Question

A ${}^{\mathbf{1000}\mathbf{kg}}$boat is traveling at${}^{\mathbf{90}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$ when its engine is shut off. The magnitude of the frictional force${}^{{\mathbf{f}}_{\mathbf{k}}}$ between boat and water is proportional to the speed v of the boat:${}^{{\mathbf{f}}_{\mathbf{k}}\mathbf{=}\mathbf{70}\mathbf{v}}$where

${}^{\mathbf{v}}$is in meters per second and${{}^{\mathbf{f}}}_{\mathbf{k}}$is in Newton. Find the time required for the boat to slow to${}^{\mathbf{45}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$.

Short answer

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

Step-by-step solution

Given

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

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

Determining the concept

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:

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

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

Determining the time required for the boat

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,

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

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