Step 1: Write the equation of motion for the decelerating boat.
After the engine is shut off, the only horizontal force on the boat is the frictional resistance $f = 35v$ (opposing motion). By Newton's second law with mass $m = 700$ kg: \[ m\frac{dv}{dt} = -f = -35v \] \[ 700\frac{dv}{dt} = -35v \] The negative sign indicates deceleration: the friction acts opposite to the direction of motion.
Step 2: Simplify and separate variables.
Dividing both sides by 700: \[ \frac{dv}{dt} = -\frac{35}{700}v = -\frac{1}{20}v \] Rearranging to separate $v$ and $t$: \[ \frac{dv}{v} = -\frac{1}{20} dt \] This is a separable first-order ODE. The friction is proportional to velocity (not constant), which leads to exponential decay of speed rather than uniform deceleration.
Step 3: Integrate both sides with appropriate limits.
The boat starts at $v = 24 \, \text{ms}^{-1}$ and we want the time $t$ when $v = 6 \, \text{ms}^{-1}$: \[ \int_{24}^{6} \frac{dv}{v} = -\frac{1}{20} \int_0^t dt \] \[ [\ln v]_{24}^{6} = -\frac{t}{20} \] \[ \ln 6 - \ln 24 = -\frac{t}{20} \] \[ \ln\!\left(\frac{6}{24}\right) = -\frac{t}{20} \] \[ \ln\!\left(\frac{1}{4}\right) = -\frac{t}{20} \]
Step 4: Solve for $t$.
Since $\ln(1/4) = -\ln 4$: \[ -\ln 4 = -\frac{t}{20} \] \[ t = 20 \ln 4 \]
Step 5: Evaluate numerically.
Using $\ln 4 = \ln(2^2) = 2\ln 2 \approx 2 \times 0.693 = 1.386$: \[ t = 20 \times 1.386 = 27.72 \, \text{s} \approx 28 \, \text{s} \]
Step 6: Reflect on the physics of velocity-dependent friction.
Unlike constant friction, velocity-dependent friction means the deceleration itself decreases as the boat slows down. The boat speed decays exponentially: $v(t) = 24 e^{-t/20}$. Setting $v = 6$ gives $e^{-t/20} = 1/4$, so $t = 20\ln 4 \approx 28$ s. This type of drag is typical for objects moving through fluids at moderate speeds. \[ \boxed{t \approx 28 \, \text{s}} \]