High School relative Velocity question

[Nicole13]

Relative Velocity & Vectors

I have the below question for Summer Physics (basically just taking physics over the summer instead of during the year)

Mee Eun and Lokesh start from the same point at the same time and travel in different directions. Mee Eun moves with a velocity of 15 m/s
at 300 N of E. Lokesh moves with a velocity of 20 m/s at 500 S of W.
a. Calculate the components of the two velocities.
b. Determine:
i. Mee Eun’s velocity relative to Lokesh.
ii. Lokesh’s velocity relative to Mee Eun.
c. After ten minutes pass:
i. Calculate the displacements of Mee Eun and Lokesh relative to the origin.
ii. Where is Mee Eun as seen by Lokesh?

I completed part A by doing Y-component for mee eun is 15sin30 and x-component is 15cos30. For Lokesh I got 20sin50 for the y and 20cos50 for the x

I got stumped at relative velocity, because there were angles I wasnt sure about how to approach this. Am I supposed to just add/subtract the velocities?

And I was stuck on C too (this was the first question, I'm not feeling good about this at all). Am I supposed to find the hypotenuse of the triangle (like conect the lines) or do something else. And what's the difference because Ci and Cii in terms of what is being asked.

Wow, I have a lot of questions and I'm so grateful to anyone who can help me

 

[ital_dj]

Sorry, I won't be able to help you as I'm studying for my physics exam tomorrow (grade 11)... but just an honest tip it's better to follow the guidelines when posting messages (the original question, relative formulas, and attempted solution)... the mods tend to like that better!

Sorry again, I got to study for my exam right now, I'm just waiting for someone to reply to my post :P

 

[dynamicsolo]

I completed part A by doing Y-component for mee eun is 15sin30 and x-component is 15cos30. For Lokesh I got 20sin50 for the y and 20cos50 for the x

OK, you have worked out Mee Eun's components correctly, since that velocity vector points into the first quadrant (positive-x, positive-y). But be careful with Lokesh's velocity vector: the magnitudes of the components are correct, but which quadrant does this vector point into? What should the signs of the components be?

I got stumped at relative velocity, because there were angles I wasnt sure about how to approach this. Am I supposed to just add/subtract the velocities?

Relative velocities are found by subtracting one vector from another. If we know the velocity vectors for A and B, as defined using a given origin (call these v_A and v_B, then the velocity of A relative to B will be given by

v_A - v_B ,

and the velocity of B relative to A will be

v_B - v_A .

But these are vectors, so it is not very convenient to subtract them when the angle between them doesn't happen to be 0º or 180º . So instead we use the expedient of subtracting the components of the vectors, and then use the "resultant" vector components to find the magnitude and direction of the "resultant vector".

So, for the velocity of A relative to B, you need to find

(v_A)_x - (v_B)_x and (v_A)_y - (v_B)_y ,

then use the appropriate trig relations to find the magnitude and direction of the vector difference, v_A - v_B .

Incidentally, how will the velocity for B relative to A compare with what you found for A relative to B? (If you notice this, you can save yourself half the work in this part of the problem...)

And I was stuck on C too (this was the first question, I'm not feeling good about this at all). Am I supposed to find the hypotenuse of the triangle (like conect the lines) or do something else. And what's the difference because Ci and Cii in terms of what is being asked.

In part (c-i), a picture may help. Both parties are starting from the same spot and moving along their respective velocity vectors. For each, the displacement will be the position each traveler reaches in ten minutes (600 seconds) as measured from the origin minus the position they started at. If you follow each line for the ten minutes, you will have a segment connecting the starting point and the point each is at ten minutes later. Does the length and direction of each segment have anything to do with where that starting point is? How far will each traveler have gone in ten minutes and which way?

For (c-ii), you will be finding a position of one traveler relative to the other. Look at the components of the displacement each traveler makes in ten minutes. These displacements will be vectors r_A and r_B. So, just as we did for the relative velocities, you can find the position of A relative to B by subtracting

(r_A)_x - (r_B)_x and (r_A)_y - (r_B)_y

to get the components, then put them together to get the direction and magnitude of the relative position.

But wait! The velocities are constant, so the components of the velocities are, too -- and so are the relative velocities. So if you know the velocity of A relative to B (and it is constant), you can get the position of A relative to B by just multiplying the velocity of A relative to B by the time interval:

(r_A) - (r_B) = { v_A - v_B } · (delta_t) .

Welcome to the wonderful world of vector algebra!