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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 ?
 one year ago
 one year 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 ?
 one year ago
 one year ago

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AravindGBest ResponseYou've already chosen the best response.0
This is one of the classic questions on this topic .But I have to leav now .can definitely answer tomorrow is that OK @mathslover ?
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
If they are perpendicular, that meas they intresect at a 90º angle. the distance between them (i think hypotenusre) = 1
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
How did I end up in physics section Lol.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
dw:1359311516476:dw Let time taken be t, then (v cosx) t = 1 ...(1) u = vsinx ..(2) Just eliminate x to solve for t
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
Oh wait, am missing on the fact that x is a variable.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
Getting too late here, I will get back to it later.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
is \(u>v,\quad u=v,\quad \text{or}\quad u<v\) ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
u<v (since dog is always heading towards cat) , I guess.
 one year ago

JamesJBest ResponseYou've already chosen the best response.0
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.
 one year ago

AravindGBest ResponseYou've already chosen the best response.0
@mathslover do you have final answer ? I wanted to check before replying
 one year ago

JamesJBest ResponseYou've already chosen the best response.0
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 vectorsor the y component of itsatisfy 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.
 one year ago

JamesJBest ResponseYou've already chosen the best response.0
But it is clear that the dog is NOT moving in a straight line. The dog is constantly changing direction.
 one year ago

harsh314Best ResponseYou've already chosen the best response.1
thanks @jamesJ i got it the time period is \[\frac{ v }{ {v ^{2}u ^{2}} }\]
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
I got it now , Thanks a lot every one...
 one year ago

harsh314Best ResponseYou've already chosen the best response.1
and once again thanks to @mathslover for giving such a nice question keep it up!!
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
Thanks @harsh314 Just hope that I get a doubt soon in such a problem :)
 one year ago

harsh314Best ResponseYou've already chosen the best response.1
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
 one year ago

JamesJBest ResponseYou've already chosen the best response.0
Wait. We still don't have a good solution to this problem. Harsh has written down a solution. But where's the proof?
 one year ago
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