How Do You Calculate a Castaway's Final Position Using Vectors?

[Anonymous1135]

Homework Statement

In an attempt to escape a desert island, a castaway
builds a raft and sets out to sea. The wind shifts a great
deal during the day and he is blown along the following
directions: 2.50 km and 45.0° north of west, then 4.70 km
and 60.0° south of east, then 1.30 km and 25.0° south of
west, then 5.10 km straight east, then 1.70 km and 5.00°
east of north, then 7.20 km and 55.0° south of west, and
finally 2.80 km and 10.0° north of east. Use a graphical
method to find the castaway’s final position relative to the
island.

Ok so I am just doing this for fun getting ready for the spring semester trying to get ahead.

So the book is giving me 7.34km and 63.5 SW I am getting after multiple attempt 9.14km and 62.50SW
What am I doing wrong

The Attempt at a Solution

[/B]
2.50km 45° NW is 135°
4.70km 60° SE is 300°
1.30km 25° SW is 205°
5.10km 0° E is 0°
1.70km 5° NE is 5°
7.20km 55° SW is 235°
2.80km 10° NE is 10°
---------------------------------------------------------
2.50cos(135) = -1.77
2.50sin(135) = 1.77

4.70cos(300) = 2.35
4.70sin (300) = 4.07

1.30cos(205) = -1.78
1.30sin(205) = -0.55

5.10cos(0) = 5.10
5.10sin(0) = 0

1.70cos(5) = 1.69
1.70sin(5) = 0.15

7.20cos(235) = -4.13
7.20sin(235) = -5.90

2.80cos(10) = 2.76
2.80sin(10) = 0.49

My answer is 4.22i - 8.11j 9.14km and 62.50SW

 

[PeroK]

You need a diagram to check that your calculations have the correct sign, at least.

The first part of the journey is NW and the second is SE. These are opposite directions, so I can see at a glance that you have made a mistake in the first four lines.

 

[Anonymous1135]

You need a diagram to check that your calculations have the correct sign, at least.

The first part of the journey is NW and the second is SE. These are opposite directions, so I can see at a glance that you have made a mistake in the first four lines.

When you say first four line are you talking about before the dotted lines or after?

 

[PeroK]

When you say first four line are you talking about before the dotted lines or after?

After the dotted line.

 

[Anonymous1135]

Can you explain for the first part what I am doing wrong just do I have an idea. I am really confuse when it comes to which quadrant and direction. For example sometime when something is in quadrant 4 some either takes 360 off the angle or add 270.

 

[PeroK]

Can you explain for the first part what I am doing wrong just do I have an idea. I am really confuse when it comes to which quadrant and direction. For example sometime when something is in quadrant 4 some either takes 360 off the angle or add 270.

I'm not sure how you made the mistake, but if you draw a diagram, you should see what is going on. You have a positive bearing for north in step one and a positive bearing for south in step two.

Now, anyone can make a mistake, but you should see that what you have is wrong.

 

[Anonymous1135]

Im not understanding. for the first part, wouldn't 2.5 @ W45N = -1.77i + 1.77j . Sorry I am still new at this but if you draw a diagram for that part 1 x would be negative and y would be positive giving NW?

 

[PeroK]

Im not understanding. for the first part, wouldn't 2.5 @ W45N = -1.77i + 1.77j . Sorry I am still new at this but if you draw a diagram for that part 1 x would be negative and y would be positive giving NW?

Okay. And SE?

 

[Anonymous1135]

Ok I see the issue lol. so 4th quadrant positive x and negative y. Now I subtracted 360 from 60° to get 300. I'm assuming that's wrong so would I need to add 270 to the 60° or make it -60° this is where the confusion comes in

 

[haruspex]

add 270 to the 60° or make it -60° this

adding 270 to 60 would make it 330, or minus 30.
Adding or subtracting some standard angle is only part of the story. Which is right depends on what you do with the result. If you add or subtract an odd multiple of 90 then you are going to flip between cos and sin, which might be confusing. Probably better to stick to 180 and 360.
These are the rules I use:

• Adding 360 clearly changes nothing, since it is the same angle;
• sin(-θ)=-sin(θ)
• cos(-θ)=cos(θ)
• sin(180+θ)=-sin(θ)
• cos(180+θ)=-cos(θ)

E.g. for sin(300°) we can write sin(360°-60°)=sin(-60°)=-sin(60°).
As a cross-check, running through the quadrants 1 to 4, the trig functions with positive sign are All, Sine, Tan, Cos. Mnemonic: all silly tom cats.

 

[Anonymous1135]

I figured it out thanks for all the help