## Problem on relative motion and vector addition

A river flows with speed $10$ m/s in the northeast direction. A particular boat can propel itself at speed $20$ m/s relative to the water. In which direction should the boat point in order to travel due west.?
• ## Solution

We introduce symbols for the three velocities in this problem.
Let $\bfv_\text{river}$ be the velocity of the river.
Let $\bfv_\text{boat}$ be the velocity due to the boat's propulsion.
Let $\bfv_\text{net}$ be the net velocity of the boat relative to the ground.
We now look for a relation between these velocities.
Recall that
That is, $$\bfv_\text{net} = \bfv_\text{river} + \bfv_\text{boat}.$$
Graphically, we depict this relationship as:
Now, we express the problem statement in terms of these vectors:
We are given that $\bfv_\text{river}$ has length 10 and is in the direction of $\bfi + \bfj$. We are given that the $\bfj$ component of $\bfv_\text{net}$ is to be 0. We are given that the length of $\bfv_\text{boat}$ is 20. We are to find the direction of $\bfv_\text{boat}$.
First, we find $\bfv_\text{river}$. Then we will characterize a general $\bfv_\text{boat}$. We will select the vector such that the sum has no $\bfj$ component.
To find $\bfv_\text{river}$ we recall how to specify a vector from its length and direction:
Using the information that $\bfv_\text{boat}$ has length 10 and is in the direction of $\bfi + \bfj$, we have that $$\bfv_\text{boat} = 10 \text{ dir } (\bfi + \bfj).$$
Using (1) to also compute that $\text{dir } (\bfi + \bfj) = \frac{\bfi + \bfj}{\sqrt{2}}$, we get that $$\bfv_\text{boat} = 10 \frac{\bfi + \bfj}{\sqrt{2}}.$$
Second, we characterize all possible boat velocities as $\bfv_\text{boat} = a \ \bfi + b \ \bfj.$
Setting the $\bfj$ component to zero, we conclude $$b = - \frac{10}{\sqrt{2}}.$$
Hence, \begin{align}
a^2 &= 350\\
a &= \pm \sqrt{350}
\end{align}
For the boat to travel west, $a$ must be negative. Hence, \begin{align}\bfv_\text{boat} &= - \sqrt{350} \ \bfi - \frac{10}{\sqrt{2}} \ \bfj\\
&\approx -18.7 \ \bfi - 7.1 \ \bfj\end{align}