The initial velocity is
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.![\frac{dv}{dt} =10-0.8v \\\\ \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \, \frac{dv}{10-0.8v} = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\ \frac{0.8v -10}{3.2} =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D10-0.8v%20%5C%5C%5C%5C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%3Ddt%20%5C%5C%5C%5C%20%5Cint_%7B16.5%7D%5E%7Bv%7D%20%5C%2C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%20%20%3D%20%5Cint_%7B0%7D%5E%7Bt%7D%20dt%20%5C%5C%5C%5C%20-%20%5Cfrac%7B1%7D%7B0.8%7D%20%5Bln%2810-0.8v%29%5D_%7B16.5%7D%5E%7Bv%7D%3Dt%20%5C%5C%5C%5C%20ln%20%5Cfrac%7B10-0.8v%7D%7B-3.2%7D%3D-0.8t%20%5C%5C%5C%5C%20%20%5Cfrac%7B0.8v%20-10%7D%7B3.2%7D%20%20%3De%5E%7B-0.8t%7D%20%5C%5C%5C%5C%200.8v%20%3D%2010%20%2B%203.2e%5E%7B-0.8t%7D%20%5C%5C%5C%5C%20v%3D12.5%2B4e%5E%7B-0.08t%7D)
The velocity function is
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
The depth of the lake is
![d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft](https://tex.z-dn.net/?f=d%3D%5Cint_%7B0%7D%5E%7B5.7%7D%20%5C%2C%20%2812.5%2B4e%5E%7B-0.8t%7D%29dt%20%5C%5C%5C%5C%20%3D%2012.5%285.7%29%2B%20%5Cfrac%7B4%7D%7B%28-0.8%29%7D%5Be%5E%7B-0.8t%7D%5D_%7B0%7D%5E%7B5.7%7D%20%5C%5C%5C%5C%20%3D71.25-5%280.0105-1%29%20%3D76.198%20%5C%2C%20ft)
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.
The velocity function is
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
The depth of the lake is
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft