# Homework Help: Vectors: Find magnitude and direction of the changes in velocity

1. Oct 16, 2011

### srsGreem

1. The problem statement, all variables and given/known data
Given the initial (u) and final (v) velocities below, find the magnitude and direction of the changes in velocity:

2. Relevant equations
(a) through (f) - I only have trouble with (e) and (f)

(e): u = 10ms east, v = 15ms N40°W
(f): u = 250ms N20°W, v = 200ms south

3. The attempt at a solution

http://dl.dropbox.com/u/14178839/img026.pdf [Broken]

As you can see, I'm trying to use trigonometry to solve the problems. I think I'm doing something wrong because the textbook answers are not matching up with the answers I have been getting.

Probably what I'd like is someone to walk me through at least one of the questions, maybe link me to relevant tutorials, and then see if I can figure out the other one.

I've been trying to figure this out on my own for about 5 hours now, and finally decided to just post up my problem on these forums. Thanks for any and all assistance.

Last edited by a moderator: May 5, 2017
2. Oct 16, 2011

### Staff: Mentor

What are the x & y components of these velocities? (+x = east; +y = north)

3. Oct 16, 2011

### srsGreem

Hey, I'm not entirely sure what you are trying to say there... but yes, that is the orientation I am working with, as shown in the pdf file.

4. Oct 16, 2011

### Staff: Mentor

Well, what are the components then?

5. Oct 16, 2011

### srsGreem

I don't know why that's relevant for this question... can you explain please.

6. Oct 16, 2011

### Staff: Mentor

One easy way to express a vector is in terms of its components. Then to find the difference between two vectors, you can just subtract the components.

7. Oct 16, 2011

### HallsofIvy

If you do not want to use "components", as Doc Al is suggesting (and using components is simpler), you can use the "sine law" and "cosine law" on the triangles formed.

In the first exercise you have a vector, u, that goes due east with "length" 10 and another vector, v, that goes 40 degrees W of N with "length" 15. If you draw those two sides of a triangle, the difference, v- u, is the third side of the triangle (directed from the tip of v to the tip of u). The length of that third side is given by the cosine law: $c^2= a^2+ b^2- 2ab \cos(C)$ where a and b are the two given lengths and C is the angle opposite side c. Here, the angle between the two given angle is 90+ 40= 130 degrees so $c^2= 10^2+ 15^2- 2(10)(15)cos(130)$. You can then use the sine law to find the other two angles in the triangle and so deduce the direction of that vector.

8. Oct 16, 2011

### srsGreem

Hey, so I've been using the sine rule to find the lengths, and I don't know enough about components to work with that method (yet). So my first question relates to (e):

my magnitude is correct, but according to the physics book, direction is wrong, can you tell me why?

and this question relates to (f):

everything was wrong, aparently the magnitude is around about 90(I don't have the book open), but in my answer, I got around 450.

Can you explain how I went a little wrong in (e) and why I went so massively wrong in (f)?

9. Oct 17, 2011

### Staff: Mentor

I'd say you are correct and the book is wrong.