Question

A canoeist wants to paddle to an island which located 38m north, and 100m east, of him. He finds that by pointing his canoe due east, and paddling at a constant speed of 2.0m/s relative to the water, he is able to reach the island in 42 seconds. Find the speed and direction of the current in the water through which the canoeist paddles.

Answer 1

Answer:

Vw = 0.98m/s due 67.16° north of east

Explanation:

The distance the canoeist wants to travel is:

$d=\sqrt{38^2+100^2}=106.98m$

And the angle of the destination point is

$\alpha =atan(\frac{38}{100} )=20.8\°$

With this trajectory and the time of 42s, we get the velocity of the canoe respect to ground:

$Vc = \frac{106.98<20.8\°}{42s}=(2.547<20.8\°)m/s$

Now we can calculate the velocity of the water:

$V_{c/w} = V_c - V_w$

$V_w = V_c - V_{c/w}=(2.547<20.8\°)-(2<0\°)$

$V_w = (0.98<67.16\°)m/s$