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Constant Acceleration of a cheetah

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data

    A cheetah is hunting. Its prey runs for 3.2 s at a constant velocity of +10.0 m/s. Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

    2. Relevant equations

    V=Vo + a(t)

    3. The attempt at a solution

    I plug in the numbers to get 3.1 m/s but i know im missing something. Help with answer would be appreciated!
     
  2. jcsd
  3. Aug 31, 2009 #2

    rl.bhat

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    What is the distance traveled by the cheetah?
    Use the kinematic equation d = vo*t + 1/2*a*t^2 to find a.
     
  4. Aug 31, 2009 #3
    this is all the problem says, doesnt give a distance.
     
  5. Aug 31, 2009 #4

    rl.bhat

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    Velocity of the cheetah is given, time taken be it is given. What is the relation between distance, velocity and time?
     
  6. Aug 31, 2009 #5
    What distance does the prey runs?

    You know the d now, t is 3.2s ... find a
     
  7. Aug 31, 2009 #6
    still have no idea what the distance is
     
  8. Aug 31, 2009 #7

    rl.bhat

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    Open any text book and find the relation between velocity, distance and time.
     
  9. Aug 31, 2009 #8
    distance= velocity x time right?
     
  10. Aug 31, 2009 #9

    rl.bhat

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    Right.
     
  11. Aug 31, 2009 #10
    but when you use d = vo*t + 1/2*a*t^2

    i get 0= 1/2*a*3.2^2

    which equals 0
     
  12. Aug 31, 2009 #11
  13. Aug 31, 2009 #12

    rl.bhat

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    d = 3.2 s*10 m/s
     
  14. Aug 31, 2009 #13
    ya but so does Vo*t on the right side of the equation so they cancel each other out
     
  15. Aug 31, 2009 #14
    it comes out to
    32=32 + 1/2*a*3.2^2
    0=1/2*a*3.2^2
    ?
     
  16. Aug 31, 2009 #15

    rl.bhat

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    Initial velocity of cheetah is zero.
     
  17. Sep 1, 2009 #16
    6.25 thank you!
     
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