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Relative Speed, Clocks and Circular Motion
by Total Gadha - Monday, 18 June 2007, 02:02 AM
  cat 2009 2010 xat 2009 relative speed clocks circular motionThe 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:


cat 2007 2008 xat 2008 relative speed clocks circular motion

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.

 

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Re: Relative Speed, Clocks and Circular Motion
by bishnu dev - Monday, 18 June 2007, 10:50 AM
 

Thanks T.G. for a wonderful post.

But can u post something for straight line relative speed problems. like when to people start from two opposite points on a straight line with different speeds . they meet for the first time at some distance and then for the 5th time at some other distance , then calculate the time taken for them to meet for the tenth time such kind of problems. it would be relly helpful.

Thanks

Re: Relative Speed, Clocks and Circular Motion
by Deep Thinker Gadha - Monday, 18 June 2007, 10:52 AM
 

Nice one...........

TG, You are not only teach topics but also you develope the line of thinking for those topics which is most crucial.

Thanx a lot!

 

-- Sandeeeeeeeeep

Re: Relative Speed, Clocks and Circular Motion
by ramesh akula - Monday, 18 June 2007, 01:18 PM
 

now all the myths were gone cool

thank u so much TG

Re: Relative Speed, Clocks and Circular Motion
by chinni kiran - Monday, 18 June 2007, 05:31 PM
 

Hi TG,

In circular motion case, when they were moving in opp direction, I think they meet in 17 diffrent places, am i right?thoughtful

i think they meet at V1-V2 diff places while in same direction and V1+V2 diff places in opp direction. is it right?

Re: Relative Speed, Clocks and Circular Motion
by KayBee Aar - Monday, 18 June 2007, 05:32 PM
 

Wonderful article.....Thanks again...

 

KayBEE

Re: Relative Speed, Clocks and Circular Motion
by roger dylan - Monday, 18 June 2007, 06:55 PM
 

TG SIR,

Thanks a lot for fulfilling your promise.

 

Regards,

Roger

Re: Relative Speed, Clocks and Circular Motion
by TG Team - Monday, 18 June 2007, 09:35 PM
  Hi TG
this is really an amazing article.Simply superb. Heartfelt gratitudes.
Re: Relative Speed, Clocks and Circular Motion
by Total Gadha - Tuesday, 19 June 2007, 12:53 AM
  Thank you all smile
Re: Relative Speed, Clocks and Circular Motion
by subh mukherjee - Tuesday, 19 June 2007, 03:42 PM
 

an awesome article by TG again.

thank u !

Re: Relative Speed, Clocks and Circular Motion
by Tempo # 911 - Wednesday, 20 June 2007, 04:12 PM
 

Good.smile

The minute and the hour hand of a watch meet every 65 minutes.
How much does the watch lose or gain time and by how much?

Re: Relative Speed, Clocks and Circular Motion
by Kunal Gupta - Wednesday, 20 June 2007, 04:23 PM
 

Waiting for Geometry lessons..

 

Thnks 4 a wonderful lesson..

Re: Relative Speed, Clocks and Circular Motion
by ramesh akula - Wednesday, 20 June 2007, 04:53 PM
 

yaaaaaaaaaaaa

waiting for Geometry lessions... smile

Re: Relative Speed, Clocks and Circular Motion
by Total Gadha - Thursday, 21 June 2007, 11:26 PM
  ok smile
Re: Relative Speed, Clocks and Circular Motion
by Saurabh Sharma - Tuesday, 26 June 2007, 03:20 PM
 

yes.. they meet 17 times..

we can get by adding their speeds..

Re: Relative Speed, Clocks and Circular Motion
by rhishabh garg - Wednesday, 4 July 2007, 12:56 PM
 

wonderful article...waiting 4 some more mindboggling articles........

gud work TG

Re: Relative Speed, Clocks and Circular Motion
by chinni kiran - Saturday, 7 July 2007, 08:47 AM
 

Hi TG,

just for confirmation....

In circular motion case, when they were moving in opp direction, I think they meet in 17 diffrent places, am i right?thoughtful

i think they meet at V1-V2 diff places while in same direction and V1+V2 diff places in opp direction. is it right?

My doubt here is if they(v1 and V2) have common factors, then should we take LCM or HCF....?? or directly should we addd...

Re: Relative Speed, Clocks and Circular Motion
by Total Gadha - Sunday, 8 July 2007, 04:23 AM
  Hi Chinni,

I did not think about it before but after seeing your question I decided to do some investigation on my own. I am sorry to say that I did not reach any conclusion. sad

Let's take two persons, A and B, running with a speed of 18 m/s and 30 m/s, respectively, on a circular track of length d.

Same Direction:
B catches A every d/12 s. The distances traveled by A and B in d/12 s are equal to 18 × (d/12) = 3d/2 (d + d/2) and 30 × (d/12) = 5d/2 (2d + d/2). Therefore, they meet halfway the first time. The second time, they will meet at the starting point. Therefore, they are meeting only at two different places. If the distance d is even, say 1000 m, they will meet only at the starting point every time.

Opposite Direction:
They meet every d/48 s. The distances traveled by A and B in d/48 s are equal to 18 × (d/48) = 3d/8 and 30 × (d/48) = 5d/8. Therefore, they meet at 8 places if d is not even. They meet at 4 places if d is a multiple of 2 but not 4. They meet at 2 places if d is a multiple of 4 but not 8. They meet only at the starting point if d is amultiple of 8.

Total Gadha
Re: Relative Speed, Clocks and Circular Motion
by Amishe 800 - Monday, 9 July 2007, 09:37 PM
 

when the two are moving in opp directions , you pointed out that they will meet every 1000/7 ("Thus A and B will meet after 1000/7 s")

Isn't it 1000/17 and not 1000/7 ....

Please correct me if I am wrong

Re: Relative Speed, Clocks and Circular Motion
by Total Gadha - Tuesday, 10 July 2007, 06:03 AM
  Thanks Amishe, corrected the typo smile
Re: Relative Speed, Clocks and Circular Motion
by clenched fist - Thursday, 12 July 2007, 11:58 AM
  Yet another great article from TG!

However there is a gap in my understanding. Could you please solve the following using relative speed funda.

At what time between 5 and 6 o'clock will the minute and hour hand:
(a) make an angle of 34 degrees with each other?
(b) are perpendicular to each other?
(c) are on the same straight line but facing opposite directions?

Cheers,
CF.
Re: Relative Speed, Clocks and Circular Motion
by clenched fist - Thursday, 12 July 2007, 01:11 PM
  Also,

A watch loses 1 second in 10 minutes was set right at 10pm. When the watch indicated 58 minutes past 7 o'clock, the next morning, the true time is?

Ans: 8am

This is how I went about it, but I'm not arriving at the answer. Where am I going wrong?

Total minutes = 9 hours x 60 minutes + 58 = 598 minutes. So the watch loses 598 / 10 = 59.8 seconds. So that should be approximately 7 o' clock and 59 minutes.

Cheers,
CF.
Re: Relative Speed, Clocks and Circular Motion
by clenched fist - Thursday, 12 July 2007, 02:32 PM
  I tried to solve At what time between 5 and 6 o'clock will the minute and hour hand:
(a) make an angle of 34 degrees with each other?

Here is how I went about it...
Let the hour hand travel 'x' degrees after 5 o'clock. So the time taken by the minute hand for travel = (150 + x + 34) / 6 and the time taken for the hour hand = 2x

=> 184 + x = 12x
=> 11x = 184
=> x = 184/11

Therefore to get ahead by 34 degrees, the minute hand has to travel 150 + 184 / 11 + 34 degrees = = 2208 / 11 degrees

Hence the time taken to get ahead = 4416 / 121 = 36 60 / 121 minutes.

Hence the answer is 5:36 60 / 121.

But the actual answer is 5:21 1/11 and again at 5:33 5/11.
Where am I going wrong?

Sorry for so many consecutive posts.

Cheers,
CF.
Re: Relative Speed, Clocks and Circular Motion
by DEEPAK KANSAL - Monday, 30 July 2007, 05:49 PM
 

a)   (150-34)/(11/2) = 232/11  and (150+34)/(11/2) = 368/11 after 5 pm

b)  (150-90)/(11/2) = 120/11  and (150+90)/(11/2) = 480/11 after  5 pm

c) (150+180)/(11/2) = 660/11 = 60 min after 5 pm i.e at 6 pm

 

Re: Relative Speed, Clocks and Circular Motion
by DEEPAK KANSAL - Monday, 30 July 2007, 05:50 PM
  yes u r right
Re: Relative Speed, Clocks and Circular Motion
by Bhaskar T - Tuesday, 7 August 2007, 01:32 PM
 

Hi TG,

  Your  contribution to help  CAT Aspirants like me is  marvelous . I am really very thankfull to you for  providing us a such a collection of problems, concepts etc.

Regds

Bhaskar

 

Re: Relative Speed, Clocks and Circular Motion
by Bhaskar T - Tuesday, 7 August 2007, 01:33 PM
 

Hi TG,

  Your  contribution to help  CAT Aspirants like me is  marvelous . I am really very thankfull to you for  providing us a such a collection of problems, concepts etc.

Regds

Bhaskar

 

Re: Relative Speed, Clocks and Circular Motion
by Sandeep Raina - Wednesday, 8 August 2007, 01:35 AM
  5. A train after travelling 50 km from A meets with an accident and proceeds at 4/5th of the former speed and reaches B, 45 min late. Had the accident happened 20 km further on , it would have arrived 12 min sooner. Find the orginal speed and the distance.
Re: Relative Speed, Clocks and Circular Motion
by Ranvijay Singh - Wednesday, 8 August 2007, 01:58 PM
  Hi Sandeep,

Please find below the solution:

Let the original speed of the train be S km/h

"Had the accident happened 20 km further on, it would have arrived 12 min sooner" means:
20/(4S/5) - 20/S = 12/60
=> 20/S (4/5 - 1) = 1/5
=> S = 25 km/h

i.e., the original speed, S = 25 km/h

Let the total distance be D km.
The train after traveling 50 km meets an accident and proceeds with 4/5th of the original speed (i.e., 20 km/h) and
reaches 45 minutes late. This means,

(D-50)/20 - (D-50)/25 = 3/4
=> (D-50)/5 (1/4 - 1/5) = 3/4
=> (D-50)/5 * (1/20) = 3/4
=> D-50 = 300/4
=> D = 125

i.e., the total distance, D = 125 km


Vijay
Re: Relative Speed, Clocks and Circular Motion
by Ranvijay Singh - Wednesday, 8 August 2007, 03:53 PM
  Hi CF,

Please find below the solution to your first question. The other three have already been solved by Deepak.

According to me such problems should be approached by taking the actual time in consideration and then computing the loss to finally deduce the time being shown on the watch not the other way roundsmile

Ok...coming to the question. I don't think that '8 AM' is the correct answer here. Since, the watch loses 1 sec in every 10 minutes. Therefore, it'll lose 6 * 10 = 60 seconds = 1 minute in 10 hours after 10:00 PM. Evidently, the watch will show 07:59 AM (not 7:58 AM) when the actual time is 08:00 AM.

For 7:58 AM on the watch, the actual time will be less than 08:00 AM. Let's check for 07:59 AM: total loss = 9 * 6 + (59/60) * 6 = 59.9 seconds, so the watch will show 07:59 - 59.9 seconds i.e., 07:58+ AM (little more than 07:58 AM). Try for 07:58:59 AM (HH:MM:SS time format): total loss = 9 * 6 + (58/60) * 6 + (59/600) = 54 + 5.8 + 0.09 = 59.89 seconds, so the time shown on watch will be 07:58:59 - 59.89 seconds = 07:57:59+ AM (little more than 07:57:59 AM).

Hence, the actual time will be somewhere between 07:58:59 AM and 07:59:00 AM for the time on the watch to be 07:58 AM.


Vijay
Re: Relative Speed, Clocks and Circular Motion
by lavika gupta - Sunday, 12 August 2007, 09:35 PM
  hi tg.thanks 4 the wonderful article.keep up the good work.
Re: Relative Speed, Clocks and Circular Motion
by garima jain - Tuesday, 18 September 2007, 09:28 PM
  please add boats and streams.
Re: Relative Speed, Clocks and Circular Motion
by parag kumar - Friday, 5 October 2007, 07:45 PM
 

Thanks TG

Not only for this article but all that you have posted on this site. With every article that I read, I want to read the next one as soon as possible. Infact for all that I have gained from your teachings I can't thank you enough.

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). The smallest integer with which this distance should be multiplied to yield a multilpe of the track length is the number of different meeting places.

For example, if track length is 800 metres and two guys are running with 20 m/s and 7 m/s in same direction. The first meeting point will be  800*7/(20-7)= 5600/13 metres. Now see that every subsequent point will be 5600/13 m ahead of the previous one. 5600/13 has to be multiplied with 13 to get a multiple of 800. After 13 rounds ( of 7 m/s guy), the whole meeting places would be repeated all over again. Hence, 13 different places.

If the speeds had been 20 m/s and 12 m/s then the first meeting point would be at  1200 metres from the starting point. Now 1200 in simply 2 rounds can become a factor of 800 (track length). Hence there would be only two meeting places (despite the difference in speed being 4).

I know that the quality of my explanation is not comparable to that of TG, but I guess that's what you will have to bear with for now.

Have a great CAT#

 

Re: Relative Speed, Clocks and Circular Motion
by Total Gadha - Saturday, 6 October 2007, 09:17 AM
  Hi Parag,

I learnt something today. Thank you. smile

I think your explanation was very good. You have a good command over English also. smile

Let me chew on this concept for a while. I would like to explore it further.

Total Gadha

Re: Relative Speed, Clocks and Circular Motion
by lakshmi iyer - Tuesday, 16 October 2007, 02:39 PM
 

approve heyy TG....looooooove ur way of puttin things...vv lucid....thanx...

thoughtful...nw fr my question...wrt to parag's postin...we sw dat whn da faster person catches up wd da slower one he makes a round extra. so in case of da example of 1000m n 7 n 10 m/s speedes, deyll meet after 1000/3 secs...dats is da slower person wdve covered a distance of 7000/3 n on multiplyin wd da integer 3...he wdve been at da startin point n da fster person simply wdve coverd one round extra rt?? n so der wd b 3 meetings bfore dey meet up at startin point???

n if say da speeds are 12 n 6 ms/s n da dis is 800 mts n dey run in same directions,dell meet after 800/6 secs...so slower person wd cover 800/6 mts..nw multiplyin by 3 wd give 400..a multiple of 800..so dey wd meet 3 times bfore dey meet agn at da startin point rt?????   lemme knw please...thanx...

Re: Relative Speed, Clocks and Circular Motion
by naga kiran kosuru - Thursday, 25 October 2007, 11:44 AM
 

@lakshmi iyer

12m/s , 6m/s n the distance to be covered is 800m.

now, they wil meet once after 800m/12-6(m/s)=800/6 sec.

in this time , dist cov by slower guy is (800/6)*6=800m and the faster guy is 1600m.....since , after the first meeting itself , the slower person coverd 800m which is a multiple of thr trach length , so there is only one ditinct meeting point , i.e., the staring point.

Re: Relative Speed, Clocks and Circular Motion
by naga kiran kosuru - Thursday, 25 October 2007, 11:45 AM
 

@lakshmi iyer

12m/s , 6m/s n the distance to be covered is 800m.

now, they wil meet once after 800m/12-6(m/s)=800/6 sec.

in this time , dist cov by slower guy is (800/6)*6=800m and the faster guy is 1600m.....since , after the first meeting itself , the slower person coverd 800m which is a multiple of thr trach length , so there is only one ditinct meeting point , i.e., the staring point.

Re: Relative Speed, Clocks and Circular Motion
by lakshmi iyer - Thursday, 25 October 2007, 04:23 PM
  hey...thanx kiran...dis concept shd b handy in td probs....big grin
Re: Relative Speed, Clocks and Circular Motion
by Parikshit Bhattacharjee - Monday, 5 November 2007, 08:18 PM
  TG, this is regarding ur post where u describe 2 runners with speeds 18 and 30 running on a track of length d. You have worked out that they meet halfway the first time at a dist d/2 from the starting point. So there are 2 distinct meeting points- one at the starting point, and the other at a distance of d/2 from it. This is irrespective of the value of d. So please explain this statement of yours -
"If the distance d is even, say 1000 m, they will meet only at the starting point every time."

They will meet first at a dist of 500 m from the starting pt, and for the 2nd time, they will meet at the starting pt- So 2 distinct points !!!!

Also, in the case when the runners run in opp directions, these constraints are not very clear. Please explain-
"distances travelled are 3d/8 and 5d/8..they meet at 8 places if d is not even. They meet at 4 places if d is a multiple of 2 but not 4. They meet at 2 places if d is a multiple of 4 but not 8. They meet only at the starting point if d is amultiple of 8. "
Re: Relative Speed, Clocks and Circular Motion
by manish garg - Wednesday, 14 November 2007, 08:18 PM
  there's a direct formula to calculate the time at which a  particular angle will be subtended b/w any time

for the first time :

6p-(angle+p/2)=angle to be formed

for the 2nd time (in certain cases)

(angle+p/2)-6p=angle to be formed

e.g=

as in the above question
what time b/w 5-6 will there be an angle of 34 degree

we know at 5 the angle subtended(b/w hour hand and min hand)
is 360/12 * 5= 150 degree

6p-(150+p/2)=34
11p/2=184            or p=368/11

p=33.4

hence at 5:33:4/11 there will be an angle of 34 degree
similarly at 5:17:11/13 there will be an angle of 34 degree

Re: Relative Speed, Clocks and Circular Motion
by Anuprita Malik - Friday, 16 November 2007, 07:32 AM
 

Hii TG..

Gr8 work... Thanks soo much..  

Re: Relative Speed, Clocks and Circular Motion
by sunny mongia - Sunday, 30 March 2008, 12:15 AM
 

jhakaaaz

Re: Relative Speed, Clocks and Circular Motion
by amiable gal - Monday, 31 March 2008, 10:11 PM
 

hi manish

Can you please put some light over the formula you gave. i didn't get "the time at which a  particular angle will be subtended b/w any time".What is this 34? Kindly explain. thanks in advancesmile

Re: Relative Speed, Clocks and Circular Motion
by Vipul Patki - Thursday, 17 April 2008, 01:10 AM
 

Hi,

Another way of putting Manish's eqn would be:

x = ABS((11/2)m - 30H)

Where x = Angle between hour and minute hand, m = minute part of the time, H = hour part of time.

So we have: 34 = (11/2)m - 150

and -34 = (11/2)m - 150

Also, I guess the second answer should be 21 minutes (approx) rather than 17 min.

Hope this helps...

 

Vipul

Re: Relative Speed, Clocks and Circular Motion
by Ajit kumar - Wednesday, 23 April 2008, 06:21 PM
 

GOOD Manish, But I think the relative speed approach is simpler..

But thanks anyway.. keep posting

Re: Relative Speed, Clocks and Circular Motion
by akhil syal - Thursday, 24 April 2008, 02:35 PM
  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
by Harsh Chandra - Friday, 25 April 2008, 09:09 AM
  Please add Mixtures and solutions and boats and streams too. I really need it/.
RACE
by billi cat - Monday, 28 April 2008, 12:08 PM
 

In a Race....A wins over B by 200metres; B wins over C by 200metres; By how much distance does A win over C? Length of the track is 1000m

Pls explain the logic, if possible more than one approaches

Re: RACE
by TG Team - Monday, 28 April 2008, 06:32 PM
 

Hi Billi CAT smile

Just understand the logic.

A wins over B by 200 meters in a 1000 meters race means when A travels a distance of 1000 meters then B only manages to cover a distance of 800 meters.

=> A/B = 1000/800

Similarly B wins over C by 200 meters in a 1000 meters race means when B travels a distance of 1000 meters then C only manages to cover a distance of 800 meters.

=> B/C = 1000/800.

Now multiply both equations

=> (A/B)*(B/C) = A/C = 1000/640.

That means A wins over C by 360 meters in a 1000 meters race.smile

Re: RACE
by Lord Aragorn - Monday, 28 April 2008, 08:23 PM
  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
by Ashaley Gadha - Thursday, 15 May 2008, 10:34 AM
  cool work TG.
Re: Relative Speed, Clocks and Circular Motion
by Navneet Jha - Thursday, 29 May 2008, 02:03 PM
 

Hi Clenched fist:

Answers are:

a) At 5 o clock, diff between min and hr hand is 150 degree. So for a gap of 34 degree it has to cover 150-34=116 degree => 116/ (11/2) = 21 1/11 mins

hence time is 5:21 1/11

b) Same for 90 degree, it need to cover 150-90=60 degree

=60 / (11/2) = 10 10/11 mins => 5:10 10/11 hrs

c) for straight line i.e., 180 degree, its need 150+180 degree = 330 degree

=330/ (11/2) = 60mins

Re: Relative Speed, Clocks and Circular Motion
by sayandeep b - Thursday, 29 May 2008, 03:00 PM
 
image

The ratio of speeds of Appu and Gappu is

a. 3: 1
b. 1 + sqrt(2): 1
c. 2: 1
d. 4 - sqrt(2): 1

 

it is a Q from the Quiz on TG please can any body solve it for me?

Re: Relative Speed, Clocks and Circular Motion
by ramakant pandey - Saturday, 31 May 2008, 01:20 AM
 

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 a-b 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
by ramakant pandey - Saturday, 31 May 2008, 01:38 AM
 

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
by Amit Malik - Saturday, 31 May 2008, 10:06 AM
  @ 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  = (a-g) m /s

So time taken to catch up Gappu (assuming him to be at rest) = c/(a-g) secs.

Now the distance travelled by Gappu when he travels with speed g m/s =
                                                            g { c/(a-g)} meter

Therefore the distance to be covered to complete the Circle = c - { gc / (a-g)}                                                                              = c {(a- 2g) / a-g}     meter

And distance to be traveled by Appu  = Distance travelled by Gappu  =
                                                                                    g { c/(a-g)} meter

Time taken by them will be equal.
Therefore,   [g { c/(a-g)}] / a =  [c {(a- 2g) / a-g} ] / g

Solving it we get a/g = 3

Re: Relative Speed, Clocks and Circular Motion
by madhur arora - Tuesday, 3 June 2008, 02:09 AM
  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
by avijit mohapatra - Wednesday, 4 June 2008, 12:50 PM
  Is the answer "b"???
Re: Relative Speed, Clocks and Circular Motion
by sriram s - Saturday, 7 June 2008, 07:33 AM
 

@ 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 l-s.

Therefore s1/s2=s/l-s;

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
by rohit gangwar - Thursday, 19 June 2008, 01:12 PM
  dear sir,
escalator problem is still
a puzzle for me .... plz help
me with tht ...
Re: Relative Speed, Clocks and Circular Motion
by Ramkrishna Roy - Thursday, 19 June 2008, 01:42 PM
  thanks a lot!!!!!!!!!
Re: Relative Speed, Clocks and Circular Motion
by abhinav tripathi - Saturday, 28 June 2008, 11:30 AM
  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
by Tushar Tushar - Tuesday, 16 September 2008, 05:01 PM
 

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
by jaswinder singh - Wednesday, 1 October 2008, 03:49 PM
 

Hi,

TG you simply rocks ,its 5-6 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
by ruchika jain - Sunday, 5 October 2008, 08:08 PM
  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
by Neelesh Sethi - Sunday, 26 October 2008, 10:22 PM
  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
by shantanu rangaswami - Sunday, 9 November 2008, 08:04 PM
  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
by shantanu rangaswami - Sunday, 9 November 2008, 08:23 PM
  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
by Rajarshi Guha - Sunday, 12 July 2009, 10:30 AM
  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
by Rajesh Kumar - Sunday, 12 July 2009, 04:55 PM
 

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.smile

 Understood?

-- Rajesh Mittal

Re: Relative Speed, Clocks and Circular Motion
by William Wallace - Sunday, 12 July 2009, 06:25 PM
  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.

(d-5x)/(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
by Rajarshi Guha - Sunday, 12 July 2009, 07:05 PM
  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 vis-a-vis the hour hand..
Re: Relative Speed, Clocks and Circular Motion
by Rajesh Kumar - Sunday, 12 July 2009, 11:08 PM
 

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
by Rajarshi Guha - Monday, 13 July 2009, 03:44 PM
  Okk..now I get it..One of them is considered to be static.Thanks man!!
Re: Relative Speed, Clocks and Circular Motion
by Prateek Gupta - Friday, 14 August 2009, 05:52 PM
  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
by abhishek tripathi - Tuesday, 18 August 2009, 02:09 AM
  thx tg for dis awesome article..its of great help 4 every aspirant preapring 4 cat.
Re: Relative Speed, Clocks and Circular Motion
by Nitin Kumar - Wednesday, 19 August 2009, 12:18 AM
 

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
by skyskiers 44 - Wednesday, 19 August 2009, 01:26 AM
  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
by skyskiers 44 - Wednesday, 19 August 2009, 02:03 AM
  the answer should be a) 2...
this is because  they meet for the first time in t=100/(2-x-y) time..
and they meet together at the starting point at
L.C.M[50,100/x , 100/y]=3t =3[100/(2-x-y)]
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
by sunakshi bajpai - Wednesday, 19 August 2009, 08:21 PM
  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
by Nitin Kumar - Thursday, 20 August 2009, 04:47 PM
 

First time A and B will meet after 100/(2-x) seconds.

First time A and C will meet after 100/(2-y) seconds.

First time A, B and C will meet after LCM[100/(2-x) and 100/(2-y)]seconds.

t = 100/(2-x)(2-y)

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
by saurabh aggarwal - Friday, 4 September 2009, 08:39 PM
 

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
by rohit dwivedi - Monday, 7 September 2009, 11:00 PM
  Hi Nitin ,
I got this question in one of the papers .
Its answer is 3 sad
I am still comprehending , how it could be ....

Re: Relative Speed, Clocks and Circular Motion
by Rajesh Kumar - Friday, 25 September 2009, 02:27 PM
  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
by Rajesh Kumar - Friday, 25 September 2009, 02:29 PM
  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
by rupasree raj - Tuesday, 27 October 2009, 09:00 AM
  hi TG sir,
  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


Re: Relative Speed, Clocks and Circular Motion
by mayuri joshi - Friday, 30 October 2009, 12:17 AM
  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
by mayuri joshi - Friday, 30 October 2009, 12:31 AM
  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
by rupasree raj - Saturday, 31 October 2009, 08:14 PM
  hi mayuri....
 thank you very much.
Re: Relative Speed, Clocks and Circular Motion
by karan khetan - Friday, 6 November 2009, 07:09 PM
  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
by ROHIT K - Friday, 6 November 2009, 09:07 PM
  Hi Karan,

I think the answer is 22

1-25 even nos contribute 22 twos and 2 fives

26-200 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.smile

Rohit
Re: Relative Speed, Clocks and Circular Motion
by kumar ajit - Sunday, 30 May 2010, 09:59 PM
  thanks TG
can u give some tools to solve problem on gain and lose of clock etc
Re: Relative Speed, Clocks and Circular Motion
by stg jindal - Thursday, 29 July 2010, 11:00 PM
 

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
by nitish mittal - Saturday, 31 July 2010, 09:28 AM
  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
by sharad joshi - Monday, 2 August 2010, 11:52 PM
  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
by kumar sourav - Monday, 25 October 2010, 12:23 AM
  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
by Total Gadha - Monday, 25 October 2010, 09:57 AM
  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
by Gautam Kamath - Sunday, 7 November 2010, 12:45 PM
  Thanks a ton TG for clearing the concepts related to clock problems.
Gr8 work...Keep it up...
Re: Relative Speed, Clocks and Circular Motion
by sai deshmukh - Monday, 15 November 2010, 12:34 AM
  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... sad
Re: Relative Speed, Clocks and Circular Motion
by pranjal singh - Monday, 15 November 2010, 01:30 AM
  yes its correct
Relative Speed, Clocks and Circular Motion
by Jyothsna V - Friday, 10 June 2011, 10:24 PM
  Awesome article!! smile
Re: Relative Speed, Clocks and Circular Motion
by jigyasa tiwari - Saturday, 11 June 2011, 09:18 PM
  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... smile
Re: Relative Speed, Clocks and Circular Motion
by Jyothsna V - Monday, 13 June 2011, 08:23 PM
  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
by nw its ma turn to tame d cat .. - Tuesday, 14 June 2011, 09:42 PM
  is d ans 19 nd 20 hrs???
Re: Relative Speed, Clocks and Circular Motion
by anoop kumar pandey - Thursday, 16 June 2011, 08:50 AM
  no hcf or lcm shud not b taken.....they shud be added directly
Re: Relative Speed, Clocks and Circular Motion
by yatin hans - Tuesday, 1 November 2011, 04:43 PM
  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
by TG Team - Tuesday, 1 November 2011, 04:51 PM
 

Hi Yatin smile

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. smile

Kamal Lohia

Re: Relative Speed, Clocks and Circular Motion
by Soumya Sharma - Wednesday, 15 August 2012, 09:39 PM
  This really helped me solve relative speed questions in a jiffy, appreciate your help! big grin
Re: Relative Speed, Clocks and Circular Motion
by Anuj Asthana - Monday, 20 August 2012, 05:47 PM
  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 smallsmile

Build Concepts although there is not much time left !!!!

So I am really spellbound by TG

Re: Relative Speed, Clocks and Circular Motion
by Raksha K - Friday, 14 September 2012, 08:17 PM
  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
by Rajat Soni - Saturday, 22 September 2012, 03:19 PM
  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
by Raksha K - Saturday, 27 October 2012, 06:41 PM
  Thank you so much... smile 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
by Ajay Dhameliya - Thursday, 23 May 2013, 12:30 PM
  Great solution !!! smile smile 
Re: Relative Speed, Clocks and Circular Motion
by Nikhil P - Wednesday, 23 October 2013, 06:28 PM
  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
by dhruvika solanki - Wednesday, 5 November 2014, 08:26 AM
  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? smile