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Relative velocity of accelerating and non accelerating particles

  1. Oct 21, 2014 #1
    Q. Two particles start simultaneously from the same point and move along two straight lines, one (particle A) with uniform velocity v and other (particle B) with a uniform acceleration a. If ##\alpha## is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given by:

    (a) ##vsin\alpha/a## (b) ##vcos\alpha/a## (c) ##vtan\alpha/a## (d) ##vcot\alpha//a##

    My text gives the answer as (b) ##vcos\alpha/a## . I think they reached at the answer as follows:

    At an instant (the instant when the relative velocity is minimum; lets say) V=at, therefore ##vcos\alpha-V=0## where 0 is the shortest relative velocity (assumed). Therefore ##vcos\alpha=at## or ##t=vcos\alpha/a##.

    But this answer seems unconvincing. It is not necessary that for the relative velocity to be minimum the component of velocity of the 'A' (which attains a constant velocity) should be equal to the velocity of the 'B' (which has the constant acceleration) at a particular instant. Can anyone provide the precise solution?
     
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  3. Oct 21, 2014 #2

    gneill

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    Hello Hijaz Aslam. In future be sure to use the posting template for homework questions. It's a Physics Forum rule.

    Can you write an expression for the relative speed versus time (or more simply, the square of the relative speed)? If so, how does one usually go about minimizing a function?
     
  4. Oct 21, 2014 #3
    gneill - Sorry about that. I didn't know. My Bad.

    I am afraid, I didn't get what you said. Can you be more precise please?
     
  5. Oct 21, 2014 #4

    gneill

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    Suppose particle A's line lies along the x-axis and particle B's line is at angle α to the x-axis. Both particles start out at the origin. Can you write expressions for the x and y components of the velocities of both particles?

    If you can, then can you find the components of the relative velocity vector?
     
  6. Oct 21, 2014 #5
    Yes, I think at the instant when the relative velocity is least we have the velocity of particle 'B' as : ##(atcos\alpha)\hat{i}+(atsin\alpha)\hat{j}##

    And the relative velocity of 'B' with reference to 'A' is given by ##v_{BA}=v_{B}-v_{A}## so we have: ##v_{BA}=(atcos\alpha-v)\hat{i}+(atsin\alpha)\hat{j}##

    And the magnitude ##|v_{BA}|=a^{2}t^{2}-2avtcos\alpha+v^{2}##. Solving the quadratic to find time gives: ##t=(cos\alpha\pm isin\alpha)/av##. NOW WHAT?

    Am I correct so far?
     
  7. Oct 21, 2014 #6

    gneill

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    Actually what you have is the square of the magnitude of the relative velocity. Which is perfectly fine! When one is minimized then so it the other.

    Rather than solve for the time, what you want to do is find when this magnitude is minimized, right? How do you go about finding where a function has a minimum?
     
  8. Oct 21, 2014 #7
    Of course we differentiate it (silly me, I forgot the minimum part) , then equate it to zero ie : ##2a^2t+2av-2a^2tcos\alpha-2avcos\alpha=0##
    => ##at+v-atcos\alpha-vcos\alpha=0## , simplifying we get ##at+v=0## or ##t=-v/a##. And Wallah!! That's a wrong answer :( Am I wrong somewhere?

    (BTW I didn't consider the denominator of the differential coefficient, because of course it gets transported into the right hand side which is zero. And isn't the acceleration constant? So we don't have to consider the D.C of 'a'. Do we? ).
     
    Last edited: Oct 21, 2014
  9. Oct 21, 2014 #8

    gneill

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    I'm not following how you arrive at your derivative with respect to time of the squared magnitude.

    You have ##| V_{BA} |^2 = a^2 t^2 - 2 a v cos(\alpha) t + v^2## to start with. What's the derivative w.r.t. time?
     
  10. Oct 21, 2014 #9

    haruspex

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    I don't see how you got that from differentiating your expression for the square of the relative velocity.
    With regard to the textbook explanation, the thing to notice is that the relative velocity perpendicular to B's path is constant, so it reduces to minimising the component parallel to B's path.
     
  11. Oct 21, 2014 #10
    gneill - I have use ##|V_{BA}|=\sqrt(a^2t^2-2avtcos\alpha+v^2)## itself. See differentiating ##|V_{BA}|## w.r.t we have:

    ##(2a^2t-2avcos\alpha-2a(dv/dt)tcos\alpha+2v(dv/dt))/2\sqrt(a^2t^2-2avtcos\alpha+v^2)####(2a^2t-2avcos\alpha-2a^2tcos\alpha+2av)/2\sqrt(a^2t^2-2avtcos\alpha+v^2)## . To find the minimum of this value (which is ##|V_{BA}|##), we equate it to zero, so we have: ##2a^2t-2avcos\alpha-2a^2tcos\alpha+2av## . I have skipped the above steps. (NOTE: I've differentiated the above presuming that the acceleration is constant. Is that assumption correct?) (I've also used ##dv/dt=a##)
     
  12. Oct 21, 2014 #11

    gneill

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    You've made your life unnecessarily complicated by including the square root :smile: Minimize f(x)2 and you minimize f(x) too.

    Note that a and v are not related to each other in this problem, and that both are constants, not variables. Certainly ##dv/dt## is not a.
     
  13. Oct 21, 2014 #12
    Oh my.. blimey.....I've never made such a blunder in my learning career, I don't know what's got into my head lately. Of course v is a constant and...OMG how did I even think about it: Of course yes so you have ##2a^2t-2avcos\alpha=0## that is ##t=v/acos\alpha##. Thank you very much gneill, my brain's all shut down. Sorry for wasting your time with such a silly mistake. Thank you gneill. (I really appreciate your method of squeezing out the answer from the questioner itself, it made me understand the whole concept myself. And I like that TARDIS D.P..I am a Whovian :D ) Thanks again.


    And one more thing (I'm sorry) : Is that solution I gave in the beginning which takes ##vcos\alpha=at## and then proving that ##t=vcos\alpha/a## a reliable one? Or is it just coincidentally correct?
     
  14. Oct 21, 2014 #13
    haruspex - Do you mean that we can arrive at the answer by taking : ##vcos\alpha=at## and hence ##t=vcos\alpha/a##? Can you please explain what you've mentioned?
     
  15. Oct 21, 2014 #14

    gneill

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    You're welcome. I'm a big fan of The Doctor too.
    Offhand I can't think of a good obvious argument for assuming ##vcos(\alpha)=at## means the relative velocity is minimized, other than it follows from the correct solution :smile:
     
  16. Oct 22, 2014 #15

    haruspex

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    As I wrote in post #9, it's valid because the rel velocity perpendicular to B's path is constant, so it's just a matter of minimising the component parallel to B's path.
    In fact, the problem can be generalised in a couple of ways. If the variable speed is w=w(t) then the minimum rel velocity occurs when ##vcos(\alpha)=w(t)##.
    Neither is it specific to velocity and acceleration. The same equations apply if you drop the d/dt down a level, i.e. change velocities to displacements and accelerations to velocities (in which context, the answer is rather more obvious).
     
  17. Oct 22, 2014 #16
    lAzJ47S.png
    haruspex - Do you mean the relative velocity perpendicular to B or A? Please consider the above diagram.
     
  18. Oct 22, 2014 #17

    haruspex

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    Perpendicular to B's path, as I wrote. What is the component of the relative velocity in that direction? Is it not constant?
     
  19. Oct 22, 2014 #18
    Alright, Yes as B doesn't have any components towards its perpendicular, Component of relative velocity in its perpendicular direction is constant. Alright so the only velocity in consideration is the one parallel to B. Got it!

    So we can't say that ##atcos\alpha=v## gives the minimum relative velocity because, the perpendicular relative velocity of 'A' is not constant. Isn't it?
     
  20. Oct 22, 2014 #19
    gneill - harupex above has got an explanation for the method I've mention in the beginning. It's far simpler. Your method made the concepts clear for me though. :) Thanks
     
  21. Oct 22, 2014 #20

    haruspex

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    That's right, the situation is not symmetric.
     
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