$A$ body moves from rest with a constant acceleration of $5\,m/s^2$. Its instantaneous speed (in $m/s$) at the end of $10\,s$ is:

$A$ particle is moving in a straight line and passes through a point $O$ with a velocity of $6\;m/s$. The particle moves with a constant retardation of $2\;m/s^2$ for $4\;s$ and thereafter moves with constant velocity. How long after leaving $O$ does the particle return to $O$?

The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration 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 starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n^{\text{th}}$ second of its motion to the total distance travelled in $n$ seconds is

$A$ particle moves along the $x$-axis as $x = 4(t-2) + a(t-2)^2$. Which of the following statements is true?