Calculating Required Heading for North-East Sailing with Current at 5km/h

AI Thread Summary
To calculate the required heading for a ship sailing northeast against a current of 5 km/h, the ship's maximum speed of 30 km/h must be considered. The correct heading is determined to be 80 degrees, which translates to a direction of N 38 degrees E. Understanding the current's direction, which is south, is crucial for solving the problem accurately. A diagram can aid in visualizing the vector addition of the ship's velocity and the current's velocity. The discussion emphasizes the importance of setting up equations based on the Pythagorean theorem to find the correct heading.
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OP warned about not including efforts to solve the problem

Homework Statement


A ships captain wishes to sail his ship north -east. A current is moving his ship with a velocity of 5.0km/h . If the ship has a maximum speed of 30 km/h what is the ships required heading?

Homework Equations


Cos angle=A/H

The Attempt at a Solution


Heading=cosine (5/30)
Heading=80 degrees
The answer is [N 38 degrees E]
 
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Just out of curiosity, are you given the direction in which the current is moving his ship?
Edit: maybe I'm mistaken, but I feel like you need that information to solve this problem..
 
_N3WTON_ said:
Just out of curiosity, are you given the direction in which the current is moving his ship?
Edit: maybe I'm mistaken, but I feel like you need that information to solve this problem..
Yes it says south.
 
Ok, for problems like these it is essential to draw a diagram. You are trying to find the angle the ship is making with the x-axis (east axis). First you will need to set up an equation using the Pythagorean theorem, try that out as a start...
 
So the vector addition of ship's velocity wrt water + velocity of water = effective velocity of ship, right ?
Write out the components in direction north and in direction east to get two equations with two unknowns (heading wrt water and effective speed). You don't need the latter, so eliminate it. I see the answer appearing without any difficulty ...

Oh, and: welcome to PF :-)
 
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