tag:blogger.com,1999:blog-4115025577315673827.post8846889354349797319..comments2020-05-25T13:34:36.365+05:30Comments on CSE Blog - quant, math, computer science puzzles: Four Ants - HC VermaPratik Poddarhttp://www.blogger.com/profile/11577606981573330954noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-4115025577315673827.post-80370602429304445642012-09-29T17:30:35.141+05:302012-09-29T17:30:35.141+05:30in the reference frame of ant 3 ,ant 1's traje...in the reference frame of ant 3 ,ant 1's trajectory is straight line with speed 2v.<br />(because velocity of ant1 is perpendicular to ant2 velocty and velocity of ant2 is perpendicular to ant 3 velocty so ant 1 velocity and ant 3 velocity are in opposite direction. )<br />hence the answer is square root(2)/2.<br />i don't know why the answers are not matching..please do sort out this issue . Anonymoushttps://www.blogger.com/profile/10276651805450711325noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-44531814412795283972012-09-29T17:25:41.388+05:302012-09-29T17:25:41.388+05:30by same arguments relative velocity of ant 1 with ...by same arguments relative velocity of ant 1 with respect to ant 3 is 2v and pointing towards 3 . <br />in fact trajectory of ant 1 in reference frame of ant 3 is straight line with speed 2v..<br />so time taken to meet is (square root 2)/2v. hence answer is (square root 2)/2..i don't know which one is right .Anonymoushttps://www.blogger.com/profile/10276651805450711325noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-66560054083070527712009-11-08T12:35:40.713+05:302009-11-08T12:35:40.713+05:30i remember having solved the HC Verma problem myse...i remember having solved the HC Verma problem myself, and perhaps would have managed this one too (considering its simpler) but feeling too lazy to exercise my brains, i saw the solution. nice.suyashhttps://www.blogger.com/profile/09632474759324558691noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-83575750001518221512009-11-06T20:28:34.645+05:302009-11-06T20:28:34.645+05:30The answer is 1. :)
We want the distance traveled...The answer is 1. :)<br /><br />We want the distance traveled by each ant. Solution goes as follows.<br /><br />The ants keep constant speed throughout their motion. Their trajectory is extremely complex. So, geometrical solution will not help. We calculate the time taken for the ants to meet. Now, we see that ant 1 s always moving towards ant 2 and ant 2 towards ant 3. They initially form a square. By symmetry, they always form a square. So, ant 1 is always moving perpendicular to the line joining ant 2 and ant 3. This means, at all times, the velocity of ant 1 is perpendicular to that of ant 2. So, the relative velocity of ant 1 towards ant 2 is always v. So, the time taken to make this distance zero is 1/v.<br /><br />So, the distance traveled total is 1 m. :)<br /><br />Hope that helps!!Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-45604608389815536012009-11-06T18:41:15.787+05:302009-11-06T18:41:15.787+05:30what is the solution? is it really 1!!!
the answe...what is the solution? is it really 1!!!<br /><br />the answer I got is :<br />1- distance travelled by each ant is Pi/4<br />2- Straight line distance from A to B is 1/2Unknownhttps://www.blogger.com/profile/15654071495644996470noreply@blogger.com