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Constant acceleration of an antelope

  1. Mar 16, 2014 #1

    An antelope moving with constant acceleration covers the distance 80.0m between
    two points in time 6.70s. Its speed as it passes the second point is 14.6 m/s.
    1) What is its speed at the first point?
    2) What is the acceleration?

    First I understand that when we have (constant acceleration) that means --> Vo=0m/s

    *_______________* If that was the two points , so t1=0 , t2=6.70 and the all distance is 80
    Vx=14.6
    For 1 ) I though the answer will be zero OR can we do 80/6.70=11.9 , I really have an issue to know if the speed they are talking about is v=Δx/t or they are talking about S=D/t
    2) α= Δv/Δt --> 14.6/6.70 = 2.2m/s^2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 16, 2014 #2

    Dick

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    Constant acceleration doesn't mean v0=0m/s. It just means the acceleration 'a' is constant. What are formulas for displacement and velocity assuming initial velocity v0 and constant acceleration a? Something like x=v0*t is only true if there is no acceleration.
     
    Last edited: Mar 16, 2014
  4. Mar 16, 2014 #3
    Sorry , but I didn't get it , ok constant a , doesn't mean Vo=0
    Can you explain the Q. more for me
     
  5. Mar 16, 2014 #4

    Ray Vickson

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    You can fire a toy rocket forward with constant acceleration, either starting from rest or starting from a moving car. The acceleration is the same, but the two situations give different velocities; they differ by the velocity the car had at the launch point. It's not rocket science!
     
  6. Mar 16, 2014 #5

    Dick

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    If there is no acceleration you can write the displacement as x(t)=x0+v0*t, where x0 is the initial position and v0 is the initial velocity. There is a similar expression for the case of constant acceleration. You must have seen it before. Try to find it.
     
  7. Mar 16, 2014 #6
    Ok thank you both
     
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