Can you help with this physics question?

Question

A woman walks due west on the deck of a ship at 2 mi/h. The ship is moving north at a speed of 17 mi/h. Find the speed and direction of the woman relative to the surface of the water.

I tried taking the derivative of the formula, but it leaves me with two variables. I think I'm making it harder than it really is...

Answer 1

Solve using vectors:

V west = 2 mi/hr "woman"

V north = 17 mi/hr "ship"

Speed=sqr((2)^2+17^2) = 17.11 mi/hr

direction=atn(17/2)*180/3.14159

= 83.3 degrees north of the west direction

Answer 2

Solve it using geometry

One leg is the 17mph north the other leg s 2 mph east.  These are at right angle to each other so its easy using pythag theorem

answer = sq root (17^2 +2^2) = 17.117 mph

Answer 3

just make a right angle triangle with one side being the ship moving north and the other the wench on the deck moving west. I cant draw the diagram, but we'll end up with a right angle triangle where we are trying to find the length of the hypotenuse when the other two sides are known. We end up with a situation where we can use the old f**t Hypotenuse's theorem.

A*2+B*2=C*2

2*2+17*2= C*2

4+289=C*2

293=C*2

C=(the square root of)293

C=17.117

Therefore, that crazy woman is traveling at 17.117 Miles an hour in some sort of north westerly direction.

Now you have the three sides of the triangle sorted, you can easily work out the angle she is heading in in relation to north;

Like I said, drawing a diagram would be the way to go. Draw your triangle and use triginometry to find the angle between the 17 side and the 17.117 side. That angle is how far west she is facing from north if that makes sense