mathslover
  • mathslover
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 ?
Physics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

mathslover
  • mathslover
AravindG
  • AravindG
This is one of the classic questions on this topic .But I have to leav now .can definitely answer tomorrow is that OK @mathslover ?
abb0t
  • abb0t
If they are perpendicular, that meas they intresect at a 90ยบ angle. the distance between them (i think hypotenusre) = 1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

abb0t
  • abb0t
How did I end up in physics section Lol.
shubhamsrg
  • shubhamsrg
|dw:1359311516476:dw| Let time taken be t, then (v cosx) t = 1 ...(1) u = vsinx ..(2) Just eliminate x to solve for t
shubhamsrg
  • shubhamsrg
Oh wait, am missing on the fact that x is a variable.
shubhamsrg
  • shubhamsrg
Getting too late here, I will get back to it later.
UnkleRhaukus
  • UnkleRhaukus
is \(u>v,\quad u=v,\quad \text{or}\quad u
shubhamsrg
  • shubhamsrg
u
JamesJ
  • JamesJ
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.
1 Attachment
AravindG
  • AravindG
@mathslover do you have final answer ? I wanted to check before replying
JamesJ
  • JamesJ
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.
abb0t
  • abb0t
yoloswag
JamesJ
  • JamesJ
But it is clear that the dog is NOT moving in a straight line. The dog is constantly changing direction.
anonymous
  • anonymous
thanks @jamesJ i got it the time period is \[\frac{ v }{ {v ^{2}-u ^{2}} }\]
mathslover
  • mathslover
I got it now , Thanks a lot every one...
anonymous
  • anonymous
and once again thanks to @mathslover for giving such a nice question keep it up!!
mathslover
  • mathslover
Thanks @harsh314 Just hope that I get a doubt soon in such a problem :)
anonymous
  • anonymous
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
JamesJ
  • JamesJ
Wait. We still don't have a good solution to this problem. Harsh has written down a solution. But where's the proof?

Looking for something else?

Not the answer you are looking for? Search for more explanations.