004 #7
    So i guess i can try and use the formula and switch it around. a=2d/(t^2)
  9. Sep 6, 2004 #8
    ummm but the problems is that this in not from a dead stop.
  10. Sep 6, 2004 #9
    Either the answer in the book is wrong, or the question is flawed.

    we know that vi = 0

    so vf/2 = 64/8 = 16m/s

    a = 16m/s/8s = 2m/s^2. If the acceleration is constant then it should be 2m/s^2 the whole time. However, i don't know where the 1.6m/s^2 came from, unless it didn't start from rest....
  11. Sep 6, 2004 #10
    Well it starts from a rest but what we are looking is already moving. It start at 0 but what we are looking at is the section from 8t to 12t
  12. Sep 6, 2004 #11

    read it once more....

    and besides if it's constant acceleration, it should be the same in the second section.
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