# Relative Motion of a ship

## Homework Statement

A ship is traveling at 8.0 m/s at an angle of 50 degrees north of east relative to the water. The water moves at 3 m/s due south relative to the shore.
a- Determine the velocity of the ship relative to the shore.
b- A person running along the deck of the ship moves at 2 m/s at 45 degrees south of west. What is his velocity relative to the shore?

## Homework Equations

V= (Va cos angle a + Vb cos angle b) i-hat + (Va sin angle a + Vb sin angle b) j-hat

## The Attempt at a Solution

part a-
(8cos50 + 3 cos -90) i-hat + (8sin 50 + 3 sin -90) j-hat =

5.14 m/s i-hat + -3.12 m/s j-hat

I am confused about part b. Do I need to factor in the motion of the ship that the person is on? Or do I combine his velocity with that of the water since it is measured relative to the shore?

Well, for part a you wrote

V (ship w/res to water) + V (water w/res to shore) = V ship w/res to shore

Assuming the info for part b is a velocity w/res to the ship, where in the equation do you think you should add the new vector?

Since the man is running in a different direction on the ship than the ship is moving, I thought I would have to create a new vector for his direction.

That is correct. You need to write V (person w/res to ship).

Now, based on how you did part a, how would you use this additional vector to find the person's velocity w/res to the shore?

I created a vector for the guy, which was (2 m/s cos 225) i + (2 m/s sin 225) j and added it to the ship to shore vector, giving me 3.73 m/s i + 1.72 m/s j.

Does that seem correct?

Yes, it does.

What you've done is

V (person w/res ship) +V (ship w/res to water) + V (water w/res to shore) = V (person w/res to shore)

I see. Thanks a lot for your help.