Let both particles reach at same position in same time t then from s = `ut + (1)/(2) at ^(2)`
For `1^(st)` particle : `s = 4(t)+ (1)/(2)(1) t^(2) = 4t + (t^(2))/(2)`, For `2^(nd)` particle : `s = 2(t) + (1)/(2) (2)t^(2) = 2t + t^(2)`
Equation above equation we get `4t+ (t^(2))/(2) = 2t + t^(2) rArr t = 4s`
Subsituting value of t in above equation s = `4(4) + (1)/(2) (1)(4)^(2)= 16 +8 = 24`m