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Math Help - velocity problem, calculus?

  1. #1
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    velocity problem, calculus?

    A motorboat is moving at 30m/s when its motor suddenly quits and it is 800 m from the shore. Ten seconds later the boat has slowed to 20 m/s. Assume that the resistance it encounters while coasting is proportional to its velocity. How far will the boat coast in all? Can the boat reach the shore?
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  2. #2
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    If the resistance is proportional to the boat's velocity, then the deceleration of the boat is also proportional to its velocity. So you have the equation v(t) = kv'(t). You have to use this equation to determine an equation for the boat's distance traveled starting at the time the boat begins to slow down.
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  3. #3
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    The equation v(t) = kv'(t) is a common differential equation. The solution is v(t) = Ce^{kt}. Since v = 30 at t = 0, this yields 30 = Ce^{k(0)} = Ce^0 = C, so the equation is v(t) = 30e^{kt}. Now, since v(10) = 20, we have 30e^{10k} = 20;

    e^{10k} = \frac{20}{30} = \frac{2}{3}

    10k = \ln \left(\frac{2}{3}\right)

    k = 0.1 \ln \left(\frac{2}{3}\right)
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  4. #4
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    Thanks! how do I find out how far the boat coasts?
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  5. #5
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    The equation for position will be the antiderivative of the velocity function, which is s(t) = \frac{30}{k}e^{kt} + C. Since s(0) = 0, s(0) = \frac{30}{k}e^0 + C = \frac{30}{k} + C = 0. Hence C = -\frac{30}{k} and
    s(t) = \frac{30}{k}e^{kt} - \frac{30}{k}. The limit of this function as t approaches infinity is \lim_{t \to \infty} \left(\frac{30}{k}e^{kt} - \frac{30}{k}\right) = 0 - \frac{30}{k}, and that is the distance the boat coasts.
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