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Thread: distance travelled

  1. #1
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    distance travelled

    a car travelling in a straight line experiences a retardation which is proportional to its speed cubed. the speed of the car reduces from 20 to 10 m/s^-1 in a time of t seconds
    show that the distance travelled in this time is 40t/3 m

    so what I tried to do was let dv/dt =-kv^3 I put in the limits and I got k=1/40t

    I then let vdv/ds =-kv^3 and I put in the limits and I got 1/20 =ks and I subbed in k and i got the wrong answer.
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  2. #2
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    Re: distance travelled

    This looks pretty straightforward

    $\dot{v} = k v^3$

    $\dfrac{dv }{v^3} = k~dt$

    $-\dfrac{1}{2v^2} = k t + v_0$

    $v(t) = \dfrac{1}{\sqrt{2(k t+v_0)}}$

    $v(0) = 20,~v(t)=10$

    $k = \dfrac{3}{800t},~v_0 = \dfrac{1}{800}$

    $\displaystyle \int_0^t~v(\tau)~d\tau = \dfrac{40t}{3}$

    I've given you just the skeleton obviously, you have to fill in the meat, but the problem is legit.
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  3. #3
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    Re: distance travelled

    Quote Originally Posted by edwardkiely View Post
    a car travelling in a straight line experiences a retardation which is proportional to its speed cubed. the speed of the car reduces from 20 to 10 m/s^-1 in a time of t seconds
    show that the distance travelled in this time is 40t/3 m

    so what I tried to do was let dv/dt =-kv^3 I put in the limits and I got k=1/40t
    This makes no sense the "constant of proportion" is a constant, not a function of t.

    I then let vdv/ds =-kv^3 and I put in the limits and I got 1/20 =ks and I subbed in k and i got the wrong answer.
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  4. #4
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    Re: distance travelled

    Quote Originally Posted by HallsofIvy View Post
    This makes no sense the "constant of proportion" is a constant, not a function of t.
    terrible use of notation but in this case $t$ is a constant.

    it's the length of time it takes for the car to decelerate from 20 m/s to 10 m/s
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  5. #5
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    Re: distance travelled

    Thanks, I would have used a capital "T".
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