In the following displacement $x$ versus time $t$ graph,at which among the points $P, Q$ and $R$ is the object's speed increasing?

If $t = \sqrt{x} + 4$,then $\left(\frac{dx}{dt}\right)_{t=4}$ is:

The position of an object moving along the $x$-axis is given by $x = a + b t^{2}$,where $a = 8.5 \; m$,$b = 2.5 \; m s^{-2}$,and $t$ is measured in seconds. What is the velocity at $t = 0 \; s$ and $t = 2.0 \; s$?

For the given displacement-time graph,what will the corresponding velocity-time graph look like?

Refer to the graph in figure. Match the following:

Graph	Characteristics
$(A)$	$(i)$ has $v > 0$ and $a < 0$ throughout
$(B)$	$(ii)$ has $x > 0$ throughout and has a point with $v = 0$ and a point with $a = 0$
$(C)$	$(iii)$ has a point with zero displacement for $t > 0$
$(D)$	$(iv)$ has $v < 0$ and $a > 0$

The velocity-time and acceleration-time graphs of a particle are given. Its position-time graph may be given as: