Car `A` starts from a point `O` and moves with constant velocity `9 m//s`. After `2 s`, another car `B` begins its journey from `O` and follows car `A`. If car `B` starts from rest and moves under constant acceleration `4 m//s^(2)`, after how much time and at what distance from `O` the cars meet?

Explanation

Here the cars starts at different timings. Let the cars meet after time t (from start of A) and at distance d from `O`. Car `B` will take time `(t-2)` sec.
Car `A` (uniform motion): `d=vt=9t …(i)`
Car `B` (accelerated motion): `d=0+1/2a(t-2)^(2)`
`=(1)/(2).4(t-2)^(2) ...(ii)`
From (i) and (ii), we get
`9t=2(t-2)^(2)`
`2t^(2)-17t+8=0`
Solving we get `t =8 s,1/2` s
`t=1/2 s` is not possible, because car `B` starts `2 s` later
`d=9t=9xx8=72m`