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mathslover

  • 2 years ago

A dog chasing the cat who is running along a striaght line at constant speed u . The dog moves with constant speed v, always heading towards the cat. Initially i.e. at t = 0 , the velocity of the dog and the cat are perpendicular and then initial perpendicular distance between them is 1. The dog catches the cat at ?

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  1. mathslover
    • 2 years ago
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    @Callisto

  2. AravindG
    • 2 years ago
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    This is one of the classic questions on this topic .But I have to leav now .can definitely answer tomorrow is that OK @mathslover ?

  3. abb0t
    • 2 years ago
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    If they are perpendicular, that meas they intresect at a 90º angle. the distance between them (i think hypotenusre) = 1

  4. abb0t
    • 2 years ago
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    How did I end up in physics section Lol.

  5. shubhamsrg
    • 2 years ago
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    |dw:1359311516476:dw| Let time taken be t, then (v cosx) t = 1 ...(1) u = vsinx ..(2) Just eliminate x to solve for t

  6. shubhamsrg
    • 2 years ago
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    Oh wait, am missing on the fact that x is a variable.

  7. shubhamsrg
    • 2 years ago
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    Getting too late here, I will get back to it later.

  8. UnkleRhaukus
    • 2 years ago
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    is \(u>v,\quad u=v,\quad \text{or}\quad u<v\) ?

  9. shubhamsrg
    • 2 years ago
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    u<v (since dog is always heading towards cat) , I guess.

  10. JamesJ
    • 2 years ago
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    I am curious to see Aravind's solution. It's not hard to write down the equations of motion here. But solving them is not so obvious.

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  11. AravindG
    • 2 years ago
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    @mathslover do you have final answer ? I wanted to check before replying

  12. JamesJ
    • 2 years ago
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    Two major objections 1. The first one is the answer doesn't make sense. Suppose the velocities u and v are equal. Then it is clear the dog never reaches the cat. But according to the formula you have ended up with, the dog catches the cat in finite time of \[ t = \frac{\sqrt{v^2 + u^2}}{v^2} = \frac{\sqrt{2}v}{v^2} = \frac{\sqrt{2}}{v} \] 2. The second one is you haven't explained your reasoning very well at all and there appear to be errors in the calculation. For instance, why does the angle theta between their velocity vectors--or the y component of it--satisfy cos theta = v/sqrt(v^2 + u^2)? Or why does (v - u) cos theta = (v - u) . v / sqrt(v^2 + u^2) imply something (their relative velocity in the y direction?) equals v^2/sqrt(v^2 + u^2)? And even if that were true, just because the relative displacement of two moving objects is <0,1>, by what kinematic equation is it the case that their relative velocity in the y direction can be directly deduced from that initial relative displacement, given that it is immediately clear that their relative displacement changes over time? ***** So I am not satisfied we have the solution.

  13. abb0t
    • 2 years ago
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    yoloswag

  14. JamesJ
    • 2 years ago
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    But it is clear that the dog is NOT moving in a straight line. The dog is constantly changing direction.

  15. harsh314
    • 2 years ago
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    thanks @jamesJ i got it the time period is \[\frac{ v }{ {v ^{2}-u ^{2}} }\]

  16. mathslover
    • 2 years ago
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    I got it now , Thanks a lot every one...

  17. harsh314
    • 2 years ago
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    and once again thanks to @mathslover for giving such a nice question keep it up!!

  18. mathslover
    • 2 years ago
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    Thanks @harsh314 Just hope that I get a doubt soon in such a problem :)

  19. harsh314
    • 2 years ago
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    may god bless you with such capabilites that you never get stuck in any question but for the sake of enrichment of our knowledge'thanks

  20. JamesJ
    • 2 years ago
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    Wait. We still don't have a good solution to this problem. Harsh has written down a solution. But where's the proof?

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