The motion of a particle along a straight line is described by the function $x = (2t - 3)^2$,where $x$ is in metres and $t$ is in seconds. The acceleration of the particle at $t = 2 \,s$ is (in $\,m/s^2$)

The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in the figure. If the velocity of the particle is $0.8 \,ms^{-1}$ at $x=0$, then its velocity at $x=1.4 \,m$ is (in $ms^{-1}$)

If $v = x^2 - 5x + 4$,find the acceleration of the particle when the velocity of the particle is zero.

$A$ body starts from the origin and moves along the $x$-axis such that its position at any instant is $x = 4t^2 - 12t$,where $t$ is in seconds and $x$ is in meters. What is the acceleration of the particle in $m/s^2$?

$A$ particle has a velocity in the negative direction and a constant acceleration in the positive direction. Match the following columns:

Column $I$	Column $II$
$(A)$ Velocity-time graph	$(p)$ Slope $\rightarrow$ negative
$(B)$ Acceleration-time graph	$(q)$ Slope $\rightarrow$ positive
$(C)$ Displacement-time graph	$(r)$ Slope $\rightarrow$ zero
	$(s)$ $|\text{Slope}| \rightarrow$ increasing
	$(t)$ $|\text{Slope}| \rightarrow$ decreasing
	$(u)$ $|\text{Slope}| \rightarrow$ constant

Which physical quantity can be found by first differentiation and second differentiation of position vector?