A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 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
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 ?

This Question is Closed

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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0If they are perpendicular, that meas they intresect at a 90º angle. the distance between them (i think hypotenusre) = 1

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0How did I end up in physics section Lol.

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

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

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

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

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

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

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

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0Two 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.

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

harsh314
 one year ago
Best ResponseYou've already chosen the best response.1thanks @jamesJ i got it the time period is \[\frac{ v }{ {v ^{2}u ^{2}} }\]

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

harsh314
 one year ago
Best ResponseYou've already chosen the best response.1and once again thanks to @mathslover for giving such a nice question keep it up!!

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

harsh314
 one year ago
Best ResponseYou've already chosen the best response.1may 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
 one year ago
Best ResponseYou've already chosen the best response.0Wait. We still don't have a good solution to this problem. Harsh has written down a solution. But where's the proof?
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.