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 $3 \, s$?

The displacement of a particle moving with uniform acceleration in time $t$ is given by $S = 30t + 5t^2$. Its initial velocity is $.......$ (in $m \ s^{-1}$)

$A$ car,starting from rest,accelerates at the rate $f$ through a distance $S$,then continues at constant speed for time $t$,and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$,then:

In some appropriate units,the time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^2 + x$. The acceleration of the particle is

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

$A$ body is moving with a uniform acceleration covers $40\,m$ in the first $4\,s$ and $120\,m$ in next $4\,s.$ Its initial velocity and acceleration are