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Homework Help: Kinematics Problem

  1. Feb 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.

    Who wins the race? when and where did ophelia catch up? (both metres and time)

    2. Relevant equations


    3. The attempt at a solution
    Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)

    I got stuck trying to find out WHEN they caught up. I tried setting the equation to


    and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
  2. jcsd
  3. Feb 24, 2008 #2
    Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or [itex]\triangle D_P = \triangle D_A + 13[/itex]
    Last edited: Feb 24, 2008
  4. Feb 24, 2008 #3


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    Gold Member

    If you know where they crossed, plug that x value into Alvin's EOM to get t.
  5. Feb 24, 2008 #4


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    Staff Emeritus
    Science Advisor
    Gold Member

    Start by listing the information you have:

    df = 100 m


    di = 13 m

    v(t) = vi = 7 m/s

    a = 0

    ==> d(t) = di + vit = 13 + 7t


    di = 0 m

    v(t) = vi = 1.2 m/s

    a = 1.5 m/s2

    ==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2

    It seems that your approach for answering the first question is correct: find out how long it takes for each person to run 100 m:


    100 = 13 + 7t

    t = (87 m) / (7 m/s) = 12.43 s


    100 = 1.2t + 0.75t2 =

    (3/4)t2 + (6/5)t - 100 = 0

    using the quadratic formula, t = 10.77 s

    It looks like you're doing alright so far.

    Now, to find the the position and time at which Ophelia catches up, you equate the two trajectories (since their positions are equal at this instant)

    13 + 7t = (3/4)t2 + (6/5)t

    0.75t2 - 5.8t - 13 = 0

    Solve for t using the quadratic formula with

    a = 3/4

    b = -5.8 = -29/5

    c = -13

    t = 9.5486 s


    d(t) = d(9.55) = 13 + 7(9.55) = 79.8402 m



    d(9.55) = 1.2(9.55) + 0.75(9.55)2 = 79.8402 m

    Since your methods are right, I went over the full solution. You must have just made a mistake in setting up the final quadratic equation
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