A river flows due south with a speed of 2.00m/s. You steer a motorboat across the river; your velocity relative to the water is 4.2m/s due east. The river is 500 m wide. (a) What is your velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of your starting point will you reach the opposite bank?

The given data can be listed below.

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

In relation to body 1, the relative velocity is the speed of body 2. The relative velocity thus calculated presumes that body 1 is at rest and that body 2 may be moving in a different direction from that of body 1 rather than toward or away from it.

The diagram for relative velocities is given below.

The velocities are directed as shown in figure below.

Using Pythagoras theorem, the velocity of motorboat relative to earth is given by,

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

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,

$\begin{array}{rcl}{v}_{BE}& =& \sqrt{{\left(2m/s\right)}^2+{\left(4.2m/s\right)}^2}\\ & =& \sqrt{4{\left(m/s\right)}^2+17.64{\left(m/s\right)}^2}\\ & =& 4.65m/s\end{array}$

The direction of boat velocity is given by,

$\theta ={\tan }^{-1}\left(\frac{v_{BR}}{v_{RE}}\right)$

Substitute all the values in the above,

$\begin{array}{rcl}\theta & =& {\tan }^{-1}\left(\frac{4.20}{-2}\right)\\ & =& 64.5^o\end{array}$

Thus, the velocity of boat relative to earth and its direction are 4.65m/s and 64.5^o respectively.

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

$t=\frac{D}{v_{BR}}$

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

Substitute all the values in the above,

$\begin{array}{rcl}t& =& \frac{500m}{4.2m/s}\\ & =& 119.05s\end{array}$

Hence, the time taken by the boat to cross the river is 119.05 s.

The diagram below shows the distance towards south.

The angle is given by,

$\begin{array}{rcl}\alpha & =& {90}^o-\theta \\ & =& {90}^o-64.5^o\\ & =& 25.5^o\end{array}$

From diagram the displacement of the boat is given by,

$∆y=D\tan \alpha$

Here, D is the total distance covered by boat.

Substitute all the values in the above,

$\begin{array}{rcl}∆y& =& 800\times \tan 25.5^o\\ & =& 800\times 0.476\\ & =& 381m\end{array}$

Thus, the distance travelled by the boat in southward direction is 381 m.