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Thread: particle moving

  1. #1
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    particle moving

    the deceleration of a particle moving in a straight line with speed v has magnitude 2e^v .It's initial speed is 10 m/s

    (i)find the time t1 for the speed to half in magnitude.
    (ii)find the time t2 for the particle to come to rest.
    (iii) deduce (t2-t1)/t1 =e^5

    I can not do any of this question.

    I am letting dv/dt = -2e^v I am having trouble seperating this and I do not know how to integrate it.
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  2. #2
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    Re: particle moving

    $\dfrac{dv}{dt} = 2 e^v$

    $e^{-v} dv = 2 dt$

    $-e^{-v} = 2t + C$

    $e^{-v} = -2t + C$ (constant absorbs the negative sign)

    $-v = \ln(-2t+C)$

    $v = -\ln(-2t+C)$

    $v(0) = -\ln(C) = 10$

    $C = e^{-10}$

    $v(t) = -ln(-2t + e^{-10}) $

    $v(t_1) = \dfrac{10}{2} = 5$

    $-\ln(-2t_1 + e^{-10}) = 5$

    $-2t_1 + e^{-10} = e^{-5}$

    $-2t_1 = e^{-5}-e^{-10}$

    $t_1 = \dfrac{e^{-10}-e^{-5}}{2} = \dfrac{1-e^5}{2e^{10}}$

    see if you can figure out the rest
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  3. #3
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    Re: particle moving

    That's pretty straight forward: dv/dt= -2e^v separates to (e^{-v})dv= -2dt

    Since the derivative of e^{ax} is ae^{ax}, the integral of e^{ax} is e^{ax}/a. I presume you can integrate -2dt.

    Since the initial speed was 10, v(t1)= 5. Solve that for t1.

    v(t2)= 0. Solve that for t2.
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  4. #4
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    Re: particle moving

    thanks.

    why does the c absorb the minus and why does it not become -c
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  5. #5
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    Re: particle moving

    Quote Originally Posted by edwardkiely View Post
    thanks.

    why does the c absorb the minus and why does it not become -c
    Because it's an arbitrary constant. It can be anything at all so using a negative sign is superfluous.
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