Question

A river is 4 km wide and flows at 7km/h. Two docks are directly across the river from each other. Frank heads out from one dock in a direction perpendicular to the current in a boat which has a speed of 16km/h in still water. How far downstream from the dock will the current push the boat?

Answer 1

The velocity vectors will add as in the attached figure. The boat has a velocity vector, (0,16) and the current, (-7,0) (taking the positive x-and y-directions as per usual). Then the resultant vector is just the vector sum:

(0,16)+(-7,0)=(-7,16) = v

We can use the vector dot product between the resultant vector, v, and the boat vector to find the angle, theta, in between. We can then use trigonometry to find the distance needed.

So\[a.b=|a||b|\cos \theta \rightarrow (-7,16).(0,16)=\sqrt{(-7)^2+16^2} \times \sqrt{0^2+16^2} \cos \theta\]that is\[256=16\sqrt{305}\cos \theta \rightarrow \cos \theta = \frac{16}{\sqrt{305}}\]The river is 4km wide, so we can use what we've found and what we're given to solve for x (in the second diagram).

\[\tan \theta = \frac{x}{4} \rightarrow x = 4 \tan \theta = 4 \tan (\cos^{-1}\frac{16}{\sqrt{305}})=1.75km\]The boat will dock 1.75km downstream.