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Using differential equations find a formula for the velocity of man descending with a parachute

Let v(t) denote the speed at time t seconds after jumping from a large height of a man falling with a parachute for which the combined weight of man plus parachute is 192 pounds. The man opens the parachute after 10 seconds. Before the chute is open air resistance is given by \frac{3}{4} v(t) pounds. While the parachute is open air resistance is given by 12v(t) pounds. Assuming acceleration due to gravity is 32 feet per second squared, find explicit formulas for v(t). (Use the approximation e^{-\frac{5}{4}} = \frac{37}{128}.)


We are given that the weight of the man plus parachute is mg = 192 and that g = 32. Therefore m =6. So, for 0 \leq t \leq 10 we have v(0) = 0 and

    \begin{align*}  ma = mg - kv && \implies && 6v' &= 192 - \frac{3}{4} v \\  && \implies && v' + \frac{1}{8} v &= 32 \\  && \implies && v &= e^{-\frac{t}{8}} \int_0^t e^{\frac{s}{8}} \cdot 32 \, ds &(\text{Theorem 8.3})\\  && \implies && v &= e^{-\frac{t}{8}} \cdot 32 (8e^{\frac{t}{8}} - 8) \\  && \implies && v &= 256 (1 - e^{\frac{t}{8}}). \end{align*}

If t > 10, then we have v(10) = 182 and k = 12v, therefore,

    \[  6v' = 192 - 12v \quad \implies \quad v' + 2v = 32. \]

So, we apply Theorem 8.3 (page 310 of Apostol) with

    \[ v(10) = 182 \quad \implies \quad A(t) = \int_0^t 2 \, ds = 2t -20. \]

Therefore,

    \begin{align*}  v &= 182 e^{20-2t} + e^{20-2t} \int_{10}^t 32 e^{2s-20} \, ds \\  &= 182 e^{20-2t} + 16 - 16 e^{20-2t} \\  &= 16 + 166 e^{20-2t}. \end{align*}

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