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**.

## Expert Answer

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:

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

\[= \space 52.47 \space degree.\]

## Numeric Answer

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.