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 time taken 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 as

For motion from A to C, 

                  $\frac{s}{2}=2ut+\frac{1}{2}a\times (2t{)}^2=2ut+2a{t}^2$                    …(i)

Also, for motion from C to B,

      $\frac{s}{2}=v^{\text{'}}t+\frac{1}{2}a{t}^2=(u+2at)t+\frac{1}{2}a{t}^2$

                                        (Let initial velocity be v' = u + 2at)

         $=ut+2a{t}^2+\frac{1}{2}a{t}^2$

Now, from Eq. (i), we get

                 $2ut+2a{t}^2=ut+2a{t}^2+\frac{1}{2}a{t}^2\Rightarrow u=\frac{at}{2}$

For overall motion,

               $v=u+3at=\frac{at}{2}+3at=\frac{7at}{2}=7u \left(\because u=\frac{at}{2}\right)$