# Solving vectors

1. Feb 15, 2014

### kaalaniz

1. A person going for a walk follows a path. The total trip consists of four straight-line paths. At the end of the walk what is the person's resultant displacement measured from the starting point? The books says the answer is 240 m at 237 degrees.

Vector A is 100m East at 0°, Vector B is 300m south and vector c is 150 meters southwest and vector D is 200 meters north west. There are two angle measurements given. I dont know how to upload the exact picture but if you go to http://www.mazzworld.net/HU2Q1.html [Broken] and find the exact problem, the picture is on there as well as the answer. I dont know how they got to that answer. Please help.

3.A_x=100m
A_y=0m
B_x=0m
B_y=300m
C_x=150cos(120)=-75m
C_y=150sin(120)=129.9 m
d_x=200cos(75)=51.8 m?
d_y=200sin(75)=193.2

Im not sure how to get the correct angle measurements for vector d. I tried to figure it out from the angles given but it was very confusing.

to find the resultant I did F_x=100+0+0+-75=25m
F_y=193.2+129.9+300+0=569.1m
[itex]/sqrt{F_x^2+F_y^2}
i got 569.6m
and tan^-1(569.1/25)=87.5°

Where did I go wrong? I think it is the angle measurements I put in but Im not sure how to get the correct measurements.

Last edited by a moderator: May 6, 2017
2. Feb 15, 2014

### jackarms

You're almost correct -- I looked at the picture, and I don't think you're interpreting the angles correctly. Well, the first thing is the signs on the displacements. North should be positive y and south should be negative y, but for B_y, which is going south, you have a positive value written.

For c and d, just use the angle measurement given to determine the x and y components. Have the vector be the hypotenuse of a right triangle, and the components be the other legs. Then use sin and cos to find the components, and make sure they have the right signs.

3. Feb 25, 2014

### kaalaniz

ok that makes sense. thankyou.

4. Mar 1, 2014

### kaalaniz

ok i have another question. I tried to use 30 degrees for the angle measurement for vector c and 60 degrees for vector D but I know that is not right. what are the angle measurements used for them?

5. Mar 1, 2014

### jackarms

No, those are the correct angles to use. It's probably an issue of having the right signs on your components. Could you show your calculations with those angles?

6. Mar 2, 2014

### kaalaniz

C_x=-150cos30=-129.9
C_y=-150sin30=-75
d_x=200cos60=100
d_y=200sin60=173.21

F_x=100-129.9+100+25
F_y=-300-75+173.21=-201.79

square root of 25 squared plus -201.79 squared=203.33 meters
tan^-1(-201.79/25)=-82.94 degrees

7. Mar 2, 2014

A_x=100
A_y=0
B_x=0
B_y=-300

8. Mar 2, 2014

### jackarms

Almost -- just two things. Your components for d look right in their values, but not in the signs. Look at the diagram -- what signs should x and y be? Also, where does the 25 come from in your calculation of F_x? I'm not following it.

9. Mar 2, 2014

### kaalaniz

sorry, i meant to type F_x = 25
I looked at the graph again.
Should D_x be negative and I keep D_y positive because it is going up?

10. Mar 2, 2014

### kaalaniz

yes! I changed the d values and got the right answer. thankyou

11. Mar 2, 2014

### kaalaniz

wait one more problem. How do i get the correct direction of the displacement? I did inverse tangent of -201.79/-129.9 and i got 57.23 degrees. This isn't what the book says.

12. Mar 2, 2014

### jackarms

Remember that inverse tangent always gives you an angle between -90 and 90 degrees, so angles outside this range get converted to be inside the range (since the tangent of any angle plus or minus 180 degrees gives you the same value as the tangent of the original angle.)

So for this case the angle should be in the third quadrant, so inverse tangent instead gives you an equivalent angle in the first quadrant. Just add 180 degrees to get back to the third quadrant and you should be good.

13. Mar 2, 2014

### kaalaniz

ok thankyou. I appreciate your help.