A vehicle moving with a constant acceleration from A to B in a straight line AB, has velocities u and v at A and B, respectively. C is the mid-point of AB. If time taken to travel from A to C is twice the timetaken to travel from C to B, then the velocity of the vehicle v at B is

According to question, the given condition can be depicted asFor motion from A to C,                   s2=2ut+12a×(2t)2=2ut+2at2                    …(i)Also, for motion from C to B,      s2=v't+12at2=(u+2at)t+12at2                                        (Let initial velocity be v' = u + 2at)         =ut+2at2+12at2Now, from Eq. (i), we get                 2ut+2at2=ut+2at2+12at2⇒u=at2For overall motion,               v=u+3at=at2+3at=7at2=7u           ∵u=at2