(a) In which direction should the motorboat in Exercise 3.35 head to reach a point on the opposite bank directly east from your starting point? (The boat’s speed relative to the water remains 4.2 m/s) (b) What is the velocity of the boat relative to the earth? (c) How much time is required to cross the river.

The given data can be listed below,

• The speed of river is, v_RE = 2.0m/s.
• The velocity of boat relative to river is, v_BR= 4.2m/s
• The width of the river is, D = 500m

Relative velocity is defined as the velocity of an object relative to another observer. It is the time rate of change of relative position of one object with respect to another object.

The relative velocity is the speed of an object or observer B in relation to an object or observer A's rest frame. The universal velocity formula is given by,

v_BE = v_B - v_A

The relative velocities are given in the diagram given below,

The direction of boat relative to water is given by,

$\sin \theta =\frac{v_{RE}}{v_{BE}}$

Substitute all the values in the above,

$\begin{array}{rcl}\sin \theta & =& \frac{2m/s}{4.2m/s}\\ \theta & =& {\sin }^{-1}\left(0.4762\right)\\ & =& 28.4^o\end{array}$

Thus, the direction of motorboat to reach opposite bank is 28.4^o northeast direction.

By Pythagoras theorem, the velocity of boat relative to the earth is given by,

$v_{BE}=\sqrt{{\left(v_{BR}\right)}^2-{\left(v_{RE}\right)}^2}$

Here, v_RE is the velocity of river relative to earth and v_BR is the velocity of boat relative to river.

Substitute all the values in the above,

$\begin{array}{rcl}{v}_{BE}& =& \sqrt{{\left(4.2m/s\right)}^2-{\left(2m/s\right)}^2}\\ & =& \sqrt{17.64{\left(m/s\right)}^2-4{\left(m/s\right)}^2}\\ & =& \sqrt{13.64{\left(m/s\right)}^2}\\ & =& 3.69m/s\end{array}$

Thus, the velocity of boat relative to earth is 3.69m/s.

The time taken by the motorboat to cross the river is given by,

$\begin{array}{rcl}{v}_{BE}& =& \frac{D}{t}\\ t& =& \frac{D}{v_{BE}}\end{array}$

Here, D is the width of the river, and v_BE velocity of boat relative to the earth.

Substitute all the values in the above,

$\begin{array}{rcl}t& =& \frac{500m}{3.69m/s}\\ & =& 135.5s\end{array}$

Hence, the time taken by the boat to cross the river is 135.5 s.