## Relative speed on circular path

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### Relative speed on circular path

by akpareek » Wed Feb 13, 2013 7:59 am
In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?

A) 10 B) 20 C) 30 D) 40 E) 50

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by buzzdeepak » Wed Feb 13, 2013 8:16 am
akpareek...
Are they traveling in the same direction or opposite direction?

If opposite direction:
Sb = 4m/s
Therefore, B will cover 40 m in 10 sec.
Remaining distance = 100-40 = 60 m.

Sa = 2 m/s
Since, they are traveling in opposite direction, we can add their speeds
Sab = 4+2 = 6 m/s

Tab = 60/6 = 10 secs
Therefore, time since B started = 10 + 10 = 20 sec...which is option B in your ans choice.

If they are traveling in the same direction, it gets a bit tricky since A's speed is lower than B's speed..which would mean B will complete one circle and then catch up with A.

Hope this helps...

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by Brent@GMATPrepNow » Wed Feb 13, 2013 8:17 am
akpareek wrote:In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?

A) 10 B) 20 C) 30 D) 40 E) 50
Are these people traveling in opposite directions?
I will assume that they are.

Here's one way to visualize this.
Rather than have them on a circular track with circumference 100 meters, picture them on a straight track that's 100 meters long (all we're doing here is straightening the track).
Runner A is at one end and Runner B is at the other end.

First: Runner B runs for 10 seconds at 4 m/s. In this time, he/she travels 40 meters.
At this point, the runners are only 60 meters apart.

Second: Runner A starts.
Important: The runners are now closing the gap between each other at a rate of 6 m/s (2 m/s + 4 m/s = 6 m/s)

So, at a rate of 6 m/s, we can see that it will take 10 seconds to close the 60 meter gap between them.

This 10 seconds represents the time AFTER runner A starts.
Since Runner B had already be running for 10 seconds, the total time is 20 seconds

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Feb 13, 2013 8:46 am
akpareek wrote:In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?

A) 10 B) 20 C) 30 D) 40 E) 50
RUNNING IN THE SAME DIRECTION

If the two runners are going in the same direction, then we can see that Runner A (the slower runner) will never catch up to Runner B. Instead, the first time they meet will be when Runner B laps Runner A. When this occurs, Runner B will have run 100 meters more than Runner A.

(Distance B travels) = (Distance A travels) + 100

Let t = number of seconds that elapse after Runner B begins running.

Distance Runner B travels = (time)(rate)
= (t)(4)
= 4t

Note: Since Runner A starts running 10 seconds after Runner B, Runner A's time spent running will equal t - 10

Distance Runner A travels = (time)(rate)
= (t-10)(2)
= 2t - 20

So, we get...
(Distance B travels) = (Distance A travels) + 100
4t = (2t - 20) + 100
4t = 2t + 80
2t = 80
t = 40

Cheers,
Brent
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by sushantsahaji » Wed May 04, 2016 2:26 am
Brent@GMATPrepNow wrote:
akpareek wrote:In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?

A) 10 B) 20 C) 30 D) 40 E) 50
RUNNING IN THE SAME DIRECTION

If the two runners are going in the same direction, then we can see that Runner A (the slower runner) will never catch up to Runner B. Instead, the first time they meet will be when Runner B laps Runner A. When this occurs, Runner B will have run 100 meters more than Runner A.

(Distance B travels) = (Distance A travels) + 100

Let t = number of seconds that elapse after Runner B begins running.

Distance Runner B travels = (time)(rate)
= (t)(4)
= 4t

Note: Since Runner A starts running 10 seconds after Runner B, Runner A's time spent running will equal t - 10

Distance Runner A travels = (time)(rate)
= (t-10)(2)
= 2t - 20

So, we get...
(Distance B travels) = (Distance A travels) + 100
4t = (2t - 20) + 100
4t = 2t + 80
2t = 80
t = 40

Cheers,
Brent
Another way to solve this,
B - 4m/s
A- 2m/s
Thus extra speed B is traveling with = 2m/s
Now as A starts 10secs later, distance covered by B = 40m.
Now as B is traveling faster and started earlier than A, B only can catch A.
So, remaining distance between A and B= 60m
Relative speed of B =2m/s
So time taken = 30secs.
Total time taken= 10+30=40 secs.