$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

$A$ 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 $\,s$)

$A$ particle starts with an initial velocity of $10.0 \, m/s$ along the $x$-direction and accelerates uniformly at the rate of $2.0 \, m/s^2$. The time taken by the particle to reach the velocity of $60.0 \, m/s$ is $....... \, s$.

$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

$A$ lift is coming down from the $8^{th}$ floor and is just about to reach the $4^{th}$ floor. Taking the ground floor as the origin and the upward direction as positive for all quantities,which one of the following is correct?

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 . . . . . . $m$.