# Net Displacement

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Summary: Figuring net displacement given the angles and distances of three vectors.

You travel 4.00 miles East, followed by 5.00 miles at 50.0° N of E, followed by 6.00 miles at 20.0° S of W. What is your net displacement? Give the component form (x and y) and the polar form (r and theta).

I understand this will form a triangle based on the directions and distances I just don't understand how to answer in coordinates. I think it has something to do with total displacement based on the x-axis, and then based on the y-axis. I just don't know where to start. Any help is appreciated thanks

Mister T
Gold Member
I understand this will form a triangle based on the directions and distances
Can you show us your work on that? In other words, explain how you came to that conclusion?

Can you show us your work on that? In other words, explain how you came to that conclusion?
Given north is straight up on a graph (working in 2 dimensions) the first part describes a vector (A) starting at the origin and extending 4 units to the right, then vector (B) starting at (4,0) and extending six units to the right at an upward angle of 50 degrees (respective tot the x-axis), finally vector (C) starting where B ended and extending to the left at a negative 20 degree angle. I'm not sure if this actually makes a triangle because I forget how to check, however if it is the remaining angle is 30 degrees (because 180-50 would give the interior angle at the base of vector B).

jbriggs444
Homework Helper
I'm not sure if this actually makes a triangle because I forget how to check
That is a problem. Because this exercise is about how you check.

How about getting out that sheet of graph paper and actually drawing the lines.

Mister T
Gold Member
Given north is straight up on a graph (working in 2 dimensions) the first part describes a vector (A) starting at the origin and extending 4 units to the right,

So we would say that the tail of ##\vec{A}## is at (0,0) and the head of ##\vec{A}## is at (4,0).

then vector (B) starting at (4,0) and extending six units to the right at an upward angle of 50 degrees (respective tot the x-axis)

The tail of ##\vec{B}## is at (4,0). Where is the head of ##\vec{B}##?

Delta2
Homework Helper
Gold Member
I think this can be solved by converting each vector which is given in polar notation ##(r_i,\theta_i)## into cartesian ##x_i## and ##y_i## components $$(x_i,y_i)=(r_i\cos\theta_i,r_i\sin\theta_i)$$ and then adding all the ##x_i## together to get ##X## and all the ##y_i## together to get ##Y##.

Caution is required to carefully determine ##\theta_i##'s from the descriptiongs given. For example ##20^o## south of west translates to ##\theta_3=200^{o}##.

Then converting back to polar notation with $$R=\sqrt{X^2+Y^2},\Theta=\tan^{-1}\frac{Y}{X}$$

haruspex
Homework Helper
Gold Member
2020 Award
I think this can be solved by converting each vector which is given in polar notation ##(r_i,\theta_i)## into cartesian ##x_i## and ##y_i## components $$(x_i,y_i)=(r_i\cos\theta_i,r_i\sin\theta_i)$$ and then adding all the ##x_i## together to get ##X## and all the ##y_i## together to get ##Y##.

Caution is required to carefully determine ##\theta_i##'s from the descriptiongs given. For example ##20^o## south of west translates to ##\theta_3=200^{o}##.

Then converting back to polar notation with $$R=\sqrt{X^2+Y^2},\Theta=\tan^{-1}\frac{Y}{X}$$
Just need to be careful using ##\tan^{-1}##. It has two solutions in the range ##[0,2\pi)##.

Delta2
Anachronist
Gold Member
You travel 4.00 miles East, followed by 5.00 miles at 50.0° N of E, followed by 6.00 miles at 20.0° S of W. What is your net displacement?
Strictly speaking, there isn't sufficient information to answer the question. Your net displacement depends on your starting coordinate, because we live on a globe.

Think about it: If you started that trip 1 mile north of the south pole, traveling 6 miles east would take you in a circle around the south pole 1.27 times. But if you started on the equator, traveling 6 miles east would simply be 6 miles east.

Bystander and haruspex