# A canoe has a velocity of 0.40m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50m/s east. Find the velocity... A canoe has a velocity of 0.40m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

This is vector problem, but you can also think of this in small chunks For every 0.4m to the south east, the boat gets pushed east 0.5 m. We determine where the boat is after 1 second, find the displacement in that one second, and we will have the speed and direction.

Use the tip-to-tail method. Draw a vector to the right that represents 0.5 m/s to the east. From the tip of that vector, draw a new vector that goes down and to the right at a 45° angle from the horizontal. You should end up with an obtuse angle. (The angle between the two vectors is 135°. The resultant velocity is the vector that connects the first vector tail, to the tip of the last vector. The direction will be the angle made with the x-axis (south of east). If you draw this to scale, it is as simple as measuring the angle with a protractor and measuring the length of the resultant.

To get a more accurate result, use trigonometry. Break all of the vectors into x- and y-components. (In this case East and South components).

Vector 1: 0.5 m/s East 0 m/s South

Vector 2: 0.4m/s*cos(45) East 0.4m/s*sin(45) South

Resultant: 0.5+0.4cos(45) East and 0.4*sin(45) South

Resultant: 0.783 East and 0.283 South

Using pythagorean theorem the resultant speed is: 0.832 m/s

The direction can be found by taking the inverse tangent of the ratio: 0.283/0.783

Which gives: 19.9° ~ 20° South of East

Answer: 0.832 m/s 20° South of East

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