A `10-km` long straight road connects two towns `A` and `B`, Two cyclists dimultaneously start one from town `A` and the other from town `B` .On reaching the opposite town, a cyclists simmediaeately retrns to his starting town wheras the other cyclist takes some rest and then returns to his starting town. Both of them can ride at speed `20 km h^(-1)` in absence of wind but during their whole journey uniform wind from town `A` and `B` increase the speed of it decreases the speed of the cyclist going against the wind Both the cyclisrs meet twice, first at `2` km and then `6` km away from one of the tomns. If which town and for what perild does a cyclist rest ?

Explanation

`2=(v-v_(omega))t`
`8=(v+v_(omega))t`
.
`rArr 10=2 vt rArr t(10)/(2v)=(10)/(2xx20) =(1)/(4) h`
and `v_(omega)=12 km h^(-1)`
Now when they again meet at `C`: [first cyclist takes rest at `B` for time `t_(0)` ]
`(10)/(v-v_(omega)) +(4)/(v+vomega) =(10)/(v+v_(omega)) +t_(0) +(6)/(v-v_(omega))`
Solve to get `t_(0)=18.75 min`.