An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement $(s)-$ velocity $(v)$ graph of this object 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 $A$?

The displacement $x$ (in $m$) of a particle of mass $m$ (in $kg$) moving in one dimension under the action of a force,is related to time $t$ (in $s$) by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero,will be

$A$ car,starting from rest,accelerates at the rate $f$ through a distance $S$,then continues at constant speed for time $t$,and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$,then:

$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. What is the retardation of the car (assumed uniform in $m/s^2$)?

$A$ car passes three points $A$, $B$ and $C$ at $3 \text{ m/s}$, $6 \text{ m/s}$ and $9 \text{ m/s}$ respectively in a straight line with uniform acceleration. If distance $AB = 60 \text{ m}$, then find the distance $BC$. (in $m$)