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Math Help - Differential Equation Problem

  1. #1
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    Differential Equation Problem

    I have been struggling with this problem for a while:

    A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?

    I got this far:

    dv/dt = -kv^2
    -1/v = -kt + C
    v = 1/(kt + C)

    400 = v(0) = 1/(k(0) + C)
    400 = 1/C
    .0025 = C

    v = 1/(kt + .0025)

    From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.

    Any help would be appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by machi4velli View Post
    I have been struggling with this problem for a while:

    A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?

    I got this far:

    dv/dt = -kv^2
    -1/v = -kt + C
    v = 1/(kt + C)

    400 = v(0) = 1/(k(0) + C)
    400 = 1/C
    .0025 = C

    v = 1/(kt + .0025)

    From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.

    Any help would be appreciated. Thanks.
    I don't know about the DE, but
    (average velocity)*time = distance
    So,
    (180 +400)/2 *t = 0.1
    t = 0.1 /290 = 0.000345 sec -------------answer.

    ------------------------
    Edit.

    Ooops, sorry, that is wrong because the acceleration, dv/dt, is not constant.

    Now I have to solve that through the DE. Heck.

    ---------------------
    dv/dt = -kv^2 ---------------------(i)
    dv /v^2 = -k dt
    -1/v = -kt +C
    v = 1/(kt +C) ----------------------(ii)

    When v=400m/sec, t=0,
    So, in (i), 400 = 1/(k*0 +C)
    C = 1/400 = 0.0025

    Hence, v = 1/(kt +0.0025) -----------(iii)
    Develop that,
    kt +0.0025 = 1/v
    kt = 1/V -0.0025
    kt = (1 -0.0025v)/v
    k = (1 -0.0025v) /(vt) ---------------(iv)
    Substitute that into (i)
    dv/dt = -[(1 -0.0025v) /(vt)](v^2)
    dv/dt = [(0.0025v -1)(v)] /t
    dv/dt = (0.0025v^2 -v) /t
    Cross multiply,
    (dv)t = (0.0025v^2 -v)dt
    dv /(0.0025v^2 -v) = dt /t -------------(v)

    According to the Wolfram Integrator,
    INT[1 /(0.0025x^2 -x)]dx = ln(400 -0.1x) -ln(x), so,

    Integrate both sides of (v),
    ln(400 -0.1v) -ln(v) = ln(t) +C
    ln[(400 -0.1v)/v] = ln(t) +C -------------(vi)

    That pesky C!
    Now I'm stalled.

    Heck, suppose C = 0,
    [I don't know the consequence of letting C=0 here. Would the answers be wrong? I don't know. I just want to get out of this trap.]
    ln[(400 -0.1v)/v] = ln(t)
    (400 -0.1v)/v = t --------------------------(vii)
    When the bullet comes out, the v=180 m/sec, so,
    (400 -0.1*180)/180 = t
    t = 382/180 = 2.12 sec.

    Therefore, the bullet will pass through the wood in 2.12 seconds after it hit the wood. ----answer.
    Last edited by ticbol; January 23rd 2007 at 03:35 AM.
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  3. #3
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    Quote Originally Posted by machi4velli View Post
    I have been struggling with this problem for a while:

    A bullet passes through a piece of wood 0.1 m thick. It enters the wood at 400 m/s and leaves at 180 m/s. The velocity v of the bullet is assumed to obey the differential equation dv/dt = -kv^2 where k is a positive constant. How long does it take the bullet to pass through the wood?

    I got this far:

    dv/dt = -kv^2
    -1/v = -kt + C
    v = 1/(kt + C)

    400 = v(0) = 1/(k(0) + C)
    400 = 1/C
    .0025 = C

    v = 1/(kt + .0025)

    From here, I am not sure what to do. I tried to find k by making v = dx/dt and solving the displacement equation, which turned out to be another dead end.

    Any help would be appreciated. Thanks.
    \frac{dv}{dt} = -kv^2

    \frac{dv}{v^2} = -kdt

    \frac{-1}{v} = -kt + C

    So
    v = -\frac{1}{C - kt}

    v = \frac{1}{kt - C} <-- My C is the negative of yours, is the only difference.

    So v = 400 at t = 0, thus
    400 = \frac{1}{-C}

    C = -\frac{1}{400}

    Thus
    v = \frac{1}{kt - \frac{-1}{400}} = \frac{400}{400kt + 1}

    We still have that pesky k.

    v = \frac{dx}{dt}

    dx = v dt = \frac{400}{400kt + 1} dt

    \int_{x_0}^x dx = \int_0^y \frac{400dt}{400kt + 1} <-- y is the time at which the bullet leaves the wood.

    x - x_0 = 0.1 = 400 \int_0^y \frac{dt}{400kt + 1}

    Let T = 400kt + 1 ==> dT = 400k dt

    0.1 = 400 \int_{1}^{400ky + 1} \frac{\frac{dT}{400k}}{T}

    0.1 = \frac{1}{k} \int_{1}^{400ky + 1} \frac{dT}{T}

    0.1 = \frac{1}{k} \cdot ln(T)|_1^{400ky+1}

    0.1k = ln(400ky+1) - ln(1) = ln(400ky+1)

    So
    0.1k = ln(400ky + 1)

    We can get another equation in k and y by using the v(t) equation. We know that v = 180 when t = y, so
    180 = \frac{400}{400ky + 1}

    We have two equations in two unknowns.

    Solve the v equation for 400ky + 1:
    400ky + 1 = \frac{400}{180}

    Thus the ln equation says:
    0.1k = ln \left ( \frac{400}{180} \right )

    This gives a value for k, and you can use that in the v equation to get a value for y. I get that k = 7.98508 and y = 0.000383 s.

    -Dan
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