- **Introductory Physics Homework**
(*http://www.physicsforums.com/forumdisplay.php?f=153*)

- - **High School relative Velocity question**
(*http://www.physicsforums.com/showthread.php?t=241726*)

Relative Velocity & VectorsI have the below question for Summer Physics (basically just taking physics over the summer instead of during the year)
Quote:
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 |

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 gotta study for my exam right now, I'm just waiting for someone to reply to my post :P |

Re: Relative Velocity & VectorsQuote:
magnitudes of the components are correct, but which quadrant does this vector point into? What should the signs of the components be?Quote:
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 byv_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...) Quote:
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! |

All times are GMT -5. The time now is 01:35 PM. |

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.

© 2014 Physics Forums