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1. Oct 26, 2014

### mattyc33

1. The problem statement, all variables and given/known data
"An electron moves at 24.8m/s (N37.5E) and undergoes a constant 103m/s2 (E37.5S) acceleration for 3.02s what is the displacement?

2. Relevant equations
0.5at^2 = d
d= vt

3. The attempt at a solution
I calculated the displacements to be 74.896m (N37.5E) and 469.7006m (E37.5S). This is what I remember from high school, I can't remember the vector addition though. Would I just subtract to get 394 and then multiply by cos(37.5). Please help!

2. Oct 26, 2014

### Orodruin

Staff Emeritus
I suggest you try to draw the vectors on a piece of paper. First draw the displacement due to initial velocity and then that due to the acceleration starting from the end point of the first. The total displacement is from the starting point of the first displacement to the end point of the second. Can you make geometrical arguments for the total displacement?

3. Oct 26, 2014

### mattyc33

I can... I have triangle here with two distances of equal angles compared to the +x line of the origin. I'm actually solving this for my little brother so this is really outdated for me. But to get that distance shouldnt it just be a straight line... AKA the subtracted number multiplied by cos 37.5?

4. Oct 26, 2014

### Orodruin

Staff Emeritus
Could you show the triangle you have? The total displacement should be one of the sides of the triangle and you should be able to use trigonometry to get the final result. Without you writing out explicitly what you think is the answer, it is difficult to tell if you have the correct idea or not.

5. Oct 26, 2014

### mattyc33

Ummm I'll try to visualize it for you. I'm new to the site so i'm not sure if there is a function to draw.... Basically I have an origin I have a line (displacement of 74.896m 37.5 degrees North of East from the origin which is diagonally up and right (This comes from the velocity). I have another line of 469.7006m 37.5 degrees East of South from the origin which is diagonally right and down (from acceleration). This will result in a triangle with an angle of 37.5*2 (75) from from the origin and a large straight line which is what I need to find. You cant do sohcahtoa due to the lack of right angle if my memory isn't failing me.

6. Oct 26, 2014

### Orodruin

Staff Emeritus
It is still have difficulties imagining exactly how your triangle looks like. You could try taking a photo of it or making a drawing in a basic program like Paint.

I also suggest you check out the following
http://en.wikipedia.org/wiki/Law_of_cosines
http://en.wikipedia.org/wiki/Law_of_sines
to remind yourself a bit of triangle geometry.

I never really liked giving headings using north of east etc, but it does sound to me as if a line which is 37.5 degrees east of south should be perpendicular to one which is 37.5 degrees north of east.

7. Oct 26, 2014

### mattyc33

Oh wow I did draw it out wrong. So it is a right angle triangle with hypotenuse of 475.634m... I just have to give him the angle or degrees or something from the origin which I don't know how to do. This should make it more clear to you hopefully. Right angle triangle due to trigonometry one side (a) is 74.9m the other side (b) is 469.7m so (c) is 475.634m. Now I just need to figure out the angle or whatever from the origin... =/

8. Oct 26, 2014

### Orodruin

Staff Emeritus
Your answer should be a distance and a bearing. That way you know how far you have come and in what direction.

9. Oct 26, 2014

### mattyc33

Right im confused on what the bearing would be

10. Oct 26, 2014

### Orodruin

Staff Emeritus
For that you would have to relate the angles in the triangle to the original headings, i.e., what angle does the displacement make with the two different vectors?

11. Oct 26, 2014

### mattyc33

Would the longer vector justify that angle...? i.e. would the 469 (E37.5S) vector cause the displacement to be in that direction (E37.5S) again?