If the engine of a long train moving with constant acceleration crosses a tree with velocity $u$ and the last compartment of the train crosses the same tree with velocity $v$,then the velocity with which the middle compartment crosses the same tree is

$A$ body initially at rest is moving with uniform acceleration $a$. Its velocity after $n$ seconds is $v$. The displacement of the body in the last $2 \ s$ is

$A$ point starts moving in a straight line with a certain acceleration. At a time $t$ after the beginning of motion,the acceleration suddenly becomes a retardation of the same value. The time in which the point returns to the initial point is:

$A$ particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15) \ m$. The velocity of the body when its acceleration becomes zero is ........... $m/s$.

$A$ vehicle starts moving in a straight line with an acceleration $a = 4 \,m/s^2$, with initial velocity equal to zero. After accelerating for time $t_1$, the vehicle moves uniformly for time $t_2$, and finally decelerates for time $t_1$, eventually coming to a stop. The total time taken during the motion is $10 \,s$ and the average velocity during the motion is $5.1 \,m/s$. The time taken by the vehicle during acceleration is (in $\,s$)

The velocity $v$ of a particle as a function of its position $x$ is expressed as $v = \sqrt{c_1 - c_2 x}$,where $c_1$ and $c_2$ are positive constants. The acceleration of the particle is