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).

With [tex]v(0)=0[/tex], we have by the fundamental theorem of calculus[tex]v(t)=v(0)+\displaystyle\int_0^ta(\tau)\,\mathrm d\tau=\int_0^t\left(4-\frac{v(\tau)^2}{400}\right)\,\mathrm d\tau[/tex]Differentiating both sides wrt [tex]t[/tex] gives[tex]v'(t)=4-\dfrac{v(t)^2}{400}[/tex]This equation is separable as[tex]\dfrac{\mathrm dv}{\mathrm dt}=4-\dfrac{v^2}{400}\implies\dfrac{\mathrm dv}{4-\frac{v^2}{400}}=\mathrm dt[/tex]Integrate both sides; on the left, substitute[tex]v=40\sin u\implies\mathrm dv=40\cos u\,\mathrm du[/tex][tex]\implies\displaystyle\int\frac{40\cos u\,\mathrm du}{4-\frac{1600\sin^2u}{400}}=t+C[/tex][tex]\implies\displaystyle10\int\frac{\cos u}{1-\sin^2u}\,\mathrm du=t+C[/tex][tex]\implies\displaystyle10\int\sec u\,\mathrm du=t+C[/tex][tex]\implies10\ln|\sec u+\tan u|=t+C[/tex][tex]\implies10\ln\left|\dfrac{40+v}{\sqrt{1600-v^2}}\right|=t+C[/tex][tex]\implies\dfrac{40+v}{\sqrt{1600-v^2}}=Ce^{t/10}[/tex]Given that [tex]v(0)=0[/tex], we have[tex]\dfrac{40}{\sqrt{1600}}=C\implies C=1[/tex]So the velocity at time [tex]t[/tex] is [tex]v(t)[/tex] that satisfies[tex]\dfrac{40+v(t)}{\sqrt{1600-v(t)^2}}=e^{t/10}[/tex]When [tex]t=7\,\mathrm s[/tex], we have[tex]\dfrac{40+v(7)}{\sqrt{1600-v(7)^2}}=e^{7/10}\implies v(7)=\dfrac{40(e^{7/5}-1)}{e^{7/5}+1}\approx\boxed{24.2\dfrac{\rm ft}{\rm s}}[/tex]We can rewrite the particular solution as[tex]\implies(40+v)e^{-t/10}=\sqrt{1600-v^2}[/tex][tex]\implies(40+v)^2e^{-t/5}=1600-v^2[/tex][tex]\implies(1600+80v+v^2)e^{-t/5}=1600-v^2[/tex][tex]\implies(1+e^{-t/5})v^2+80v+1600(e^{-t/5}-1)=0[/tex]Taking the limit as [tex]t\to\infty[/tex] on both sides gives[tex]\displaystyle\lim_{t\to\infty}v(t)^2+80\lim_{t\to\infty}v(t)-1600=0[/tex](the exponential terms approach 0)[tex]\implies\displaystyle\left(\lim_{t\to\infty}v(t)\right)^2+80\lim_{t\to\infty}v(t)-1600=0[/tex]so the limiting velocity, call it [tex]V[/tex], satisfies the quadratic equation[tex]V^2+80V-1600=0\implies V=40(-1\pm\sqrt2)\approx-96.6\dfrac{\rm ft}{\rm s}\text{ or }\boxed{16.6\dfrac{\rm ft}{\rm s}}[/tex]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.