A motorboat moving in still water at v_0=20 km/h has its…
A motorboat moving in still water at $v_0=20$ km/h has its motor turned off; 20 seconds later its velocity has dropped to $8$ km/h. Water resistance is proportional to velocity. Find the boat's velocity 20 minutes after the motor is turned off.
Explanation of Method
Newton's second law with resistance $-kv$ gives an exponential-decay ODE. Use the 20-second data point to determine the decay factor, then apply it (raised to the appropriate power of elapsed time) at 20 minutes.
Step 1: Solve the ODE
With $\alpha=k/m$, $\dfrac{dv}{dt}=-\alpha v$ and $v(0)=v_0$ give $\boxed{\,v(t)=v_0e^{-\alpha t}.}$
Step 2: Determine the decay factor from the data
Working in hours, $t_1=20\text{ s}=\tfrac{1}{180}$ h. From $8=20e^{-\alpha/180}$,
\[e^{-\alpha/180}=\frac{2}{5}.\]
Step 3: Evaluate at $t=20$ min $= frac13$ h
Since $\tfrac13=60\cdot\tfrac{1}{180}$,
\[v\!\left(\tfrac13\right)=20\left(e^{-\alpha/180}\right)^{60}=20\left(\frac25\right)^{60}\approx2.7\times10^{-23}\ \text{km/h}.\]
Result
Final answer:
\[\boxed{\,v(20\text{ min})=20\left(\dfrac25\right)^{60}\approx0\ \text{km/h}}\]
— the boat has effectively stopped.