The relation between time $t$ and distance $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. The relation between acceleration $a$ and velocity $v$ is:

$A$ 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 $g = 10 \text{ m/s}^2$)

The equation of motion of a body is $\frac{dv}{dt} = -4v + 8$,where $v$ is the velocity in $m/s$ and $t$ is the time in seconds. The initial velocity of the particle was zero. Then,

During the first $18\,min$ of a $60\,min$ trip,a car has an average speed of $11\,m/s$. What should be the average speed for the remaining $42\,min$ so that the car has an average speed of $21\,m/s$ for the entire trip (in $,m/s$)?

$A$ train starting from rest first accelerates uniformly up to a speed of $80 \ km/h$ for time $t$,then it moves with a constant speed for time $3t$. The average speed of the train for this duration of journey will be (in $km/h$):

$A$ person wearing a parachute jumps off a plane from a height of $2 \ km$ from the ground and falls freely for $20 \ m$ before his parachute opens. After his parachute opens, if he continues to move uniformly with the velocity attained due to his freefall, the total time taken by the person to reach the ground is (Acceleration due to gravity $= 10 \ m/s^2$) (in $s$)