Question

A canoe has a velocity of 0.300 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.490 m/s east relative to the earth.
Find the direction of the velocity of the canoe relative to the river.

Answer 1

Concepts and reason

The concept used to solve this problem is the relative velocity.

Initially, resolve the velocities to determine the relative velocity in the horizontal and vertical directions.

Then, find the angle between the resultant of relative velocities and its horizontal component to determine the direction of velocity of the canoe relative to the river.

Fundamentals

The relative velocity is the vector difference between the velocities of two bodies. Consider two bodies X and Y moving relative to each other. The relative velocity is the velocity that the body X would appear to an observer on the body Y and vice versa.

The expression for relative velocity is,

${\vec v_{XY}} = {\vec v_X} - {\vec v_Y}$

Here, ${\vec v_{XY}}$ is the relative velocity of the body X with respect to Y, ${\vec v_X}$ is the velocity of the body X, and ${\vec v_Y}$ is the velocity of the body Y.

In the figure below, ${\vec v_{ce}}$ is the velocity of the canoe relative to earth and ${\vec v_{re}}$ is the velocity of the river relative to earth.

Horizontal component of ${\vec v_{ce}}$ is,

${\vec v_{{{\left( {ce} \right)}_x}}} = {\vec v_{ce}}\cos ( - 45^\circ )$

Substitute $0.300\,{\rm{m/s}}$ for ${\vec v_{ce}}$ .

$\begin{array}{c}\\{{\vec v}_{{{\left( {ce} \right)}_x}}} = (0.300\,{\rm{m/s)}}\left( {{\rm{cos}}\left( { - 45^\circ } \right)} \right)\\\\ = {\rm{0}}{\rm{.2121}}\,{\rm{m/s}}\\\end{array}$

Vertical component of ${\vec v_{ce}}$ is,

${\vec v_{{{\left( {ce} \right)}_y}}} = {\vec v_{ce}}\sin ( - 45^\circ )$

Substitute $0.300\,{\rm{m/s}}$ for ${\vec v_{ce}}$ .

$\begin{array}{c}\\{{\vec v}_{{{\left( {ce} \right)}_y}}} = (0.300\,{\rm{m/s}})\left( {\sin \left( { - 45^\circ } \right)} \right)\\\\ = - 0.2121\,{\rm{m/s}}\\\end{array}$

Horizontal component of ${\vec v_{re}}$ is,

${\vec v_{{{\left( {re} \right)}_x}}} = 0.490\,{\rm{m/s}}$

Vertical component of ${\vec v_{re}}$ is,

${v_{{{\left( {re} \right)}_y}}} = 0\,{\rm{m/s}}$

The expression for relative velocity is,

${\vec v_{XY}} = {\vec v_X} - {\vec v_Y}$

The expression for relative velocity in the horizontal direction is,

${\vec v_x} = {\vec v_{{{\left( {ce} \right)}_x}}} - {\vec v_{{{\left( {re} \right)}_x}}}$

Substitute $0.2121\,{\rm{m/s}}$ for ${\vec v_{{{\left( {ce} \right)}_x}}}$ and $0.490\,{\rm{m/s}}$ for ${\vec v_{{{\left( {re} \right)}_x}}}$ .

$\begin{array}{c}\\{{\vec v}_x} = 0.2121\,{\rm{m/s}} - {\rm{0}}{\rm{.490}}\,{\rm{m/s}}\\\\{\rm{ = }} - {\rm{0}}{\rm{.2779}}\,{\rm{m/s}}\\\end{array}$

The expression for relative velocity in the vertical direction is,

${\vec v_y} = {\vec v_{{{\left( {ce} \right)}_y}}} - {\vec v_{{{\left( {re} \right)}_y}}}$

Substitute $- 0.2121\,{\rm{m/s}}$ for ${\vec v_{{{\left( {ce} \right)}_y}}}$ and $0\,{\rm{m/s}}$ for ${\vec v_{{{\left( {re} \right)}_y}}}$ .

$\begin{array}{c}\\{{\vec v}_y} = - 0.2121\,{\rm{m/s}} - 0\\\\ = - 0.2121\,{\rm{m/s}}\\\end{array}$

In the figure below, ${\vec v_{cr}}$ is the velocity of the canoe relative to the river. The angle $\theta$ gives the direction of ${\vec v_{cr}}$ .

The expression for tangent of the angle $\theta$ is,

$\tan \theta = \frac{{{{\vec v}_y}}}{{{{\vec v}_x}}}$

The expression for the angle is,

$\theta = {\tan ^{ - 1}}\left( {\frac{{{{\vec v}_y}}}{{{{\vec v}_x}}}} \right)$

Substitute $- 0.2121\,{\rm{m/s}}$ for ${\vec v_y}$ and $- {\rm{0}}{\rm{.2779}}\,{\rm{m/s}}$ for ${\vec v_x}$ .

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 0.2121\,{\rm{m/s}}}}{{ - {\rm{0}}{\rm{.2779}}\,{\rm{m/s}}}}} \right)\\\\ = 37.35^\circ \\\end{array}$

From the figure, the direction of ${\vec v_{cr}}$ is $37.35^\circ$ south of west.

Ans:

The direction of the velocity of the canoe relative to the river is ${\bf{37}}{\bf{.35^\circ }}$ south of west.