Object $A$ starts from rest with a constant acceleration $a$. Object $B$ starts from the same position and moves in the same direction as $A$ with a constant velocity $v$. If both meet after time $t$,then $t =$

$A$ driver applies the brakes on seeing the red traffic signal $400 \ m$ ahead. At the time of applying brakes,the vehicle was moving with $15 \ m/s$ and retarding at $0.3 \ m/s^2$. The distance of the vehicle from the traffic light one minute after the application of brakes is: (in $m$)

If a train travelling at $72 \text{ km/h}$ is to be brought to rest in a distance of $200 \text{ m}$,then its retardation should be ............ $\text{m/s}^2$.

$A$ car moving with a velocity of $20 \,m \,s^{-1}$ is stopped in a distance of $40 \,m$. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is (in $\,m$)

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

Consider a particle moving along the positive direction of the $X$-axis. The velocity of the particle is given by $v = \alpha \sqrt{x}$ (where $\alpha$ is a positive constant). At time $t = 0$, the particle is located at $x = 0$. Find the time dependence of the velocity and the acceleration of the particle, respectively.