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DiffEQ, Snowing Rates, Distance Traveled

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  1. Feb 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Early one morning it starts to snow. At 7AM a snowplow sets off to clear the road. By 8AM, it has gone 2 miles. It takes an additional 2 hours for the plow to go another 2 miles. Let t = 0 when it begins to snow, let x denote the distance traveled by the plow at time t. Assuming the snowplow clears snow at a constant rate in cubic meters/hour:
    a) Find the DE modeling the value of x.
    b) When did it start snowing?

    2. Relevant equations
    N/A, but related course topics are for solving DiffEq's by separation of variables

    3. The attempt at a solution
    Modeling the plow as a rectangle with a certain length L, and assuming that the snow height H is across the length, and the plow speed is V, then "Assuming the snowplow clears snow at a constant rate in cubic meters/hour" can be expressed as
    ## LHv = C##
    where c is some constant. H and V are both functions of time.
    Simplify by combining constants (K= C/L) and dividing by H gives
    ##v= K/H##

    re-expressing v as the time derivative of distance x yields

    ## \frac{dx}{dt} = \frac{K}{H} ##
    separation of variables yields
    ## dx= \frac{K*dt}{H}##

    I want to integrate as I usually would in a separation of variable problem, but I can't figure out what to do with H (can it be modeled as a function of x?), or how to incorporate the initial conditions. I'm also confused about setting t= 0 when it starts to snow, because that seems like it would make it into a piecewise function. I can't tell if I'm missing any assumptions.

    To answer part B, it seems like I would have to pretend the plow was already moving by 7am (as in pretend it was continuously moving beforehand, and had some certain speed at 7am to match the initial conditions), then look at what time the speed approaches infinity from the right.
     
  2. jcsd
  3. Feb 8, 2016 #2

    Samy_A

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    As they plow 2 miles in the first hour, and then next 2 miles in the next 2 hours, it is clear that it is still snowing.
    You could assume that the amount of snow falling is constant over time (as nothing is said about that), so H=Ft (F some constant).
    The plow wasn't moving before 7am. It was snowing before 7am.

    First solve a), that will give you x(t) with two unknown constants.
    For b), if you set t=0 as the time it started to snow, then 7AM is some ##t_0## that you will have to determine using what you are told about what happens between 7AM and 8AM, and between 8AM and 10AM.
     
    Last edited: Feb 8, 2016
  4. Feb 8, 2016 #3
    If H=ft, then I have missed a very critical assumption.
    the answer to A should then be



    ## v =\frac{dx}{dt} = \frac{K}{ft} = \frac{C}{t}##

    where C is some other constant. This is a logarithm so the solution is:
    ## x(t) = C \ln(t) + D ##
    D is another constant.


    Attempt at part B:
    Start plugging in IC
    ## 0 = C\ln(t_0) + D ##
    ##2 = C\ln(t_0 + 1) + D##
    ##4 = C\ln(t_0+3) + D##

    Subtracting the 1st from 2nd eqs:
    ##2 = C\ln(\frac{t_0 + 1}{t_0} )##

    Subtracting 2nd from 3rd eqs:
    ##2 = C\ln(\frac{t_0+3}{t_0+1})##

    which means the argument of ln in each must be the same fraction. Solving I get t_0 = 1
    So it starts snowing at 6AM.

    Does that look right? I was concerned about assuming that it snows at some constant rate.
     
    Last edited: Feb 8, 2016
  5. Feb 8, 2016 #4

    Samy_A

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    Yes, that is the same result I got.
    I understand. Problem is that you have to assume something about the rate of snow falling in order to solve the exercise. Constant rate seems the obvious choice.
     
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