The initial velocity of the particle is $10 \ m/s$ and its retardation is $2 \ m/s^2$. The distance moved by the particle in the $5^{th}$ second of its motion is .......... $m$.

If $x \propto t^{5/2}$,then

$A$ 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 ........ $m/s^2$.

The velocity of an object is given by $v = kt$,where $k = 2 \, m/s^2$. What distance (in $m$) will it cover in the first $3 \, s$?

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

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