$A$ particle starts from rest. Its acceleration $(a)$ versus time $(t)$ graph is as shown in the figure. The maximum speed of the particle will be.....$m/s$.

At any instant,the velocity and acceleration of a particle moving along a straight line are $v$ and $a$. The speed of the particle is increasing if

$A$ 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

Match the following columns.

Column $I$	Column $II$
$(A)$ $\frac{dv}{dt}$	$(p)$ Acceleration
$(B)$ $\frac{d|v|}{dt}$	$(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$	$(r)$ Velocity
$(D)$ $\left|\frac{dr}{dt}\right|$	$(s)$ Magnitude of velocity

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 $ms^{-1}$)

$Assertion$ : Retardation is directly opposite to the velocity.
$Reason$ : Retardation is equal to the time rate of decrease of speed.