Question #3fc43

`"elapsed time from point O to A :" t=(OA)/v_x" "t=40/5=8" " s`

`"AB="v_y*t=2*8=16" meters"`

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, `5m//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 `2m//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=5cos alpha`

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

`t=x/v_x = 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 * t = (5 sin alpha - 2) 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_(yo) * t = (5 sin alpha) s /(5 cos alpha)`

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

`d = (5 sin alpha) s /(5 cos alpha)-(5 sin alpha - 2) s /(5 cos alpha)`

which simplifies to

`d = (2s)/(5 cos alpha)`