$A$ particle moves along the $x$-axis in such a way that its $x$-coordinate varies with time according to the equation $x = 4 - 2t + t^2$. The velocity of the particle will vary with time as:

The velocity-time graphs of a car and a scooter are shown in the figure. $(i)$ The difference between the distance travelled by the car and the scooter in $15\, s$ and $(ii)$ the time at which the car will catch up with the scooter are,respectively:

$A$ particle of mass $m$ is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example,when the particle is to the left of $E$,$F$ points to the right. The magnitude of $F$ is a constant $F_0$ except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards the left from the equilibrium position $E$ and released from rest at $t = 0$. The velocity-time graph of the particle is:

The velocity-time graph of a body moving in a straight line is shown in the figure. The ratio of displacement to distance travelled by the body in time $t = 0$ to $t = 8 \ s$ is:

$A$ particle covers a distance from $A$ to $B$ over a period of time; the distance versus time plot is shown below. Then which of the following is true for the motion of the particle?

The position of a particle moving along the $x$-axis is given by $x = 10t - 2t^2$. Then the time $(t)$ at which it will momentarily come to rest is .......... $s$.