# Relative motion of two ships

1. Sep 28, 2004

### quick

Ships A and B leave port together. For the next two hours, ship A travels at 35.0 mph in a direction 25.0 degrees west of north while the ship B travels 80.0 degrees east of north at 30.0 mph.
What is the distance between the two ships two hours after they depart?
What is the speed of ship A as seen by ship B?

basically i tried using trig to find the x component of both vectors and then added them. it wanted distance at t = 2 so i did 70 * sin (25) + 60 * sin(80) but it was incorrect. i tried this method with the example in the book and it was fairly close. so im guessing i did it incorrectly but got an answer that was close by coincidence? and for the second part im really unsure where to begin, i just know reference frames have something to do with it and maybe some sort of vector addition/subtraction. any suggestions will be much appreciated.

Last edited: Sep 28, 2004
2. Sep 28, 2004

### quick

3. Sep 28, 2004

### Pyrrhus

try using

$$v = \frac{d}{t}$$

4. Sep 28, 2004

### quick

i got it using the law of cosines. thanks for your suggestion though.

5. Sep 28, 2004

### NateTG

When you're calculating the lenght of a component-wise vector it's
$$\sqrt{x^2+y^2}$$
from the way you describe what you did, you might have forgotten the $$y$$ component.

The second part looks like straightforward vector addition as well.