- #1
delcypher
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Hi this seems like a simple question (and probably is) but I keep getting the answers wrong.
Two ships, A and B, leave port P at the same time. Ship A travels due north at a steady speed of 15kmh^{-1} and ship B travels N 60^{o} E at a speed of 10kmh^{-1}.
i) What is the distance and direction from A to B after 1 hour?
ii) What is the velocity of B relative to A?
c^{2} = a^{2} + ^{2} - 2.a.b.COSC
COS 60^{o} = 0.5; SIN 30^{o} = 0.5; cos 30^{o} = (\sqrt{3})/2
The workings can be seen here (sorry I prefer to write):
http://www.unicyclist.com/gallery2/main.php?g2_view=core.DownloadItem&g2_itemId=551574&g2_serialNumber=1"
The answer given in the book for the distance between A and B in i) should be 13.33Km, my calculations make it 13.23Km, it's close but not enough for me to be certain I've done it correctly.
The direction for i) is correct they've (in my opinion) written it a strange way.
I wrote 139.11^{o} (from North clockwise) or E 49.11^{o} S. They wrote S 40^{o} 54' E - this means S 40.9^{o} E
My answer for ii) is very close 13.23kmh^{-1}, but they give 13.22kmh^{-1}.
Can anyone shed any light on what I'm doing wrong?
Homework Statement
Two ships, A and B, leave port P at the same time. Ship A travels due north at a steady speed of 15kmh^{-1} and ship B travels N 60^{o} E at a speed of 10kmh^{-1}.
i) What is the distance and direction from A to B after 1 hour?
ii) What is the velocity of B relative to A?
Homework Equations
c^{2} = a^{2} + ^{2} - 2.a.b.COSC
COS 60^{o} = 0.5; SIN 30^{o} = 0.5; cos 30^{o} = (\sqrt{3})/2
The Attempt at a Solution
The workings can be seen here (sorry I prefer to write):
http://www.unicyclist.com/gallery2/main.php?g2_view=core.DownloadItem&g2_itemId=551574&g2_serialNumber=1"
The answer given in the book for the distance between A and B in i) should be 13.33Km, my calculations make it 13.23Km, it's close but not enough for me to be certain I've done it correctly.
The direction for i) is correct they've (in my opinion) written it a strange way.
I wrote 139.11^{o} (from North clockwise) or E 49.11^{o} S. They wrote S 40^{o} 54' E - this means S 40.9^{o} E
My answer for ii) is very close 13.23kmh^{-1}, but they give 13.22kmh^{-1}.
Can anyone shed any light on what I'm doing wrong?
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