The velocity of a particle moving in the positive direction of the $x$-axis varies as $v = 5 \sqrt{x}$. Assuming that at $t = 0$,the particle was at $x = 0$,what is the acceleration of the particle in $m/s^2$?

The distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as $x^{2}=at^{2}+2bt+c$. If the acceleration of the particle depends on $x$ as $x^{-n}$,where $n$ is an integer,the value of $n$ is

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 displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) s$ are $125 \ m$ and $50 \ m/s$,respectively. The distance travelled by the particle in $(t+2)^{th} s$ is . . . . . . $m$.

$A$ particle is moving along a straight line with constant acceleration. At the end of the $10^{th}$ second,its velocity becomes $20 \, m/s$,and in the $10^{th}$ second,it travels a distance of $10 \, m$. Then the acceleration of the particle will be ........ $m/s^2$.

$A$ particle initially at rest starts moving from reference point $x=0$ along the $x$-axis,with velocity $v$ that varies as $v=4 \sqrt{x} \ m/s$. The acceleration of the particle is . . . . . . $m/s^2$.