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Math Help - A question about this velocity/acceleration question

  1. #1
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    A question about this velocity/acceleration question

    I'm not sure if it's even solvable.

    "A deer jumps in front of a car. The driver immediately slams on the breaks, leaving a skid mark .020 km long. If a nearby policeman clocked the driver at a constant velocity of 50 km/hr what was the deceleration after the breaks were applied."

    Now I have:

    D= .020 KM
    V0= 50 km/hr
    V= 0
    A= ?
    T= ?

    and as there is no T value, I'm not exactly sure how to proceed. Seeing as the velocity formula is v= v0+at.

    Any help would be greatly appreciated.

    Edit: Am I to assume that "immediately" means 0 seconds? If that is the case, there wouldn't be any skid marks, as he would have stopped perfectly at 0.00 seconds...
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  2. #2
    Member
    Joined
    Jan 2009
    Posts
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    Quote Originally Posted by Caturdayz View Post
    I'm not sure if it's even solvable.

    "A deer jumps in front of a car. The driver immediately slams on the breaks, leaving a skid mark .020 km long. If a nearby policeman clocked the driver at a constant velocity of 50 km/hr what was the deceleration after the breaks were applied."

    Now I have:

    D= .020 KM
    V0= 50 km/hr
    V= 0
    A= ?
    T= ?

    and as there is no T value, I'm not exactly sure how to proceed. Seeing as the velocity formula is v= v0+at.

    Any help would be greatly appreciated.

    Edit: Am I to assume that "immediately" means 0 seconds? If that is the case, there wouldn't be any skid marks, as he would have stopped perfectly at 0.00 seconds...
    Convert everything into SI units. Then solve for A using 2aD = V^2 - V_0^2.
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