$A$ particle moving in a straight line covers half the distance with speed $6 \, m/s$. The other half is covered in two equal time intervals with speeds $9 \, m/s$ and $15 \, m/s$ respectively. The average speed of the particle during the motion is: (in $ \, m/s$)

$A$ gas balloon is going up with a constant velocity of $10 \text{ m/s}$. When this balloon reached a height of $75 \text{ m}$, a stone is dropped from it and the balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

$A$ body of mass $0.40 \;kg$ moving initially with a constant speed of $10 \;m s^{-1}$ to the north is subject to a constant force of $8.0 \;N$ directed towards the south for $30 \;s$. Take the instant the force is applied to be $t=0$,the position of the body at that time to be $x=0$,and predict its position at $t=-5 \;s, 25 \;s, 100 \;s$.

$A$ point traversed half of the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time and with velocity $v_2$ for the other half of the time. The mean velocity of the point averaged over the whole time of motion is

$Assertion$: $A$ body with constant acceleration always moves along a straight line.
$Reason$: $A$ body with constant acceleration may not speed up.

The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. The retardation of the particle is: