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

This question aims to find the direction and magnitude of the velocity of the canoe with respect to the river.This question uses the concept of velocity. The velocity of an object has both direction and magnitude.  If the object is moving towards the right, then the direction of velocity is also towards the right.

We are given the following information:

$Vc \space = \space 0.4 \space \frac{m}{s}$

which is the magnitude of the canoe going towards the southeast while:

$Vr \space= \space0.5 \space \frac{m}{s}$

which is the magnitude of the river  going towards the east.

$Vr \space= \space 0.5 x$

We have to find the direction and magnitude of the velocity the canoe which is going with respect to the river. So:

$V_c \space = \space 0.4cos \space( \space -45 \space)x \space + \space 0.4sin \space( \space -45 \space)y$

Where $sin(-45)$ is equal to $-0.7071$ and $cos(-45)$ is equal to $0.707$.

$V_c \space = \space 0.4 \space( \space 0.707\space)x \space + \space 0.4 \space( \space -0.707 \space)y$

Multiplying $0.4$ will result in:

$V_c \space = \space 0.2828x \space + \space 0.4 \space( \space -0.707 \space)y$

$V_c \space = \space 0.2828x \space – \space 0.2828y$

So:

$V \space = \space V_c \space – \space V_r$

By putting values, we get:

$V\space = \space -0.2172x \space – \space 0.2828y$

The magnitude of $V$ will result in:

$V\space = \space 0.36 \space \frac{m}{s}$

And the direction is:

$= \space tan^{-1} \frac{- \space 0.2828}{- \space 0.2172 }$

$= \space 52.47 \space degree.$

The magnitude and direction of the velocity of the canoe with respect to the river are $0.36 \frac {m}{s}$ and $52.47$ degrees, respectively.

## Example

Find the direction and magnitude of the velocity of the canoe with respect to the river while its velocity is $0.5$ \frac{m}{s} towards the southeast and $0.50$ \frac{m}{s} towards the east.

The given information in the question is as follows:

$Vc \space = \space 0.5\space \frac{m}{s}$

Which is the magnitude of the canoe going towards the southeast, while:

$Vr \space= \space 0.5 \space \frac{m}{s}$

Which is the magnitude of the river going towards the east.

$Vr \ space= \space 0.5 x$

So:

$V_c \space = \space 0.5cos \space( \space -45 \space)x \space + \space 0.5sin \space( \space -45 \space)y$

Where $sin(-45)$ is equal to $-0.7071$ and $cos(-45)$ is equal to $0.707$.

$V_c \space = \space 0.5 \space( \space 0.707\space)x \space + \space 0.5 \space( \space -0.707 \space)y$

Multiplying $0.5$ will result in:

$V_c \space = \space 0.2535x \space + \space 0.5 \space( \space -0.707 \space)y$

$V_c \space = \space 0.3535x \space – \space 0.3535y$

So:

$V \space = \space V_c \space – \space V_r$

By putting values,we get:

$V\space = \space -0.2172x \space – \space 0.3535y$

The magnitude of $V$ will result in:

$V\space = \space 0.4148 \space \frac{m}{s}$

And the direction is:

$= \space tan^{-1} \frac{- \space 0.3535}{- \space 0.2172 }$

$= \space 58.43 \space degree.$

The magnitude and direction of the velocity of the canoe with respect to the river are $0.4148 \frac {m}{s}$ and $58.43$ degrees, respectively.