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March 31, 2015

March 31, 2015

Posted by **epifanio** on Saturday, January 19, 2013 at 9:56pm.

- phisics -
**Reiny**, Sunday, January 20, 2013 at 1:36amAfter doing all the calculations below, I just realized you just needed a scale drawing.

Oh, well, no harm done

Get a nice sheet of graph paper, scale the units to the following

150 m --> 15/2 = 7.5 cm

270 m --> 27/2 = 13.5 cm

280 m --> 28/2 = 14 cm , to fit on a standard sheet of paper.

You will need a compass and a protractor.

Most of your sketch will be in the third and fourth quadrant of the grid, so put your origin near the top part of the graph paper about 1/2 of the way over.

1. from your origin, draw a horizontal line 7.5 cm long

2. at the end of that line draw a line down at S 45°E.

3. set your compass to 13.5 cm and mark off that line

4. from that end point, draw a line making an angle of 75° with the previous line

5. set your compass to 14 cm and mark off this new line.

6. Join this point to the origin

7. measure that line, it should be appr. 9.2 cm (if my work below does not contain any stupid arithmetic errors).

I think we can solve this using vectors.

any vector (x,y) can be described by (rcosŲ, rsinŲ), where r is its magnitude and Ų is the angle it forms with the positive x-axis

so we have to convert our different paths

150m straight west --> (150cos180, 150sin180) = (-150, 0)

270m in a direction 45^\circ east of south

= (270cos315°, 270sin315) = (190.919 , -190.919)

280 {\rm m} at 30^\circ east of north

= (280cos60, 280sin60) = (140 , 242.49)

so (-150, 0) + (190.919 , -190.919) + (140 , 242.49) + (x,y) = (0,0)

-150 + 190.919 + 140 + x = 0

x = -180.919

0 -190.919 + 242.49 + y = 0

y = -51.571

(x,y) = (-180.919, -51.571)

magnitude = √(x^2 + y^2) = 188.13 m or 188 m to the nearest metre

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