The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,then what is the retardation?

Acceleration of a body whose displacement follows the equation $3 s = 9 t + 5 t^2$ is in $m s^{-2}$. (in $/3$)

$A$ body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n^{\text{th}}$ second of its motion to the total distance travelled in $n$ seconds is

$A$ body starts from rest and acquires a velocity of $10 \ m \ s^{-1}$ in $2 \ s$. What is the acceleration of the body and the distance travelled?

$A$ particle starts with an initial speed $u$ and retardation $a$ to come to rest in time $T$. The time taken to cover the first half of the total path travelled is .......

$A$ particle of mass $m$ is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example,when the particle is to the left of $E$,$F$ points to the right. The magnitude of $F$ is a constant $F$ except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards the left from the equilibrium position $E$ and released from rest at $t = 0$. What is the period of the motion?