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Drag racer velocity and acceleration

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data

    A drag racer, starting from rest, speeds up for 402 m with an acceleration of +23.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.70 m/s2. How fast is the racer moving 3.35 102 m after the parachute opens?

    2. Relevant equations

    ?

    3. The attempt at a solution

    ?

    I have no clue where to begin this problem. Please help!
     
  2. jcsd
  3. Sep 17, 2007 #2

    G01

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    In order to get help here you have to show some work. It's the rules. You can't possibly know NOTHING about this problem. Give me your thoughts at least. What equations will be involved? What principles are involved?
     
  4. Sep 17, 2007 #3
    It is clearly a kinematics problem, so I am guessing it has to do with acceleration and velocity. All the variables are given in terms of acceleration and the final answer has to be in terms of velocity, so I assume the starting equation will be: vf = vi + a*t. The only thing is that any of the equations I look at, they are all missing two or more variables.
     
  5. Sep 17, 2007 #4

    G01

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    Ok, now do you know the time the parachute opens or is this not given?

    If your not given the time, I suggest looking for a different equation.
     
  6. Sep 17, 2007 #5
    It is not given. It does not give initial velocity either.
     
  7. Sep 17, 2007 #6

    G01

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    Ok, then, can you find the time that the car reaches the 402m mark from the information given?

    EDIT: Also, if the car starts at rest, what is the initial velocity?
     
    Last edited: Sep 17, 2007
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