Finding Angle of Man Walking on Ship Relative to Water

[missashley]

Homework Statement

A ship cruises forward at Vs=5 m/s relative to the water. On deck, a man walks diagonally toward the bow such that his path forms an angle theta = 22 degrees with a line perpendicular to the boat's direction of motion. He walks at Vm = 2 m/s relative to the boat

http://img407.imageshack.us/img407/4973/boatum7.th.jpg

The speed he walks relative to the water is 5.749 m/s.

At what angle to his intended path does the man walk with respect to the water? Answer in degrees.

Homework Equations

I was thinking maybe of using Tan-1(opposite/adjacent)
or some thing to that effect

The Attempt at a Solution

Tan-1( but what is the opposite and adjacent?

Would it be 22 degrees? But 22 degrees is perpendicular to the boat's direction.
hmmm help please

 

[Dick]

First express the total velocity of the man by adding the velocity of the boat to his velocity relative to the boat. Break the two vectors into xy components and add them. Then we can talk about angles.

 

[Architeuthis Dux]

To find the angle at which the man is walking relative to the water, we can use the concept of vector addition. The man's velocity can be broken down into two components: one parallel to the boat's direction of motion and one perpendicular to it. The component parallel to the boat's direction is equal to Vm=2 m/s, while the component perpendicular to the boat's direction is equal to Vm*Tan(22) = 0.782 m/s.

Since the boat is moving at 5 m/s relative to the water, the man's total velocity relative to the water is the vector sum of these two components. Using the Pythagorean theorem, we can calculate the magnitude of this total velocity to be 5.749 m/s.

Now, to find the angle at which the man is walking relative to the water, we can use the inverse tangent function as you suggested. However, the opposite and adjacent sides of the triangle we are considering are not the same as the components of the man's velocity. Instead, the opposite side is equal to the component perpendicular to the boat's direction (0.782 m/s) and the adjacent side is equal to the component parallel to the boat's direction (2 m/s).

Therefore, the angle at which the man is walking relative to the water can be calculated as Tan^-1(0.782/2) = 21.8 degrees. This means that the man is walking at an angle of 21.8 degrees with respect to the water, which is slightly smaller than the angle he is walking with respect to the boat's direction (22 degrees).

I hope this helps! If you have any further questions, feel free to ask.