$A$ car moving with a velocity $6.25 \,ms^{-1}$ is decelerated with $2.5 \sqrt{v} \,ms^{-2}$ (where $v$ is the instantaneous velocity). The time taken by the car to come to rest is: (in $\,s$)

The acceleration $a$ (in $m/s^2$) of a body,starting from rest,varies with time $t$ (in seconds) according to the relation $a = 3t + 4$. The velocity of the body at time $t = 2 \ s$ will be $........ \ m/s$.

$A$ 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 displacement $x$ of a particle along a straight line at time $t$ is given by $x = a_0 + a_1t + a_2t^2$. The acceleration of the particle is

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...........$m/s$.

$A$ body starts from rest with uniform acceleration and its velocity at a time of $n$ seconds is $v$. The total displacement of the body in the $n^{\text{th}}$ and $(n-1)^{\text{th}}$ seconds of its motion is: