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 $\dots \dots \; \text{m/s}^2$.

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?

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

The acceleration of a particle is increasing linearly with time $t$ as $a = 6t$. The particle starts from the origin with an initial velocity $u = 10 \ m/s$. The distance travelled by the particle after $t = 2 \ s$ will be: (in $m$)

The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$, then at $t = 2 \ s$, its position is: (in $m$)

$A$ car passes three points $A$, $B$ and $C$ at $3 \text{ m/s}$, $6 \text{ m/s}$ and $9 \text{ m/s}$ respectively in a straight line with uniform acceleration. If distance $AB = 60 \text{ m}$, then find the distance $BC$. (in $m$)