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Bike and car acceleration physics

  1. Jan 20, 2007 #1
    1. The problem statement, all variables and given/known data
    a car accelerates at 1.5ms^-2 from 0 ms^-1 to 30 ms^-1. It then travels at a constant speed of 30 ms^-1.

    2.A bike accelerates at 2.5ms^-2 from 0ms^-1 to 20 ms^-1 and then travels at a constant speed of 20 ms^-1

    find the instant where there together.
    (Both have the same displacement)

    2. Relevant equations
    all constant acceleration equations

    3. The attempt at a solution

    for part one, i can represent the distance as 300+30t.

    S=1/2(20)(30) + 30t= 300 +30t

    for part 2.


    S=1/2(8)(20) + 20t= 80 + 20t


    why is my answer coming in negative. My answer -22 is right but its negative.
  2. jcsd
  3. Jan 20, 2007 #2


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    Dearly Missed

    This is totally wrong!
    Your flaw lies in equating the times when both vehicles start moving with constant speed.
    Now, if we let t=0 represent the time when the CAR reaches constant speed, then during the twelve seconds from the BIKE reached constant speed to t=0, the bike has travelled an additional 12*20=240 meters.
    Thus, you are to solve
    320+20t=300+30t, meaning t=2.
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