Answer:

(a) \(\displaystyle v=\frac{5(4-e^{-40t})}{4+e^{-40t}}\). (b) The limiting value is \(5\,\mathrm{m\,s}^{-1}\).

Newton's second law gives

\(0.5\frac{dv}{dt}=50-2v^2.\)

Therefore

\(\frac{dv}{dt}=100-4v^2=4(25-v^2).\)

Separate variables:

\(\frac{dv}{25-v^2}=4\,dt.\)

Using

\(\frac{1}{25-v^2}=\frac1{10}\left(\frac{1}{5+v}+\frac{1}{5-v}\right),\)

we obtain

\(\frac1{10}\ln\left(\frac{5+v}{5-v}\right)=4t+C.\)

When \(t=0\), \(v=3\), so

\(C=\frac1{10}\ln\left(\frac{8}{2}\right)=\frac1{10}\ln4.\)

Thus

\(\frac1{10}\ln\left(\frac{5+v}{5-v}\right)=4t+\frac1{10}\ln4.\)

Multiply by 10 and rearrange:

\(\ln\left(\frac{5+v}{4(5-v)}\right)=40t.\)

So

\(\frac{5+v}{4(5-v)}=e^{40t}.\)

Solving for \(v\),

\(5+v=20e^{40t}-4ve^{40t},\)

\(v(1+4e^{40t})=20e^{40t}-5.\)

Hence

\(v=\frac{20e^{40t}-5}{4e^{40t}+1}.\)

Dividing numerator and denominator by \(e^{40t}\),

\(v=\frac{5(4-e^{-40t})}{4+e^{-40t}}.\)

As \(t\to\infty\), \(e^{-40t}\to0\), so

\(v\to\frac{5(4)}{4}=5.\)

The limiting value of the speed is \(5\,\mathrm{m\,s}^{-1}\).