In the shown diagram `m_(1)=m_(2)=4kg` and `m_(3)=2 kg`. Coefficient of friction between `m_(1) and m_(2)` is 0.5. The mass `m_(1)` is given a velocity `v` and it just stops at the other end of the mass `m_(1)` in 1 sec. Let `a_(1) and a_(2) and a_(3)` be the acceleration `m_(1),m_(2)and m_(3)` respectively, then :
A. for `t lt 1 sec, a_(1) = 5m//s^(2),a_(2)=a_(3)=(1)/(3)m//s^(2)`
B. for `t lt 1 sec,a_(1)=5 m//s^(2),a_(2)=a_(3)=0`
C. the value of `v` is `5 m//s`
D. for `t gt 1 sec,a_(1) = a_(2) =a_(3) = 2m//s^(2)`
