The acceleration-displacement graph of a particle moving in a straight line is as shown in figure, initial velocity of particle is zero. Find the velocity of the particle when displacement of the particle is `s=12 m`.

Explanation

`v^(2)-u^(2)=2ax`
`impliesax=(v^(2)-u^(2))/(2)=(v^(2))/(2)-(u^(2))/(2)` `because(u=0)`
`impliesv=sqrt(2("area under a-x graph"))`
Area under `a-x` graph
Area of `DeltaOAE+` area of reactangle `ABEF`
`+` area of trapenzium BFGC + Area of `DeltaCGD`
Area `=(1)/(2)(2)(2)+6xx2+(1)/(2)(2+4)(2)+(1)/(2)xx2xx4=24`
`impliesv=sqrt(2xx24)=4sqrt(3) m//s`