Answer to Question #101727 in Physics for Rosie

Question

A kayaker is traveling down a river at 7.5 knots due south when it encounters a current that is 4.3 knots at 43° east of north what is the new speed and direction of the boat

Answer 1

$$\bold{v}=(0,7.5)knots$$

$$\bold{u}=(4.3\sin{43},-4.3\cos{43})knots$$

$$\bold{v'={v+u}}=(4.3\sin{43},7.5-4.3\cos{43})=(2.93,4.36)knots$$

$$v=\sqrt{2.93^2+4.36^2}=5.3\ knots$$

The direction:

$$\theta=\arctan{\frac{2.93}{4.36}}=34\degree\ east\ of \ south$$