(a) A particle moving with constant acceleration from `A` to `B` in a straight line `AB` has velocities `u` and `v` at `A` and `B` respectively. Find the velocity of the particle at the midpoint of `AB`.
(b) If the time taken by the particle to go from `A` to the midpoint of `AB` is two times that from the midpoint of `AB` to `B` then find the value of `v//u`.

Explanation

(a)
C: Midpoint of `AB`
`v_(1)`: Velocity at `C`
`A` to `C: v_(1)^(2)=u^(2)+2a. d/2 …(i)`
`A` to `B: v^(2)=u^(2)+2ad (ii)`
From (i) and (ii), we get
`(v_(1)^(2)-u^(2))/(v^(2)-u^(2))=(ad)/(2ad)=1/2`
`2v_(1)^(2)=v^(2)+u^(2)`
`v_(1)=sqrt((v^(2)+u^(2))/2)`
(b) `A` to `C: v_(1)=u+a.2t_(0)impliesv_(1)-u=2at_(0) ...(i)`
Ato `B: v=u+a.3t_(0)impliesv-u=3at_(0) ...(ii)`
Dividing (i) by (ii), we get
`(v_(1)-u)/(v-u)=2/3implies3v_(1)-3u=2v-2u`
`v_(1)=(2v+u)/3implies sqrt((v^(2)+u^(2))/2)=(2v+u)/3`
squaring we get
`(v^(2)+u^(2))/2=(4v^(2)+u^(2)+4vu)/9`
`9v^(2)+9u^(2)=8v^(2)+2u^(2)+8uv`
`v^(2)-8uv+7u^(2)=0`
`(v-u)(v-7u)=0impliesv=u`, not possible
`v=7uimpliesv/u=7`