Answer:
[tex]t = 1.5\cdot \frac{m}{b\cdot v_{o}^{2}}[/tex]
Explanation:
The equation of equilibrium for the boat is:
[tex]\Sigma F = - F_{D} = m\cdot a[/tex]
[tex]-b\cdot v^{3} = m \cdot a[/tex]
[tex]m\cdot \frac{dv}{dt} = - b \cdot v^{3}[/tex]
This is a first-order linear differential equation with separable variables:
[tex]-\frac{m}{b}\cdot \frac{dv}{v^{3}} = dt[/tex]
[tex]-\frac{m}{b} \int\limits^{v_{f}}_{v_{o}} {v^{-3}} \, dv = t[/tex]
[tex]-\frac{m}{b}\cdot \left(-0.5\cdot v^{-2} \right)|_{v_{o}}^{v_{f}} = t[/tex]
[tex]0.5\cdot \frac{m}{b}\cdot \left(v_{f}^{-2} - v_{o}^{-2} \right) = t[/tex]
The time taken to decrease the speed of the boat to half of its initial value is:
[tex]t = 0.5\cdot \frac{m}{b}\cdot \left[\left(\frac{1}{2} \right)^{-2} - 1 \right]\cdot v_{o}^{-2}[/tex]
[tex]t = 1.5\cdot \frac{m}{b\cdot v_{o}^{2}}[/tex]
