Relative Speed Calculation for Man Walking on a Moving Boat

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SUMMARY

The discussion focuses on calculating the speed of a man walking diagonally on a moving boat relative to the water. The ship moves at a speed of Vs = 5 m/s, while the man walks at Vm = 2 m/s at an angle of theta = 22 degrees to the perpendicular of the boat's motion. To find the man's speed relative to the water, one must resolve the man's velocity into x and y components using trigonometric functions, specifically sine and cosine, and then combine these with the boat's velocity.

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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.

Here is the image

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

At what speed does he walk relative to the water? Answer in units of m/s.


Homework Equations



I'm unsure at what equations to use.

The Attempt at a Solution



Do I find the horizontal movement then minus that number to 5 m/s? Hmm.

Would I do cos 68 = adj/hypo. I got 68 because 90 - 22 = 68 and, I couldn't do cos 22 because it's not a right triangle when I draw in the "imaginary" lines. But what is adj and the hypo and the opposite lengths?

I'm at a lost. Help please
 
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You need to add the velocity of the boat relative to water... to the velocity of the man relative to the boat...

take x to be the direction of the boat's motion..

take y perpendicular to x...

You know that the boat's velocity relative to water is:

vxboat = 5m/s
vyboat = 0m/s

Now, get the x and y components of the man's velocity relative to the boat:

what is vxman and vyman?

The hypoteneuse of your right triangle is 2m/s. The angle is 22 degrees. calculate the two legs of the right triangle.
 

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