Finding the Magnitude and Direction of the Displacement Vector for a Ship's Trip

[Brendan Webb]

1. A ship sails 130 km due north from island A to island B and then 92 km, in the direction 22° south of east, to island C. The ship after that returns directly to island A. Calculate the magnitude and direction of the displacement vector in the last trip. Draw appropriate diagrams.

Homework Equations

[/B]

The Attempt at a Solution

I think I would find the total displacement and that would be equivalent to the last trip.

D1 = X component = 0
Y component = 130sin(90) = 130

D2 = X component = 92cos(22) = 85.30
Y component = 92sin(22) = 34.46 (-) as it is the - Y direction

Adding the component together I get:

Dx = 85.30
Dy = 95.54

Using the Pythagoras Theorem I get 128 km in a tan-1 (95.54/85.30) = 48.24 south of west direction. (Total displacement?)

This makes sense to me when drawing it out. However, the only two answers I can find online give a way different solution that I can't comprehend, it is 56.44 km at 37.8 south of west. I figure i must be understanding the problem wrong but I can't for the life of me figure out how. Any help would be much appreciated.

Thanks
Brendan[/B]

 

[TSny]

Hello, Brendan. Welcome to PF!

Your answer looks correct for the question as stated. The other solution that you found online appears to be for a slightly different question where "22^o south of east" is replaced by "22^o east of south". But in that case, they should have stated the direction of the answer as 37.8^o west of south rather than south of west.

 

[Brendan Webb]

Hello, Brendan. Welcome to PF!

Your answer looks correct for the question as stated. The other solution that you found online appears to be for a slightly different question where "22^o south of east" is replaced by "22^o east of south". But in that case, they should have stated the direction of the answer as 37.8^o west of south rather than south of west.

Thank you very much, I was confident in my answer but not confident enough. Cheers :)