A canoe has a velocity of 0.40m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50m/s east relative to the earth. Find the velocity(magnitude and direction ) of the canoe relative to the river.

**Discussion:**

We apply the relative velocity relation. The relative velocities are \(\vec{v_{CE}}\), the canoe relative to the earth, \(\vec{v_{RE}}\), the velocity of the river with respect to Earth and \(\vec{v_{CR}}\), the velocity of the canoe relative to the earth.

$$\vec{v_{CE}}=\vec{v_{CR}}+\vec{v_{RE}}$$

Hence $$

\vec{v_{CR}}=\vec{v_{CE}}-\vec{v_{RE}}$$

The velocity components of \(

\vec{v_{CR}}\) are $$ -0.5+\frac{0.4}{\sqrt{2}}=-0.217m/s$$( in the east direction)

Now, for the velocity component in the south direction

$$ \frac{0.4}{\sqrt{2}}=0.28$$ (in the south direction)

Now, the magnitude of the velocity of canoe relative to river $$ \sqrt{(-0.217)^2+(0.28)^2}=0.356m/s$$

If we consider \(\theta\) as the angle between the canoe and the river,the direction of the canoe with respect to the river can be given by

$$ \theta=52.5^\circ$$ ( in south west direction)

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