$A$ body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ starts from the same point and moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:

Derive the equations of uniformly accelerated motion by the graphical method.

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^3 - 6t^2 + 3t + 4$ meters. The velocity when the acceleration is zero is ........ $m s^{-1}$.

$A$ particle starts from rest at $x=0 \, m$ with an acceleration of $1 \, m/s^2$. At $t = 5 \, s$,it receives an additional acceleration in the same direction as its motion. At $t = 10 \, s$,its speed and position are $v$ and $x$,respectively. Had the additional acceleration not been provided,its speed and position would have been $v_0$ and $x_0$,respectively. It is found that $x - x_0 = 12.5 \, m$. Then one can conclude that $v - v_0$ is .............. $m/s$.

The displacement of a particle moving with uniform acceleration in time $t$ is given by $S = 30t + 5t^2$. Its initial velocity is $.......$ (in $m \ s^{-1}$)

The relation between time $t$ and displacement $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,the retardation is: