Relative Speed, Clocks and Circular Motion  
The concept of relative
speed is not a new or extra concept that CAT 2009 or 2010 aspirants need to
learn. It is still the basic time, speed and distance formula applied to
distance between two moving bodies. In a simple case of the distance formula, a
body traveling with a speed of 50 km/h is reducing the gap between its starting
point and the finish point by 50 km every hour. In the relative speed case of
the distance formula two moving bodies, traveling at a relative speed of 50
km/h towards/away from each other, are reducing/increasing the gap between them
by 50 km every hour. Let me explain this through an example: I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this in the CBT Club this week.

Re: Relative Speed, Clocks and Circular Motion  
now all the myths were gone thank u so much TG 
Re: Relative Speed, Clocks and Circular Motion  
Wonderful article.....Thanks again...
KayBEE 
Re: Relative Speed, Clocks and Circular Motion  
TG SIR, Thanks a lot for fulfilling your promise.
Regards, Roger 
Re: Relative Speed, Clocks and Circular Motion  
Hi TG this is really an amazing article.Simply superb. Heartfelt gratitudes. 
Re: Relative Speed, Clocks and Circular Motion  
Thank you all 
Re: Relative Speed, Clocks and Circular Motion  
an awesome article by TG again. thank u ! 
Re: Relative Speed, Clocks and Circular Motion  
Good. The minute and the hour hand of a watch meet every 65 minutes. 
Re: Relative Speed, Clocks and Circular Motion  
Waiting for Geometry lessons..
Thnks 4 a wonderful lesson.. 
Re: Relative Speed, Clocks and Circular Motion  
yaaaaaaaaaaaa waiting for Geometry lessions... 
Re: Relative Speed, Clocks and Circular Motion  
ok 
Re: Relative Speed, Clocks and Circular Motion  
yes.. they meet 17 times.. we can get by adding their speeds.. 
Re: Relative Speed, Clocks and Circular Motion  
wonderful article...waiting 4 some more mindboggling articles........ gud work TG 
Re: Relative Speed, Clocks and Circular Motion  
Thanks Amishe, corrected the typo 
Re: Relative Speed, Clocks and Circular Motion  
yes u r right 
Re: Relative Speed, Clocks and Circular Motion  
hi tg.thanks 4 the wonderful article.keep up the good work. 
Re: Relative Speed, Clocks and Circular Motion  
please add boats and streams. 
Re: Relative Speed, Clocks and Circular Motion  
hey...thanx kiran...dis concept shd b handy in td probs.... 
Re: Relative Speed, Clocks and Circular Motion  
Hii TG.. Gr8 work... Thanks soo much.. 
Re: Relative Speed, Clocks and Circular Motion  
jhakaaaz

Re: Relative Speed, Clocks and Circular Motion  
GOOD Manish, But I think the relative speed approach is simpler.. But thanks anyway.. keep posting 
Re: Relative Speed, Clocks and Circular Motion  
hi T.G , thanks for the post . could u put some material related to linear motion . and also some material on races and trackes. thanks Akhil 
Re: Relative Speed, Clocks and Circular Motion  
Please add Mixtures and solutions and boats and streams too. I really need it/. 
Re: RACE  
Hi TG Sir, As the linear TSD, the circular TSD too was awesome. I would like you to explain the concept of meeting points of two runners. 
Re: Relative Speed, Clocks and Circular Motion  
cool work TG. 
Re: Relative Speed, Clocks and Circular Motion  
Dear TG Thanx for the wonderful leson , just wanna ask like if bodies moving in a circle in same direction has speed ratios a : b , would they meet at ab points on the circumference. Eg if speeds are in ratios 9:4 , they would meet at 5 different points ? and what happens when they move in different directions...Plz help

Re: Relative Speed, Clocks and Circular Motion  
Dear Sayan Is the answer is D ie 2.61:1 .Wel wht i ve done if we interpret the problem physically like clock options a & c can be ruled out. Now two options are left one is 2.6: 1 and 2.41:1. if we take the phhysicaly interpetaion of 2.5:1 we see that gappu crosse the finsihing point before appu , so 2.41: 1 can be ruled out baecause by that gappu would reach more sooner. So the correct answer is D. Please rply me whether I am right

Re: Relative Speed, Clocks and Circular Motion  
@ Sayandeep b Let the Circumference of the circle = c meters Let the speed of Appu = a m/s and Gappu be = g m/s Relative speed of Appu and Gappu = (ag) m /s So time taken to catch up Gappu (assuming him to be at rest) = c/(ag) secs. Now the distance travelled by Gappu when he travels with speed g m/s = g { c/(ag)} meter Therefore the distance to be covered to complete the Circle = c  { gc / (ag)} = c {(a 2g) / ag} meter And distance to be traveled by Appu = Distance travelled by Gappu = g { c/(ag)} meter Time taken by them will be equal. Therefore, [g { c/(ag)}] / a = [c {(a 2g) / ag} ] / g Solving it we get a/g = 3 
Re: Relative Speed, Clocks and Circular Motion  
Deepak why dont u take into consideration the distance travelled by hour hand which is also travelled by minute hand 
Re: Relative Speed, Clocks and Circular Motion  
Is the answer "b"??? 
Re: Relative Speed, Clocks and Circular Motion  
@ above yes,i also got the ans as b : sqrt(2)+1:1 The explaination is Let l be the length of track and s be the distance frm starting pt where they meet. Since they both travel for the same time,the distances travelld by both are prop to speeds.If s1,s2 r their respective speeds,then s1/s2=(l+s)/s; Also in the next meeting,appu travels s backwards while gappu travels ls. Therefore s1/s2=s/ls; Applying componendo,dividendo and solvin,we get l=sqrt(2)*s; substitutin it in any of the eqn,we get the ans. 
Re: Relative Speed, Clocks and Circular Motion  
dear sir, escalator problem is still a puzzle for me .... plz help me with tht ... 
Re: Relative Speed, Clocks and Circular Motion  
thanks a lot!!!!!!!!! 
Re: Relative Speed, Clocks and Circular Motion  
THANKS TG .............!!! you r 2 gud.......................i used to ecape from clocks n circular motion quetions now i will chase dem.......... 
Re: Relative Speed, Clocks and Circular Motion  
It was a very good article.. Having problems in questions involving clocks gaining and loosing time .... Please help TG.

Re: Relative Speed, Clocks and Circular Motion  
Hi, TG you simply rocks ,its 56 days i gone through this site and I rate it as the best site I have ever found ..................the material are simply superb.............Thanks a lot . You are definately doing a social service, if any how i am able to get into IIMS its 60% share goes to you(simply you deserve this ) Jaswinder Singh 
quant phobia  
hii TG i want 2 ask u something as i m preparing for cat08 n i m not able 2 cope up with mainly quant section n day by day its getting tougher for me . now there r only 5 weeks left for cat08 n i m really feeling under pressure due 2 quant section , plz help me how should i proceed further as this is affecting my whole performance. thanks hope u ll soon reply me. 
Re: quant phobia  
About the number of different meeting points along any closed track, it depends on the distance of the first meeting point from the starting point (actually travelled by the slower person). What if they are moving in opposite direction will it still depend on distance travelled by slower person or distance travelled by faster person. 
Re: Relative Speed, Clocks and Circular Motion  
hey,,,,can just tel me how they meet 17 times in whole time span before they meet at starting poin......thanx in advance 
hi,TG  
I m new 2 this group nd m not able to make myself comfortable on topic circular motion .......i m not able 2 figure out points of their meet ....hope u can help me out ......waitin for reply. 
Re: Relative Speed, Clocks and Circular Motion  
Hiii TG, I have a question pertaining to the time interval between two successive meetings of the hour and the minute hands of a clock. As the minute hand travels a distance of 360 degrees,the hour hand would also have moved from its original position by 30 degrees.So effectively for the minute hand to catch up with the hour hand,it now has to cover a distance of 360+30=390 degress at a rate of 11/2 degress/min.In that case the answer would be different.. Am I correct.Kindly clear my doubt. 
Re: Relative Speed, Clocks and Circular Motion  
hello Rajarshi dost 11/2 deg/min is the relative speed means the difference b/t the speeds of hour and minute hand as both are moving in difference direction. Understood?  Rajesh Mittal 
Re: Relative Speed, Clocks and Circular Motion  
Hi Friends, Please try to solve this question and pour your thoughts on it. There are two cities A and B .Two trains start from A and B towards B and A respectively(on parallel tracks) at 1am and 6am respectively.At 10:20 am they meet each other.They both reach their destination at the same time.At what time will they reach their destination? My approach: Let d be the distance between stations and x and y be the speeds of trains. (d5x)/(x+y) = 13/3 x*(t+5)=y*t=d got stuck after this.Can someone throw some light on it. 
Re: Relative Speed, Clocks and Circular Motion  
Precisely Rajesh.But that is not what my question is.I am not talking about the speeds of the two hands.I am talking about the distance covered by the minute hand visavis the hour hand.. 
Re: Relative Speed, Clocks and Circular Motion  
Rajarshi, if u look closely at the initial examples of CARs again then you will come to know that after calculating relative speed of 2 cars we consider one of the 2 cars as static(means it is not moving) .Same is here in case of clock problem when we have calculated relative speed of Minute hands w.r.t Hour Hand, we are considering Hour hand as static. Hope this will help you. Rajesh Mittal 
Re: Relative Speed, Clocks and Circular Motion  
Okk..now I get it..One of them is considered to be static.Thanks man!! 
Re: Relative Speed, Clocks and Circular Motion  
Ans : option (b) Consider the distance from the starting point when A and G first meet and equate the time taken by each of them to reach the starting point henceforth. 
Re: Relative Speed, Clocks and Circular Motion  
thx tg for dis awesome article..its of great help 4 every aspirant preapring 4 cat. 
Re: Relative Speed, Clocks and Circular Motion  
Can anyone plz help me on solving below prob? Three people A, B & C start moving around a circular track of 100m simultaneously with speeds of 2, x & ym/s respectively in clockwise direction. they meet first time after t seconds and meet their starting point after 3t seconds. Then which of the following can never be the value of x? a. 2 b. 3 c. 5 d. 6 e. 8 
Re: Relative Speed, Clocks and Circular Motion  
the watch gains 5/12 mins every hour.. if someone could tell me whether it is rt or wrong..?!!?? 
Re: Relative Speed, Clocks and Circular Motion  
the answer should be a) 2... this is because they meet for the first time in t=100/(2xy) time.. and they meet together at the starting point at L.C.M[50,100/x , 100/y]=3t =3[100/(2xy)] if x takes a value of 2 then the lcm will be negative which is not possible... tell me if im wrong.. 
Re: Relative Speed, Clocks and Circular Motion  
there are no explainations given to the answers of the quizzes we attempt.if u cn provide with some explainations then it wud b a gr8 help.it really bcms a prblm if we dnt gt an answer to a numerical and then thr is no solution also provided with it. 
Re: Relative Speed, Clocks and Circular Motion  
First time A and B will meet after 100/(2x) seconds. First time A and C will meet after 100/(2y) seconds. First time A, B and C will meet after LCM[100/(2x) and 100/(2y)]seconds. t = 100/(2x)(2y) I think from here only it's clear that x can never attain value as 2. Let me know if I am right!!!! 
Re: Relative Speed, Clocks and Circular Motion  
thank u so much....i almost thought that i would leave circular motion concept for cat prep.....but after reading this i got the concept in no time.... plz do give the concept of 3 persons running along the circular track.. 
Re: Relative Speed, Clocks and Circular Motion  
Hi Nitin , I got this question in one of the papers . Its answer is 3 I am still comprehending , how it could be .... 
Re: Relative Speed, Clocks and Circular Motion  
Hi TG For the problems related to find the number of different points A and B meet on a circular track, i thought of one approach. Suppose speed of A = V1, speed of B = V2 and the circular track is of length D. Let T be the time after which A and B meet at the starting point. T = LCM of D/V1 and D/V2. i.e. LCM of the time taken to complete the circle by A and B. In time T, A will make T/(D/V1)= ( T x V1 )/ D = r1 (say) rounds of circle, and B will make T/(D/V2) = ( T x V2 )/D = r2 (say) rounds of circle. Case 1: A and B running in the same direction. Number of different points at which they will meet = r1 – r2 Case 2: A and B are running in the opposite direction. Number of different points at which they will meet = r1 + r2 Please tell me whether they are correct or not. Thanks Rajesh 
Re: Relative Speed, Clocks and Circular Motion  
Hi TG I am a new member of this community. Your posts are really awesome. But whenever there was any mistake in your post and you corrected them, they are not reflected in the original post. For example in the last problem the “A and B will meet every 1000/17 s.”has not been corrected to 1000/7 s. I saw in the reply that you have corrected the type mistake. But this change has not been reflected in the main post (available to me). Please tell me where to find the updated post. Thanks Rajesh 
Re: Relative Speed, Clocks and Circular Motion  
sir pls help me wid dis: Qs) 3 persons A,B,C start runnin simultaneously on 3 concentric circular tracks from 3 collinear points P,Q,R respectively which r collinear wid de centre and r on de same side of de centre. de speed of A,B,C r 5m/s,9m/s,8m/s respectively. de lengths of de tracks on which A, B, C r running r 400m,600m,800m respectively. 1) If A and B run in de clockwise direction and C in de anti clockwise direction, after how much time’ll de positions of A,B,C be collinear wid de centre and on de same side of de centre, for de first time after dey start? a) 200 sec b) 400 sec c) 600 sec d) 800 sec
2) 2) If A runs in de clockwise direction and C in de anti clockwise direction, after hw much time ‘ll de positions of A and C be collinear wid de centre (and on de same side), for forth time after dey start? a) 1600/9 sec b) 160 sec c) 1600/11 sec d) 200 sec  hi TG sir,
Re: Relative Speed, Clocks and Circular Motion  
hi rupasree this is how i think u do the first problem  1. consider a and b first. let n be the number of rounds taken by a. then b gains one complete round over a when  600(n+1)/9 = 400(n)/5 (n+1)/n = 6/5 i.e. a takes 5 and b takes 6 rounds. when a takes 5 rounds time = 400*5/5 = 400 secs similarly when b takes 6 rounds time = 600*6/9 = 400 secs 2.now consider b and c. then b gains one complete round over c when  800(n)/8 = 600(n+1)/9 i.e. b takes 3 and c takes 2 rounds. (n+1)/n = 3/2 when b takes 3 rounds time = 600*3/9 = 200 secs. when c takes 2 rounds time = 800*2/8 = 200 secs. b gains one round over a every 400 secs and over c every 200 secs. hence they all meet for the first time at 400 secs which is the LCM of the above two times. ans .. 400 secs. P.S please note that b meets a for the first time at 400 secs since they are moving in same directions. however at 200 secs b only gains one round over c but it is not the first time that they have met since they are moving in opposite directions. 
Re: Relative Speed, Clocks and Circular Motion  
rupasree, for the next problem  when a and c travel in opposite directions. Consider if both were travelling in opposite directions on the same track. Then we would have calculated the first time they meet as  divide the circle into parts equal to the sum of the ratio of their speeds. i.e 5+8 = 13 parts. hence when a moves 5 parts clockwise b moves 8 anticlockwise and they meet for the first time. 400/13 * 5 would give u distance traveled by a. and (400/13 * 5)/5 secs would give u time when they meet for the first time. considering the same logic but this time both individuals are on concentric tracks. so the above logic can be modified as  when c moves 8 divisions on its track its actual effect as mapped onto the track of c is = 8 * 800m/400m ie. 4 divisions. hence when a moves 5 divisions, c moves 4 (on the mapped track). dividing track 400m into 5+4=9 parts we get  (400/9 * 5)/5 secs. this is time that c and a meet first time. i.e. 400/9 secs. when the meet the 4th time  400/9 * 4 = 1600/9 secs. hence ans > 1600/9 secs. 
Re: Relative Speed, Clocks and Circular Motion  
hi mayuri.... thank you very much. 
Re: Relative Speed, Clocks and Circular Motion  
u have to select even numbers from 1 to 25 and odd numbers from 26 to 200 and if u take a product of them how many zeros will be there in the product...can someone answer this question...??? 
Re: Relative Speed, Clocks and Circular Motion  
Hi Karan, I think the answer is 22 125 even nos contribute 22 twos and 2 fives 26200 odd nos contribute 0 twos and 21 fives. So total while taking product we will have 22 twos and 23 fives Number of zeroes = 22 Hope this helps. Rohit 
Re: Relative Speed, Clocks and Circular Motion  
thanks TG can u give some tools to solve problem on gain and lose of clock etc 
Re: Relative Speed, Clocks and Circular Motion  
awesome article TG sir! it would be great if we get some brush up on 'clocks gaining and loosing time'. 
Re: Relative Speed, Clocks and Circular Motion  
hi sir i disagree sir you see when A 18 m/s, B 30 m/s... and track lenth =1000 m will meet 1st at starting pt. for the 1st time at 1000/6 s and they meet for the 1st time when B increases distance b/w them by 1000 m i.e. at 1000/12 s they'll b at half the track only. and they will meet at two different points on the track. Same goes with the other example when they move in opposite direction. I think m clear,correct me if m wrong. Regards Nitish Mittal 
Relative Speed, Clocks and Circular Motion  
hello friends this formula i found somwhere hope u ll find it useful To find the angle between the hands of a clock.... if "H" is the hours and "M" is the mins...then angle can be directly obtained by ANGLE = 30H  (11/2)M 
Re: Relative Speed, Clocks and Circular Motion  
Page and Plant are running on a track AB of length 10 metres. They start running simultaneously from the ends A and B respectively. The moment they reach either of the ends, they turn around and continue running. Page and Plant run with constant speeds of 2m/s and 5m/s respectively. How far from A (in metres) are they, when they meet for the 23rd time? (a) 0 (b) 10 (c)40/7(d)60/7 
Re: Relative Speed, Clocks and Circular Motion  
Speed ratio is 2: 5 so when the first takes 2 rounds the second takes 5 rounds. They are together in this case. And when the first takes 4 rounds, the second takes 10 rounds. They are again at opposite ends in this case. Calculate the number of meets when first has taken 4 rounds and the second has taken 10 rounds. Proceed from there. 
Re: Relative Speed, Clocks and Circular Motion  
Thanks a ton TG for clearing the concepts related to clock problems. Gr8 work...Keep it up... 
Re: Relative Speed, Clocks and Circular Motion  
Two bugs are climbing a slippery wall. At time = 0, both are at the bottom of the wall. Bug A climbs at the rate of 3 inches a minute. Bug B climbs at the rate of 4 inches a minute. Due to the slippery wall, however, Bug A slips back 1 inch for every 2 inches climbed and bug B 1.5 inches for every 2 inches. Moreover, bug A takes a rest for 1 minute after every 2 minutes and Bug B, a rest for 1 minute after every 3 minutes. Assume that slipping occurs continuously when climbing. If Bug B does not have rest periods, how many times do the bugs meet in the first 10 minutes. m stuck up wid dis thing!!! plz Tg help me... 
Re: Relative Speed, Clocks and Circular Motion  
yes its correct 
Relative Speed, Clocks and Circular Motion  
Awesome article!! 
Re: Relative Speed, Clocks and Circular Motion  
sir i have a doubt regarding dis..u said if d iis 100 dey ll meet nly at starting pt..bt sir..here a will travel d+d/2..so he'll cover 1500 mt nd b will cover 2d+d/2 dat means 2500..means even den dey r meeting at mid pt..crrect me if i am wrng... 
Re: Relative Speed, Clocks and Circular Motion  
TG, Sir please explain this question Two trains A and B start from stations X and Y towards each other. B leaves station Y after half n hr after A leaves station X. Two hrs after train A has started the distance between trains A and B is 19/30 th of the distance between the stations X and Y. How much time would it take each train A and B to cover the distance to X to Y if train A reaches half an hr later to its destination compared to B. I tried it using the conventional method n its too lengthy.. Please explain the short cut.. 
Re: Relative Speed, Clocks and Circular Motion  
is d ans 19 nd 20 hrs??? 
Re: Relative Speed, Clocks and Circular Motion  
no hcf or lcm shud not b taken.....they shud be added directly 
Re: Relative Speed, Clocks and Circular Motion  
At what time between 5.30 and 6 will the hands of a clock be at right angles? 
Re: Relative Speed, Clocks and Circular Motion  
Hi Yatin Read the article carefully and you will be able to answer it easily. The required time is 90/5.5 = 180/11 minutes earlier than 6 i.e. 5:43(7/11) exactly. Kamal Lohia 
Re: Relative Speed, Clocks and Circular Motion  
This really helped me solve relative speed questions in a jiffy, appreciate your help! 
Re: Relative Speed, Clocks and Circular Motion  
Dear TG SIR The Only thing that takes you ahead in Competitive exams is strong hold on concepts You can read lots of Encyclopedias in Maths or Verbal Ability but even a small problem of 10th class can trick us. That is the beauty of Concepts and Total Gadha has made me love concepts. I really adore the LCM thing it comes to my aid always when I solve some problems be it Relative Speed , Number System or Circular motion. It is such a wonderful thing yet infintesmally small Build Concepts although there is not much time left !!!! So I am really spellbound by TG 
Re: Relative Speed, Clocks and Circular Motion  
Hi... Can any one please help me in solving this problem... Rahul and Vinay run a 9km race on a circular race of length 900m. They complete one round in 9 seconds and 15 seconds respectively.After what time from the start will the faster person overtake the slower person for the last time? 1)90 sec 2) 150sec 3) 48sec 4)114 sec 
Re: Relative Speed, Clocks and Circular Motion  
SPEED OF RAHUL=90m/s SPEED OF VINAY=60m/s TIME TAKEN BY RAHUL=9s TIME TAKEN BY VINAY=15s L.C.M(9,15)=45s 1ST MEETING TIME=45s last meeting time=90s 
Re: Relative Speed, Clocks and Circular Motion  
Thank you so much... I understood till the process of getting the 1st meeting time but I didn't get how the last meeting time is obtained.Is it twice the time taken for the 1st meeting.. why is it so... Please help me out 
Re: Relative Speed, Clocks and Circular Motion  
Great solution !!! 
Re: Relative Speed, Clocks and Circular Motion  
Let d be 90m (90/12) * x = 30 ==> same direction (90/48) * x = 30 ==> opp direction where (x  1) is the number of times they meet before meeting for the first time at the starting point @ 30s Is the approach right? 
Re: Relative Speed, Clocks and Circular Motion  
thanks for the post, TG sir. It was helpful. I had a question regarding the sum where 2 people move towards each other and we have to find where they are 1 min before collision. In such sums, the initial distance between them doesn't matter. But, what if the gap reduced per unit time turns out to be greater than the initial distance between them? Will it still not matter? And how should we approach that sum? 
Re: Relative Speed, Clocks and Circular Motion  
ya thats true.but i haven't been able to identify how to find those 17 points.basically if speeds are in the ratio a:b then total disticnt points at which they meet is a+b.can u help me how to find those points 
Re: Relative Speed, Clocks and Circular Motion  
Hi everyone,Hope it will be helpful I found that if speeds are in the ratio x:y. Then there are xy distinct meeting points will be there 