Question

Two cyclists start from the middle of a 200m track and move in opposite directions towards the end of the track at 5m/s and 10m/s respectively. At the time when a cyclist first reaches the end, both cyclists turn around and start moving towards each other with their initial speeds. During this motion, a fly moves at a speed of 17 m/s, between the 2 cyclists, landing on one cyclist and then moving to the next continuously. How much distance has the fly covered when the two cyclists meet again?

Answer 1

Answer:

$\sf{The \ fly \ covers \ the \ distance \ of \ 340 \ m}$

$\sf{until \ cyclists \ meet \ again.}$

Given:

• Two cyclists start from the middle of a 200m track and move in opposite directions towards the end of the track at 5m/s and 10m/s respectively.

• At the time when a cyclist first reaches the end, both cyclists turn around and start moving towards each other with their initial speeds.

• During this motion, a fly moves at a speed of 17 m/s, between the 2 cyclists, landing on one cyclist and then moving to the next continuously.

To find:

• How much distance has the fly covered when the two cyclists meet again.

Solution:

$\sf{Let \ cyclist_{1} \ speed \ be \ 10 \ m \ s^{-1}}$

$\sf{and \ cyclist_{2} \ speed \ be \ 5 \ m \ s^{-1}}$

$\sf{They \ start \ from \ middle \ of \ 200 \ m \ track.}$

$\sf{\therefore{Distance \ to \ cover=\dfrac{200}{2}}}$

$\sf{\therefore{Distance \ to \ cover=100 \ m}}$

$\boxed{\sf{Time=\dfrac{Distance}{Speed}}}$

$\sf{For \ cyclist_{1},}$

$\sf{Time=\dfrac{100}{10}}$

$\sf{\therefore{Time \ taken \ by \ cyclist_{1} \ to}}$

$\sf{reach \ the \ end \ of \ the \ track=10 \ s}$

$\sf{In \ same \ distance \ cover \ by \ cyclist_{2}}$

$\boxed{\sf{Distance=Speed\times \ Time}}$

$\sf{Distance \ cover=5\times10}$

$\sf{\therefore{Distance \ cover \ by \ cyclist_{2}}}$

$\sf{in \ same \ time=50 \ m}$

$\sf{Distance \ between \ both \ cyclists}$

$\sf{=Distance \ cover \ by \ cyclist_{1}+}$

$\sf{Distance \ cover \ by \ cyclist_{2}}$

$\sf{=100+50}$

$\sf{\therefore{Distance \ between \ them=150 \ m}}$

$\sf{Time \ taken \ by \ cyclists \ to \ meet \ again}$

$\sf{=\dfrac{Distance \ between \ cyclists}{Speed \ of \ cyclists}}$

$\sf{=\dfrac{150}{10+5}}$

$\sf{=\dfrac{150}{15}}$

$\sf{=10 \ s}$

$\sf{Total \ time=10+10=20 \ s}$

$\sf{\therefore{Distance \ covered \ by \ fly=17\times20}}$

$\sf{\therefore{Distance \ covered \ by \ fly=340 \ m}}$

$\sf\purple{\tt{\therefore{The \ fly \ covers \ the \ distance \ of \ 340 \ m}}}$

$\sf\purple{\tt{until \ cyclists \ meet \ again.}}$