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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.
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The speed for object A in relation to object C is the that is the sum of the velocity of object A relative to object B as well as the velocity of object B relative towards object C. This velocity for object A in relation towards object B exactly the same as the velocity negative of object B in relation to object A.♦ Relevant knowledge
This problem can be solved using the concept of relative velocity.
First, determine the relative velocity between the vertical and horizontal directions by resolving the velocities.
Find the angle between relative velocities (or its horizontal component) to determine the direction of the canoe’s velocity relative to the river.
The vector difference between two bodies’ velocities is called the relative velocity. Consider two bodies X and Y that are moving relative to one another. The relative velocity refers to the speed at which the body X appears to an observer looking at the body Y, and vice versa.
Relative velocity can be described as:
[katex]{\vec v_{XY}} = {\vec v_X} – {\vec v_Y}[/katex]
Here, [katex]{\vec v_{XY}}[/katex]
The figure below shows [katex]{\vec v_{ce}}[/katex]
Horizontal component of [katex]{\vec v_{ce}}[/katex]
[katex]{\vec v_{{{\left( {ce} \right)}_x}}} = {\vec v_{ce}}\cos ( – 45^\circ )[/katex]
[katex]\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}[/katex]
Vertical component of [katex]{\vec v_{ce}}[/katex]
[katex]{\vec v_{{{\left( {ce} \right)}_y}}} = {\vec v_{ce}}\sin ( – 45^\circ )[/katex]
[katex]\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}[/katex]
Horizontal component of [katex]{\vec v_{re}}[/katex]
[katex]{\vec v_{{{\left( {re} \right)}_x}}} = 0.490\,{\rm{m/s}}[/katex]
Vertical component of [katex]{\vec v_{re}}[/katex]
[katex]{v_{{{\left( {re} \right)}_y}}} = 0\,{\rm{m/s}}[/katex]
Relative velocity can be described as:
[katex]{\vec v_{XY}} = {\vec v_X} – {\vec v_Y}[/katex]
For relative velocity in horizontal direction, the expression is
[katex]{\vec v_x} = {\vec v_{{{\left( {ce} \right)}_x}}} – {\vec v_{{{\left( {re} \right)}_x}}}[/katex]
[katex]\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}[/katex]
For relative velocity in vertical direction, the expression is
[katex]{\vec v_y} = {\vec v_{{{\left( {ce} \right)}_y}}} – {\vec v_{{{\left( {re} \right)}_y}}}[/katex]
[katex]\begin{array}{c}\\{{\vec v}_y} = – 0.2121\,{\rm{m/s}} – 0\\\\ = – 0.2121\,{\rm{m/s}}\\\end{array}[/katex]
The expression is for the tangent of an angle
[katex]\tan \theta = \frac{{{{\vec v}_y}}}{{{{\vec v}_x}}}[/katex]
The angle expression is:
[katex]\theta = {\tan ^{ – 1}}\left( {\frac{{{{\vec v}_y}}}{{{{\vec v}_x}}}} \right)[/katex]
[katex]\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}[/katex]
The direction of [katex]{\vec v_{cr}}[/katex]Ans can be seen in the figure:
The velocity of the canoe relative the river is [katex]{\bf{37}}{\bf{.35^\circ }}[/katex]