1. ## vectors problem

1)Adam can swim at the rate of 2 km/h in still water.At what angle to the bank of a river must he head if he wants to swim directly across the river and the current in the river moves at the rate of 1 km/h?
2) A boat heads 15degrees west of of north with a water speed of 12 m/s. Determine its resultant velocity , relative to the ground, when it encounters a 5 m/s current from 15 degrees north of east.

2. Originally Posted by anna12345
1)Adam can swim at the rate of 2 km/h in still water.At what angle to the bank of a river must he head if he wants to swim directly across the river and the current in the river moves at the rate of 1 km/h?
2) A boat heads 15degrees west of of north with a water speed of 12 m/s. Determine its resultant velocity , relative to the ground, when it encounters a 5 m/s current from 15 degrees north of east.

1.) We can set up a triangle. His velocity of 2 km/h is the hypotenuse of the triangle and the current's velocity of 1 km/h is the opposite height. Thus the angle we're looking for is $\displaystyle \sin^{-1}(1/2) = \boxed{30^o}$

2.) The 15 degrees is the direction of the vectors and the 12 m/s and 5 m/s are the magnitudes of the vectors. If you draw out the vectors, you will see that they intersect at a right angle. Thus, using the "tip-to-tail" method of adding vectors is just the Pythagorean Theorem: $\displaystyle \sqrt{12^2+5^2}=13$. So it is going 13 m/s. To find its angle with the origin, you need to start by taking $\displaystyle \tan^{-1}(5/12)\approx 22.62^o$. However, this gives you the one of the angles of the 5-12-13 triangle, not the angle the 13 m/s vector makes with the origin. To find this angle, subtract $\displaystyle 15^o$ from the angle you found above to get $\displaystyle 7.62^o$. So the boat travels at 13 m/s in the direction $\displaystyle 7.62^o$ east of north.

I'm sure that was confusing but if you draw out the picture it should be easier to understand.