The relation between time $t$ and displacement $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,the retardation is:

Two particles $A$ and $B$ start from rest and move for equal time on a straight line. Particle $A$ has an acceleration of $2\,m/s^2$ for the first half of the total time and $4\,m/s^2$ for the second half. The particle $B$ has acceleration $4\,m/s^2$ for the first half and $2\,m/s^2$ for the second half. Which particle has covered a larger distance?

An automobile,travelling at $40\, km/h$,can be stopped at a distance of $40\, m$ by applying brakes. If the same automobile is travelling at $80\, km/h$,the minimum stopping distance,in metres,is (assume no skidding)..........$m$

$A$ bus moving along a straight highway with a speed of $72 \,km/h$ is brought to a halt within $4 \,s$ after applying the brakes. The distance travelled by the bus during this time (assume the retardation is uniform) is . . . . . . $m$.

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be

$A$ vehicle starts moving in a straight line with an acceleration $a = 4 \,m/s^2$, with initial velocity equal to zero. After accelerating for time $t_1$, the vehicle moves uniformly for time $t_2$, and finally decelerates for time $t_1$, eventually coming to a stop. The total time taken during the motion is $10 \,s$ and the average velocity during the motion is $5.1 \,m/s$. The time taken by the vehicle during acceleration is (in $\,s$)