Net Displacement: Calculate x,y & r,θ

[otirik]

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

 

[Herman Trivilino]

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?

 

[otirik]

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]

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.

 

[Herman Trivilino]

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]

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]

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 $[0,2\pi)$.

 

[Anachronist]

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.