If the velocity of a particle is given by $v = (180 - 16x)^{1/2} \text{ m/s}$,then its acceleration will be ....... $\text{m/s}^2$.

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

$A$ car moving with a speed of $40 \, km/h$ can be stopped by applying brakes after at least $2 \, m$. If the same car is moving with a speed of $80 \, km/h$,what is the minimum stopping distance in meters?

$A$ 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 $\,s$)

$A$ body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ starts from the same point and moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:

Two cars $P$ and $Q$ are moving on a road in the same direction. The acceleration of car $P$ increases linearly with time,whereas car $Q$ moves with a constant acceleration. Both cars cross each other at time $t = 0$ for the first time. The maximum possible number of crossings (including the crossing at $t = 0$) is . . . . . . .