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Snow Plow Problem

  1. Feb 5, 2017 #1
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    Hi guys.

    I am currently stuck on the classic snow plow problem. I have the following differential equation and initial conditions:
    @ 7am plow starts off to clear snow at a constant rate
    By 8am, plow has gone 4mi
    By 9am, plow has gone an additional 3mi
    Let t=0 when it started to snow, when did it start to snow. They gave the answer as 4:27am

    k(dx/dt)=1/t

    It wants me to find the constant k.

    Here is my work:

    ∫k dx= ∫dt/t
    kx +C = ln (t) At this point I am assuming that x is the independent variable and t is the dependent variable.

    e^(kx) +C = t
    Using the first initial condition of when it starts to snow of x(0)=0, I get the C is 0, therefore

    e^(kx) = t. Using the second initial condition @8, x(t+1)=4

    e^(4k)= t+1.

    This is the point I get stuck. I cant seem to solve for k that has a real number.

    Any help would be appreciated.
     
  2. jcsd
  3. Feb 6, 2017 #2

    haruspex

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    Right.
    Wrong.
     
  4. Feb 6, 2017 #3
    Ok, looking back then the c would be e^c which would then be Ae^(kx)=t? But having initial condition of t(0)=1 would make A=0.
     
  5. Feb 6, 2017 #4

    haruspex

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    At t=0, the snow plow was safely in its garage.
     
  6. Feb 6, 2017 #5
    I understand that but will always result in a zero answer if A is 0.
     
  7. Feb 6, 2017 #6

    haruspex

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    The equation you used to specify its motion is only valid when.... what?
     
  8. Feb 6, 2017 #7
    At t=0. So is A not a factor? Leaving e^kx=t? Then would I just plug in the next conditions and solve for k?
     
  9. Feb 6, 2017 #8

    haruspex

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    No. t=0 is when it started snowing. At that time, the plow was going nowhere. The plow set off at 7 am. (We do not know when that was as a value of t.) The equation only applies while the plow is moving.
     
  10. Feb 7, 2017 #9

    pasmith

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    Actually it doesn't; it wants you to find [itex]T[/itex], which is the time from when it started snowing until 7am. You do not need to know [itex]k[/itex] in order to find this.

    However I think you may have copied the question incorrectly, because as far as I can tell you end up with a sixth-order polynomial for [itex]T[/itex] which doesn't have a root anywhere near the alleged answer of [itex]T = 2.55[/itex].
     
  11. Feb 7, 2017 #10

    haruspex

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    Do you mean a sixth order polynomial for ek?
    The root of interest is about 1.09, giving T about 2.55.
     
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