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:

$A$ body moves from rest with a constant acceleration of $5\,m/s^2$. Its instantaneous speed (in $m/s$) at the end of $10\,s$ is:

$A$ 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 distance travelled by a particle is directly proportional to $t^{1/2}$,where $t =$ time elapsed. What is the nature of motion?

The velocity of a body moving with a uniform acceleration of $2\,m/s^2$ is $10\,m/s$. Its velocity after an interval of $4\,s$ is...........$m/s$.

If the velocity of a body related to displacement $x$ is given by $v = \sqrt{5000 + 24x} \; \text{m/s}$,then the acceleration of the body is $\dots \dots \; \text{m/s}^2$.