# Basic Vector problems

1. Mar 25, 2013

### cdahal

I am currently reading this "the easy way" physics text book. I have this problem to be kind of confusing. The confusion I have is posted below:

Question A: You drive a car 45 miles in a direction north 30 Degree West. How much farther west are you ?
Explanation for answer is given like this: 45cos(60deg) = 22.5 = 23 m farther to west.

Question B: A sailor's compass says that the ship is traveling North 55 degree West and the ship sextant says that at the end of 6.0 hours the ship is 35 miles farther north. How fast is he going?
Explanation for answer is given like this: displacement = 35cos(55deg) mi = 61 mile

So my question is why is there 45cos(60deg) ? why not 45cos(30deg) ? and why in the second question, there is 35cos(55deg). For the first one 30 deg is given not 60 and second one it uses same 55 deg as given why? can you clarify this to me? thank you so much everything. This is my first question in this fourm

2. Mar 26, 2013

### Simon Bridge

Welcome to PF;
Which easy way text?
Because the direction of the car is not 30 degrees to the north direction, which is 60 degrees from the west direction.
Because the questions are referring to different directions.

It is clearer if you draw the displacement vector.

For the first one - the car is went 45miles in direction 30deg west of north.
The car's displacement vector makes a 30 degree angle to the NORTH direction and a 90-30=60 degree angle to the WEST direction. The question wants the distance along the WEST direction so you use the 60 degree angle with cosine.

For the second one, the ship is travelling 55deg west of north.
The displacement vector makes 55 degree angle to the NORTH direction and a 90-55=35 degrees to the WEST direction. (Also 90+55=145 degrees anticlockwise from the EAST direction... and so on.)

The question gives you the distance travelled in the NORTH direction.
They want the speed: v=d/t, you know that t=6hrs
This means that 35=d.cos(55) so d=35/cos(55).

In general, the amount of a vector v in direction of vector u is v.u/|u| = |v|cosθ

3. Mar 26, 2013

### xAxis

I think none of these kind of problems would exist if students were taught trigonometric circle in high school, which they were not (at least in the UK).

4. Mar 26, 2013

### cdahal

Thank you Simon. I finally understood it.

5. Mar 26, 2013

### Simon Bridge

I know what you mean, I teach both ways - the different approaches appeal to different people.

Great.

Being able to sketch things out is a core skill in physics - see that you get lots of practice.
BTW: if you have trouble with trig, have a look through this:
http://mathmistakes.info/facts/TrigFacts/learn/uc/uc.html