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Drag care race, finding distance and time

  1. Oct 15, 2008 #1
    EDIT***: sorry I figured out how to do this problem and don't need help anymore...:\
    Disregard and even delete this thread.

    1. The problem statement, all variables and given/known data
    A race-car driver buys a car (CAR A) that can accelerate at +5.9 m/s2 . The racer decides to race against another driver with a car (CAR B) that can move with a constant acceleration of +3.6 m/s2. Both start from rest, but CAR B leaves 1.0 s before the driver of CAR A.

    Find the time it takes CAR A to overtake CAR B

    Find the distance the two drivers travel during this time.

    Find the velocities of both cars when they are side by side.


    2. Relevant equations

    I'm not sure if this is what to use, but
    Final velocity = Initial velocity + at
    or maybe
    d = (initial velocity)(t) + 1/2 (a)(t)2

    3. The attempt at a solution

    Because car B has a 1 second lead, it will have established a 3.6 m lead on car A and already entered acceleration when car A gets off the line. I would think then that car B has an initial velocity of 3.6 while car A's is 0. I'm trying to find t where d1=22

    So d1 = 0 + 1/2 (5.9)(t)2
    d2 = 3.6 + 1/2 (3.6)(t)2

    if these are set equal to equal to each other and the first is solved for t, I get t2=2.95

    then I plugged this number into the t2 in the other equation and get 3.6 + 1/2(3.6)(2.95)

    = 8.91s

    Did I do this right?

    vvvvv

    EDIT***: sorry I figured out how to do this problem and don't need help anymore...:\
    Disregard and even delete this thread.
     
    Last edited: Oct 15, 2008
  2. jcsd
  3. Oct 15, 2008 #2
    You can do it this way if you like it.
    The distance traveled by the first car in 1 s is not 3.6 m. Look again at your second formula.
    Then, when you write again the distance traveled by the first car, it has an initial velocity of 3.6 m/s. You should have this term too (x=xo+vo*t + 1/2 a*t^2)
     
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