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Math Help - Big Problem :/

  1. #1
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    Big Problem :/

    This problem is kinda big and just cant seem to start it off :/.

    A new motion detector is used to graph the position of a sports car. If the themotion detector is not callibrated properly, it might give unrealistic graph data. The position of the car is described by the function S= 2t^3-21t^2 + 60, tE [0,6], where s is measured in meters and t in seconds. it is known the largest acceleration is 12 m/s ^2.

    a) WHat is its initial velocity?
    b)When is the car at rest?
    c)When is the car at rest?
    d)draw a diagram
    e)Find the total distance traveled by the car in the first 6 seconds
    f)Find the max acceleration of the car?



    and i did a problem that i thought was to easy to be right just wondering if someone can second my answer, i doubt its right.

    A stone is tossed down at 8 m/s from a edge of a cliff 63 m high how long will it take to hit the ground at the foot of the cliff?

    3.8 seconds is what i got :/


    Thanks for the help TJ
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  2. #2
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    Quote Originally Posted by tnkfub
    This problem is kinda big and just cant seem to start it off :/.

    A new motion detector is used to graph the position of a sports car. If the themotion detector is not callibrated properly, it might give unrealistic graph data. The position of the car is described by the function S= 2t^3-21t^2 + 60, tE [0,6], where s is measured in meters and t in seconds. it is known the largest acceleration is 12 m/s ^2.

    a) WHat is its initial velocity?
    b)When is the car at rest?
    c)When is the car at rest?
    d)draw a diagram
    e)Find the total distance traveled by the car in the first 6 seconds
    f)Find the max acceleration of the car?
    So you have s=2t^3-21t^2+60,0\leq t\leq 6
    To find its initial velocity is when t=0, but not in the formula above because that is for distance, but in its derivative (because derivative of distance is speed),
    v=6t^2-42t thus the initial velocity is zero.
    The car is at rest when it have no speed, thus, v=0 solving 6t^2-42t=0 we factor 6t(t-7)=0 only solution is t=0 because 0\leq t\leq 6.
    The total distance traveled is \int^6_0|6t^2-42t|dt which is, 324.
    The acceleration is the derivative of velocity thus,
    12t-42, it is maximum when t is as large as possible, that happens when t=6 and its acceleration is 30.
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