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Math Help - [SOLVED] Mechanics Calculus - Average Speed

  1. #1
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    Cool [SOLVED] Mechanics Calculus - Average Speed

    Q) A motorcyclist starts from rest at A and travels in a straight line. For the first part of the motion, the motorcyclist’s displacement x metres from A after t seconds is given by x = 0.6t^2 − 0.004t^3.

    (i) Show that the motorcyclist’s acceleration is zero when t = 50 and find the speed V ms^-1 at this time.

    For t ≥ 50, the motorcyclist travels at constant speed V ms^-1.

    (ii) Find the value of t for which the motorcyclist’s average speed is 27.5ms^-1


    My Attempt

    I've already done part (i) to get v=30ms^-1
    I'm stuck at part (ii). By using calculus or any other easy method kindly show me how to solve this.

    Thanks in advance!
    Last edited by unstopabl3; May 17th 2010 at 10:46 AM.
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  2. #2
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    As we see that the velocity is positive for the first 50 seconds of journey therefore net displacement=net distance traveled.

    calculating displacement for first 50 seconds from x=0.6t^2-0.004t^3 we get 900 meters.

    then assuming that it travels for t more seconds at 30 m/sec therefore displacing 30t meters.

    Average speed=(Total Distance)/(Total time)

    on solving: 27.5=(900+30t)/(50+t)

    t=190 seconds....
    so net time time is 240 seconds from the starting place to get an avearge speed of 27.5m/sec
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  3. #3
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    Thanks for the quick reply! The distance for the first 50s is 1000 not 900. I think you made an error in a hurry

    But I've got the concept behind it. Therefore, thanks!
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