# Question #3fc43

May 19, 2016

$t = 8 s$
$A B = 16 m$

#### Explanation: $\text{elapsed time from point O to A :" t=(OA)/v_x" "t=40/5=8" } s$

$\text{AB="v_y*t=2*8=16" meters}$

May 19, 2016

The time is $t = \frac{s}{5 \cos \alpha}$ and the distance from point $A$ to point $B$ is $d = \frac{2 s}{5 \cos \alpha}$ where $s$ is the distance to shore and $\alpha$ is the angle which the boat points.

#### Explanation:

Sometimes, when we get a question like this, we don't have all of the information that is needed. In that case, we can put a place-holder (variable) in for that information and proceed with a solution. In this case, a diagram helps us decide what we have and what we need: We are looking for the distance between points $A$ and $B$, let's call that $d$. We know the speed of the boat, $5 m / s$, and we know that it starts by pointing itself at point $A$. Let's call the angle that the boat is pointed in $\alpha$ with respect to the direction to shore, $x$.

We also know that the river is moving at $2 m / s$ which adds to the velocity of the boat giving the resultant velocity, ${v}_{r}$, of the boat shown in the green arrow. This resultant velocity is what causes the boat to arrive at point $B$.

Finally, we need to know the distance to the shore, let's call this $s$. The remainder is just geometric constructions.

The x-velocity of the boat is simply:

${v}_{x} = 5 \cos \alpha$

We can calculate the time to get to shore from this and the distance to shore as:

$t = \frac{x}{v} _ x = \frac{s}{5 \cos \alpha}$

The y-velocity of the boat is given by:

${v}_{y} = 5 \sin \alpha - 2$

From this and the time we can get the distance traveled upstream as

$c = {v}_{y} \cdot t = \left(5 \sin \alpha - 2\right) \frac{s}{5 \cos \alpha}$

If the river wasn't moving, the boat would have reached point $A$ in the same amount of time. So we can use the same approach to find the distance it would travel in this case as

$c + d = {v}_{y o} \cdot t = \left(5 \sin \alpha\right) \frac{s}{5 \cos \alpha}$

We can now subtract $c$ from this total to get $d$

$d = \left(5 \sin \alpha\right) \frac{s}{5 \cos \alpha} - \left(5 \sin \alpha - 2\right) \frac{s}{5 \cos \alpha}$

which simplifies to

$d = \frac{2 s}{5 \cos \alpha}$