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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.
    Thanks from edwardkiely
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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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    Re: distance travelled

    Thanks, I would have used a capital "T".
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    Re: distance travelled

    romsek, I am not sure where you get your last line from.

    I have worked out k=-3/800 and c (constant) =-1/800

    I am trying to use this line to get the answer (v^-2)/-2 =kt + c

    I am not sure what T stands for. I thought we were just working with 1 t.

    what are you doing to get 40t/3

    thanks
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  7. #7
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    Re: distance travelled

    Quote Originally Posted by edwardkiely View Post
    romsek, I am not sure where you get your last line from.

    I have worked out k=-3/800 and c (constant) =-1/800

    I am trying to use this line to get the answer (v^-2)/-2 =kt + c

    I am not sure what T stands for. I thought we were just working with 1 t.

    what are you doing to get 40t/3

    thanks
    there is no $T$

    I just plugged the values of $k$ and $v_0$ into $v(\tau)$ and integrated with respect to $\tau$
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    Re: distance travelled

    When I sub your values for k and v0 into the expression and sub in 10 for v I get t=3/5

    I am not sure where you get 40t/3 from?
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    Re: distance travelled

    Your expression was already integrated.

    You integrated it in the their line of your workings. So you already have an expression
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  10. #10
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    Re: distance travelled

    I have to use $\tau$ for the time variable as you unfortunately chose to use $t$ to stand for a constant

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

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

    Integrate both sides including the constant of integration $v_0$ on the right hand side

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

    solve for $v(\tau)$

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

    I did make a mistake here

    $v^2 = -\dfrac{1}{2(k\tau + v_0)}$

    $v(\tau) = \pm \sqrt{-\dfrac{1}{2(k\tau + v_0)}}$

    we can ignore the negative solutions

    $v(0) = 20$

    $20 = \sqrt{-\dfrac{1}{2(k(0) + v_0)}}$

    $\dfrac{1}{400} = -2v_0$

    $v_0 = -\dfrac{1}{800}$

    $v(t) = 10$

    $10 = \sqrt{-\dfrac{1}{2(k t -\dfrac{1}{800})}}$

    $\dfrac{1}{100} = -2(k t -\dfrac{1}{800})$

    $-\dfrac{1}{200}+\dfrac{1}{800} = k t$

    $k =-\dfrac{3}{800t}$

    so $v(\tau) = \sqrt{-\dfrac{1}{2(-\dfrac{3}{800t}\tau-\dfrac{1}{800})}} = \sqrt{\dfrac{800t}{6\tau+2t}}= 20\sqrt{\dfrac{t}{3\tau+t}}$

    $\begin{align*}
    &s = \displaystyle \int_0^t ~v(\tau)~d\tau = \\ \\

    &\displaystyle \int _0^t 20\sqrt{\dfrac{t}{3\tau+t}} ~d\tau = \\ \\

    &\left. \dfrac{40}{3} \sqrt{t^2+3 \tau t}\right |_0^t = \\ \\

    &\dfrac{80t}{3} - \dfrac{40t}{3} = \dfrac{40t}{3}

    \end{align*}$
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    Re: distance travelled

    thanks. I understand it better now, but...

    Time is always variable and there should only be one version of it so I do not understand why you are using T and t.?
    time can never be a constant in these questions I thought. In all the worked examples in my text book there was only one type of t ever used.

    I am also unsure of how you integrate v(T)dT which rule are you using when you do this?
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  12. #12
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    Re: distance travelled

    Quote Originally Posted by edwardkiely View Post
    T and t.?

    where do you see $T$?

    You specified in the problem $t$ was a constant, see post #4.

    Thus I had to find some other variable to represent the unknown quantity time.

    I used $\tau$. That's all there is to it.

    As far as the integration well I leave you to figure that out, it's a standard application of the rule for a variable raised to a power.
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