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Math Help - Differential equation-

  1. #1
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    Differential equation-

    A girl lives 500 metres from school. She sets out walking at 2m/s, but when she has walked a distance of x metres her speed has dropped to (2-x/400)m/s. How long does she take to get to school.

    Can you do the working-

    Just if you are interested this is my working for the question-
    dx/dt=(2-x/400)
    dx/(2-x/400)+k=dt
    -400in(2-x/400)+k=t
    Is this right?
    So after this do I substitute 500 into x or something to find the time- what occurs from here-
    I got a really strange answer- it is an incorrect answer. Urgent!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kingkaisai2
    [SIZE=5]A girl lives 500 metres from school. She sets out walking at 2m/s, but when she has walked a distance of x metres her speed has dropped to (2-x/400)m/s. How long does she take to get to school.

    Can you do the working-

    Just if you are interested this is my working for the question-
    dx/dt=(2-x/400)
    dx/(2-x/400)+k=dt
    -400 ln(2-x/400)+k=t
    This is OK. Now at time t=0, x=0, so: k=400 ln(2).

    Now you want to find the time t_500, when x=500, so plug 500 in for x to get:

    t_500=400 ln(2) -400 ln(2-500/400)=392.3s

    RonL
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  3. #3
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    Hello, kingkaisai2!

    A girl lives 500 metres from school.
    She sets out walking at 2m/s, but when she has walked a distance of x metres
    her speed has dropped to 2-\frac{x}{400} m/s.
    How long does she take to get to school?

    If you hate fractions as much as I do, you'd eliminate them immediately.

    We have: . \frac{dx}{dt}\:=\:2 - \frac{x}{400}

    Multiply by 400: . 400\frac{dx}{dt}\:=\:800 - x\quad\Rightarrow\quad \frac{400\,dx}{800 - x} \:=\:dt

    Integrate: . \text{-}400\ln|800 - x| \:=\:t + C

    . . When t=0,\:x=0:\;\;\text{-}400\ln|800 -0| \:=\:0 + C\quad\Rightarrow\quad C \:=\:\text{-}400\ln800

    The equation becomes: . \text{-}400\ln|800 - x| \:=\:t - 400\ln800

    Divide by \text{-}400:\;\;\ln|800 - x|\:=\:\text{-}\frac{t}{400} + \ln800

    Then we have: . 800 - x \;=\;e^{\text{-}\frac{t}{400} + \ln800} \;=\;e^{-\frac{t}{400}}\cdot e^{\ln800} \;=\;e^{\text{-}\frac{t}{400}}\cdot800

    . . Then: . \text{-}x \;=\;\text{-}800 + 800e^{-\frac{t}{400}}\quad\Rightarrow\quad \boxed{x\:=\:800\left(1 - e^{\text{-}\frac{t}{400}}\right)} **


    When x = 500, we have: . 500 \:=\:800\left(1 - e^{\text{-}\frac{t}{400}}\right)\quad\Rightarrow\quad \frac{5}{8}\:=\:1 - e^{\text{-}\frac{t}{400}}

    Then: . e^{\text{-}\frac{t}{400}} \:=\:\frac{3}{8}\quad\Rightarrow\quad \text{-}\frac{t}{400} \:=\:\ln\left(\frac{3}{8}\right)

    Therefore: . t \:=\:\text{-}400\ln\left(\frac{3}{8}\right)\:=\:392.331201 \text{ seconds }\:\approx\: \boxed{6.5\text{ minutes}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    In order to answer the question, I did not have to solve for x.

    I did it to show how it is done (in case anyone was interested).

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