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Arithmetic II

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Time & Speed
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CAT 2025 Lesson : Time & Speed - Circular Track

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8. Distance Covered on a circular track

8.1 Running in the same direction from the same starting point

If two people A and B are running in the same direction around a circular track of ll metres at speed of s1s_1 and s2s_2 respectively (where s1s_1 > s2s_2 ), their relative speed is s1−s2s_1- s_2. The first time they meet after starting is when A overlaps B. So, the distance covered is the length of the track.

Time taken to meet for the first time =ls1−s2= \dfrac{l}{s_1 - s_2}

Time taken to meet for the second time =2ls1−s2= \dfrac{2l}{s_1 - s_2}

Time taken to meet for the nthn^{\text{th}} time = nls1−s2 \dfrac{nl}{s_1 - s_2}

8.2 Running in the opposite direction from the same starting point

If two people A and B are running in the opposite directions around a circular track of ll metres at speed of s1s_1 and s2s_2 respectively, their relative speed is s1+s2s_1 + s_2. The first time they meet after starting is when a complete lap is covered by the two. So, the distance covered is the length of the track.

Time taken to meet for the first time =ls1+s2= \dfrac{l}{s_1 + s_2}

Time taken to meet for the second time =2ls1+s2= \dfrac{2l}{s_1 + s_2}

Time taken to meet for the nthn^{\text{th}} time =nls1+s2= \dfrac{nl}{s_1 + s_2}

Note that the ratio of the distance covered by them individually will be the ratio of their speeds.

8.3 Time taken to meet for the first time at the Starting Point

When two people are running in the same or opposite direction, the time taken to meet for the first time at the starting point == LCM of the individual time taken to complete a lap

== LCM of {ls1,ls2 \dfrac{l}{s_1}, \dfrac{l}{s_2}}

Example 26

A and B start running, in the same direction, around a circular track of length 400 metres at speeds of 25 m/s and 15 m/s respectively. When A and B meet for the 20th20^{th} time, what is the distance covered by A (in km)?

Solution

As they are running in the same direction, relative speed =25−15=10= 25 - 15 = 10 m/s

Relative distance covered when they meet for the 20th20^{\text{th}} time is the 2020 overlaps of A. So, the distance is 20×400=800020 \times 400 = 8000 metres.

So, time taken =800010=800= \dfrac{8000}{10} = 800 seconds.

Distance covered by A =25×800=20000= 25 \times 800 = 20000 metres = 2020 km

Answer: 20

Example 27

Parul and Ramnique take 3 minutes and 5 minutes to go around a circular track. How many minutes does it take for them to meet for the fourth time if they started running from the same point and in the same direction?

Solution

Given the length of the track being constant for both, ratio of speeds is the ratio of reciprocal of the times taken.

Ratio of times taken by Parul and Ramnique =3:5= 3 : 5
Ratio of speeds of Parul and Ramnique =5:3= 5 : 3

Let the length of track be 1515 metres and the speeds of Parul and Ramnique be 55 metres/minute and 33 metres/minute respectively.

Time taken to meet for the first time =155−3=7.5= \dfrac{15}{5 - 3} = 7.5 minutes

Time taken to meet for the fourth time =4×7.5=30= 4 \times 7.5 = 30 minutes

Answer: 30 mins

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