The figure shows the force (F) versus displacement (s) graph for a particle of mass $m=2kg$ initially at rest:
(a) The maximum speed of the particle occurs at $x=......m$.
(b) The maximum speed of the particle occurs is ........$m{s}^{-1}$.
(c) The particle once again has its speed zero at $x=...........m$.

The correct option is A 4.1833, 3, 6
The velocity of the particle will keep on increasing until it starts retarding

So, the maximum velocity of the particle will occur at $x$=$3$ $m$

As after that it starts deaccelerating

Now work done=${\int }_0^x\overrightarrow{F}\cdot \overrightarrow{dx}=areaundertheforcedisplacementgraphtillx=3m$

$\frac{1}{2}\times (5+10)+\frac{1}{2}\times 2\times 10=10+7.5=17.5$

Let the maximum speed attained by the particle at $x$=$3$ $m$ is '$v$'

$Accordingtoworkenergytheorem,$

Work done on the system=change in the kinetic energy of the system=$\frac{1}{2}\times 2\times v^2=17.5\Rightarrow v^2=17.5$

$\Rightarrow v=\sqrt{17.5}$

Area of the graph under force displacement curve from $x=3mtox=6mis\frac{15+20}{2}=17.5$

$So.workdonebytheforcebetweenthatintervalis-17.5So,totalworkdonebytheforcetillx=6mis17.5-17.5=0So,kineticenergyoftheparticleatx=6mwillbezeroSo,velocityoftheparticlewillbezeroatx=6m$