$A$ particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15) \ m$. The velocity of the body when its acceleration becomes zero is ........... $m/s$.

$A$ body is moving along a straight line with initial velocity $v_0$. Its acceleration $a$ is constant. After $t$ seconds,its velocity becomes $v$. The average velocity of the body over the given time interval is

$A$ car moving at a speed $V$ is stopped in a certain distance $s$ when the brakes produce a deceleration $a$. If the speed of the car is $nV$,what must be the deceleration of the car to stop it in the same distance $s$ and in the same time $t$?

The displacement of a particle moving in a straight line is given by the expression $x = A t^3 + B t^2 + C t + D$ in metres, where $t$ is in seconds and $A, B, C$ and $D$ are constants. The ratio between the initial acceleration and initial velocity is

Two cars $A$ and $B$ are initially at rest at the same point. If car $A$ starts with a uniform velocity of $40\, m/s$ and car $B$ starts in the same direction with a constant acceleration of $4\, m/s^2$,then after how much time will car $B$ catch car $A$?

For a vehicle moving at $80 \, km/hr$, how many times is the stopping distance compared to a vehicle moving at $40 \, km/hr$, assuming both have the same deceleration (in $\times$)?