A body moving along a straight line traversed one third of the total distance with a velocity `4 m//sec` in the first stretch. In the second stretch, the remaining distance is covered with a velocity `2 m//sec` for some time `t_(0)` and with `4 m//s` for the remaining time. If the average velocity is `3 m//sec`, find the time for which body moves with velocity `4 m//sec` in second stretch:

Answer

(i) `(t_(0))/(2)` (ii) `4.0m//s le v_(av) le 2. 4m//s`

Explanation

(i) For second stretch: `t_(1) =(s//3)/(4)=(s)/(12)`
For second stretch :
Let the body moves with velocity `4 m//s` with the time `kt_(0)`.
`(2s)/(3)=2(t_(0))+ 4(kt_(0))=t_(0)(2+4k)`
Hence, `t_(0)=(2s)/(3(2+4k))`
Average velocity `=(s)/(t_(1)_t_(0)+t_(0)k)`
`=(s)/((s)/(12)+(2s(1+k))/(3(2+4k))) =(6(2+4k))/((5+6k))`
`v_(av)(5+6k)=12+24 k` gives `k=(1)/(2)`
Required time `=kt_(0)` =t_(0)/(2)` .
(ii) We have, `v_(av)(5+6k)=12+24k`
Obviously `4.0m//s le 2.4 m//s`.