Question

A motorboat starts from rest (initial velocity v(0)=v₀=0). Its motor provides a constant acceleration of 4 ft /s², but water resistance causes a deceleration of v²/400 ft /s². Find v when t=7 s, and also find the limiting velocity as t → +[infinity] (that is, the maximum possible speed of the boat).

Answer 1

With
`v(0)=0`, we have by the fundamental theorem of calculus

`v(t)=v(0)+\displaystyle\int_0^ta(\tau)\,\mathrm d\tau=\int_0^t\left(4-(v(\tau)^2)/(400)\right)\,\mathrm d\tau`

Differentiating both sides wrt
`t` gives

`v'(t)=4-(v(t)^2)/(400)`

This equation is separable as

`(\mathrm dv)/(\mathrm dt)=4-(v^2)/(400)\implies(\mathrm dv)/(4-(v^2)/(400))=\mathrm dt`

Integrate both sides; on the left, substitute

`v=40\sin u\implies\mathrm dv=40\cos u\,\mathrm du`

`\implies\displaystyle\int(40\cos u\,\mathrm du)/(4-(1600\sin^2u)/(400))=t+C`

`\implies\displaystyle10\int(\cos u)/(1-\sin^2u)\,\mathrm du=t+C`

`\implies\displaystyle10\int\sec u\,\mathrm du=t+C`

`\implies10\ln|\sec u+\tan u|=t+C`

`\implies10\ln\left|(40+v)/(√(1600-v^2))\right|=t+C`

`\implies(40+v)/(√(1600-v^2))=Ce^(t/10)`

Given that
`v(0)=0`, we have

`(40)/(√(1600))=C\implies C=1`

So the velocity at time
`t` is
`v(t)` that satisfies

`(40+v(t))/(√(1600-v(t)^2))=e^(t/10)`

When
`t=7\,\mathrm s`, we have

`(40+v(7))/(√(1600-v(7)^2))=e^(7/10)\implies v(7)=(40(e^(7/5)-1))/(e^(7/5)+1)\approx\boxed{24.2(\rm ft)/(\rm s)}`

We can rewrite the particular solution as

`\implies(40+v)e^(-t/10)=√(1600-v^2)`

`\implies(40+v)^2e^(-t/5)=1600-v^2`

`\implies(1600+80v+v^2)e^(-t/5)=1600-v^2`

`\implies(1+e^(-t/5))v^2+80v+1600(e^(-t/5)-1)=0`

Taking the limit as
`t\to\infty` on both sides gives

`\displaystyle\lim_(t\to\infty)v(t)^2+80\lim_(t\to\infty)v(t)-1600=0`

(the exponential terms approach 0)

`\implies\displaystyle\left(\lim_(t\to\infty)v(t)\right)^2+80\lim_(t\to\infty)v(t)-1600=0`

so the limiting velocity, call it
`V`, satisfies the quadratic equation

`V^2+80V-1600=0\implies V=40(-1\pm\sqrt2)\approx-96.6(\rm ft)/(\rm s)\text{ or }\boxed{16.6(\rm ft)/(\rm s)}`

Realistically, the boat won't speed up enough for the resistance to be so strong as to reverse the boat's direction, so the limiting velocity should be positive.