Vector word problem

• Mar 13th 2011, 08:06 AM
Smokinoakum
Vector word problem
Hi all...I have some homework that I'm having some problems with.

Here is the word problem that I'm confused about...

A ship travels 18.5 km on a course of 31.2° south of east, then 12.7 km due south, then 21.5 km on the course of 61.3° west of south, where it lands. Find the displacement and the angle traveled from the starting point to the landing point.

Here is what I did to try and solve it...

18.5cos31.2 + 12.7cos58.8 + 21.5cos61.3 = 32.73
18.5sin31.2 + 12.7sin58.8 + 21.5sin61.3 = 39.3

32.73^2 + 39.3^2 * 1/2 = 51.15

tan^-1(39.3 / 32.73) = 50.22
°

so...51.15 at 50.22

For some reason, this just does not look right. Can anyone check my logic, and straighten me out here?

Neil

• Mar 13th 2011, 09:04 AM
HallsofIvy
Where did you get the "58.8" from? It looks like you are calculating the "x" (east west) and "y" (north south) components separately. But the middle leg, 12.7 km, is "due south". There is no "east west" movement. Also, the third leg, at "61.3° west of south", since it is "west of south" should be -21.5 sin(61.3) For the "north south" movement, the middle leg is 12.7 km and the third leg 21.5 cos(61.3).

You should have, for "x", 18.5 cos31.2 + 0 - 21.5 sin 61.3 (which is negative, so to the west) and, for "y", 18.5 sin(31.2)+ 12.7+ 21.5 cos(61.3) to the south.
• Mar 13th 2011, 09:32 AM
Smokinoakum
Quote:

Originally Posted by HallsofIvy
Where did you get the "58.8" from? It looks like you are calculating the "x" (east west) and "y" (north south) components separately. But the middle leg, 12.7 km, is "due south". There is no "east west" movement. Also, the third leg, at "61.3° west of south", since it is "west of south" should be -21.5 sin(61.3) For the "north south" movement, the middle leg is 12.7 km and the third leg 21.5 cos(61.3).

You should have, for "x", 18.5 cos31.2 + 0 - 21.5 sin 61.3 (which is negative, so to the west) and, for "y", 18.5 sin(31.2)+ 12.7+ 21.5 cos(61.3) to the south.

Wow...I was way off in my assumptions.

So what your saying is this...

X = 18.5 cos(31.2) + 0 - 21.5 sin(61.3) = -3.034403707
Y = 18.5 sin(31.2) + 12.7 + 21.5 cos(61.3) = 15.00936432

-3.034403707^2 + 15.00936432^2 * 1/2 = 14.69943575

Then... tan^-1(15.00936432 / -3.034403707) = -78.5707125

I'm not sure what to do with the -78.5707125 answer for the angle. I'm guessing I need to decide what sector it's in? Can you help me a bit more?
• Mar 13th 2011, 10:18 AM
skeeter
sure does help to make a sketch ...

$\Delta x = 18.5\cos(31.2) - 21.5\cos(28.7)$

$\Delta y = -18.5\sin(31.2) - 12.7 - 21.5\sin(28.7)$

$|r| = \sqrt{(\Delta x)^2 + (\Delta y)^2} \approx 32.7 \, km$

$\theta = \arctan\left(\dfrac{\Delta y}{\Delta x}\right) \approx 84.7^\circ + 180^\circ$ or about $5.3^\circ$ W of S
• Mar 13th 2011, 10:43 AM
Smokinoakum
Quote:

Originally Posted by skeeter
sure does help to make a sketch ...

$\Delta x = 18.5\cos(31.2) - 21.5\cos(28.7)$

$\Delta y = -18.5\sin(31.2) - 12.7 - 21.5\sin(28.7)$

$|r| = \sqrt{(\Delta x)^2 + (\Delta y)^2} \approx 32.7 \, km$

$\theta = \arctan\left(\dfrac{\Delta y}{\Delta x}\right) \approx 84.7^\circ + 180^\circ$ or about $5.3^\circ$ W of S

Ahhhh...The picture helps a lot. Especially the angles! Thanks a LOT.