# Vector help

• September 28th 2012, 04:09 PM
Oldspice1212
Vector help
Hey guys I need some help on this question...

A spelunker is surveying a cave. She follows a passage 180m straight west, then 230m in a direction 45 degrees east of south, and then 280m at 30 degrees east of north. After a fourth unmeasured displacement, she finds herself back where she started.

1. Use a scale drawing to determine the magnitude of the fourth displacement. 2. Determine the direction of the fourth displacement. Two sig digs.

So where should I start I'm terrible at vectors.
• September 28th 2012, 04:54 PM
skeeter
Re: Vector help
Quote:

Originally Posted by Oldspice1212
Hey guys I need some help on this question...

A spelunker is surveying a cave. She follows a passage 180m straight west, then 230m in a direction 45 degrees east of south, and then 280m at 30 degrees east of north. After a fourth unmeasured displacement, she finds herself back where she started.

1. Use a scale drawing to determine the magnitude of the fourth displacement. 2. Determine the direction of the fourth displacement. Two sig digs.

So where should I start I'm terrible at vectors.

acquire some graph paper and a protractor for the scale drawing ... sounds to me like this is a exercise in actual physical measuring using scale to come up with a solution.
• September 28th 2012, 06:22 PM
Soroban
Re: Vector help
Hello, Oldspice1212!

I hope you're familiar with the basics of vectors.

Quote:

A spelunker is surveying a cave.
She follows a passage 180m straight west,
then 230m in a direction 45 degrees east of south,
and then 280m at 30 degrees east of north.
After a fourth unmeasured displacement, she finds herself back where she started.

1. Use a scale drawing to determine the magnitude of the fourth displacement.
What a clumsy way to get a measurement!

2. Determine the direction of the fourth displacement. (Two sig. digits)

Let North and East be positive.

Going from A to B: 180 m west.
. . $\overrightarrow{AB} \;=\; \langle 180\cos180^o,\:180\sin180^o\rangle \;=\;\langle \text{-}180,\:0\rangle$

Going from B to C: 230 m, 45o east of south.
. . $\overrightarrow{BC} \;=\;\langle 230\cos(\text{-}45^o),\:230\sin(\text{-}45^o)\rangle \;=\; \langle 115\sqrt{2},\:\text{-}115\sqrt{2}\rangle$

Going from C to D: 280 m, 30o east of north.
. . $\overrightarrow{CD} \;=\;\langle 280\sin30^o,\:280\cos30^o\rangle \;=\; \langle 140,\:140\sqrt{3}\rangle$

. . $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} \;=\;\langle 115\sqrt{2} - 40,\:140\sqrt{3}-115\sqrt{2}\rangle \;\approx\;\langle 122.63,\:79.85\rangle$

Point D is 122.63 m east and 79.85 m north of point A.

Code:

* D
*  *
*  θ *
*        * 79.85
*          *
*              *
A *  *  *  *  *  *  *
122.63

$\tan\theta \:=\:\frac{122.63}{79.85} \:=\:1.53575454$

. . . $\theta \:=\:59.93000891^o$

The direction of the last displacement is: 60o west of south.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

By the way, the final magnitude is:

. . . $\sqrt{122.63^2 + 79.85^2} \;\approx\;146.3\text{ m.}$
• September 28th 2012, 10:04 PM
Oldspice1212
Re: Vector help
Thanks a ton so it would be 33 degrees south of west :P