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Math Help - Velocity and Acceleration

  1. #1
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    Post Velocity and Acceleration

    The distance traveled by a car in a time t seconds afer the brakes are applied is given by:

    y= 10t-1.8t^2


    Where y is the distance in metres and t is time in seconds.

    Determine equations for velocity and acceleration of the car.

    What is the velocity at the instant the brakes are applied?

    How many seconds does it take the car to stop?

    -----------------------------------

    10t

    1.8t^2 = 3.24t

    10t - 3.24t = 6.76

     Time = 6.76 Seconds?


    Whats the next step? and is that correct so far? help welcomed please.


    I know that  V= dx/dt
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  2. #2
    Member SengNee's Avatar
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    From
    Pangkor Island, Perak, Malaysia.
    Posts
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    Quote Originally Posted by ForgottenMemorie View Post
    The distance traveled by a car in a time t seconds afer the brakes are applied is given by:

    y= 10t-1.8t^2


    Where y is the distance in metres and t is time in seconds.

    Determine equations for velocity and acceleration of the car.

    What is the velocity at the instant the brakes are applied?

    How many seconds does it take the car to stop?

    -----------------------------------

    10t

    1.8t^2 = 3.24t

    10t - 3.24t = 6.76

     Time = 6.76 Seconds?


    Whats the next step? and is that correct so far? help welcomed please.


    I know that  V= dx/dt

    y= 10t-1.8t^2
    dy/dt=10-3.6t
    v=10-3.6t
    dv/dt=-3.6
    a=-3.6

    v=10-3.6t  ,t=0
    v=10

    v=10-3.6t
    0=10-3.6t
    t=2.778
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  3. #3
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    Dec 2007
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    Okies that really does help. Thanks for thte responce.
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