91Ó°ÊÓ

Mass of the boat, ${}^{\mathrm{m}=1000\mathrm{kg}}$

Frictional force, ${}^{{\mathrm{f}}_{\mathrm{k}}=70\mathrm{v}}$where${}^{\mathrm{v}}$is the speed of the boat.

Initial speed of the boat when engine is shut off,${}^{{\mathrm{v}}_0=90\mathrm{km}/\mathrm{h}}$

Final speed of the boat after time t, is

$$\begin{gathered} \mathrm{v}=45\mathrm{km}/\mathrm{h} \\ =\frac{{\mathrm{v}}_0}{2}\mathrm{km}/\mathrm{h} \end{gathered}$$

In this problem, the frictional force is not constant, but instead proportional to the speed of the boat. Integration is required to solve for the speed. Also, it involves Newton’s second law. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\underset{}{\overset{}{∑ma}}$

where, F is the net force, m is mass and a is an acceleration.

Take the direction of the boat’s motion to be positive. Newton’s second law gives,

$$\begin{gathered} F=ma \\ -f_k=m\left(\frac{dv}{dt}\right) \\ -70v=m\left(\frac{dv}{dt}\right) \\ \frac{dv}{dt}=-\frac{70}{m}v \\ \frac{dv}{v}=-\frac{70}{m}dt \end{gathered}$$

Taking integration to both sides,

$$\begin{gathered} {∫}_{vo}^v\frac{dv}{v}=-\frac{70}{m}{∫}_{t-0}^{t}dt \\ In\left(\frac{v}{v_0}\right)=-\frac{70}{m}t \\ t=-\frac{m}{70}In\left(\frac{v_0/2}{v_0}\right) \\ =-\frac{1000}{70}\left(-0.693\right) \\ =9.9 \end{gathered}$$

Hence, the time required for the boat to slow to${}^{45\mathrm{km}/\mathrm{h}}$is${}^{9.9\mathrm{s}}$