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 :approve:
 

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  • #2
Mister T
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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?
 
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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).
 
  • #4
jbriggs444
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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.
 
  • #5
Mister T
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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}##?
 
  • #6
Delta2
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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 $$R=\sqrt{X^2+Y^2},\Theta=\tan^{-1}\frac{Y}{X}$$
 
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haruspex
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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)##.
 
  • #8
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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.
 
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