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 January 29th, 2018, 09:26 AM #1 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 Applied integrals find the time When Bob runs a 10km marathon he generates the power (P). $\displaystyle \displaystyle P(t) = 600e^{-\frac{t}{5000}}$ Where the time is in seconds. A sugerpience contains about 40kJ energy. How long time does it take for Bob to run a 10 km if he consumes 30 sugarcubes during the marathon. [IMG]Link to problem http://www.mediafire.com/view/u68i8n8zvmcn3hk/IMG_20180129_181555.jpg[/IMG] What do I need to do to solve for t (the time)? I tried doing this first but that didn't work out well. $\displaystyle \displaystyle P(t) = 600e^{-\frac{t}{5000}}\\ \displaystyle E_{total} = E_s \cdot t\\ \displaystyle \int P(t) \ dt = E_s \cdot t\\ \displaystyle \int 600e^{-\frac{t}{5000}} \ dt = 1200\\$ No solutions for real numbers.
 January 29th, 2018, 10:39 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 $\displaystyle 600\int_0^t e^{-\frac{\tau}{5000}}d\tau= \left[-3000000 e^{-\frac{\tau}{5000}}\right]_0^t= -300000\left(1- e^{\frac{t}{5000}}\right)= 1200$. $\displaystyle 1- e{\frac{t}{5000}}= -\frac{1200}{300000}= -\frac{12}{3000}= -0.004$ $\displaystyle e^{-\frac{t}{5000}}- 1= 0.004$ $\displaystyle e^{-\frac{t}{5000}}= 1.004$ $\displaystyle -\frac{t}{5000}= ln(1.004)= 0.00399$ $\displaystyle t= -20$ Is it that this is negative that bothers you? Last edited by Country Boy; January 29th, 2018 at 10:52 AM.
 January 29th, 2018, 11:19 AM #3 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 I get t = -1.99600107. That doesn't seem realistic that Bob ran the 10 km marathon in about 2 seconds. I think you missed a 0 in -1200 / 30 0000 should be -1200 / 3 000 000 if I'm not mistaken. Last edited by DecoratorFawn82; January 29th, 2018 at 11:22 AM.
 January 29th, 2018, 11:33 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 $P = \dfrac{dW}{dt} = 600e^{-t/5000}$ $\displaystyle W = 600 \int_0^T e^{-t/5000} \, dt$ $30 \cdot 40\, kJ = 1200 \, kJ$ $\displaystyle 1200000 = 600 \int_0^T e^{-t/5000} \, dt$ $\displaystyle 2000 = \int_0^T e^{-t/5000} \, dt$ $2000 = 5000\left[1-e^{-T/5000}\right]$ $e^{-T/5000} = \dfrac{3}{5}$ $T = 5000 \log(5/3) \approx 2554 \, sec \approx 43 \, min$ reasonable ... about a 7 minute per mile pace Thanks from greg1313 and Country Boy

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