A partical is moving along the x-axis and its velocity-time graph is shown in following diagrams. Sketch displacement-time graph.

Explanation

(a)
Area of the v-t graph gives displacement.
`Area (1) =20`
`Area (2) =-20`
`t=0` to `t=2 s`, Displacement`=20`
`t=0` to `t=4 s`, Displacement=0
`t=0` to `t=2 s`, is +ve and constant and hence the slope of the s-t graph is positive and constant, graph will be a straight line of positive slope.
`t=2` to `t=4 s`, v is -ve and constant and the slope of the s-t graph is -ve and constant, graph will be a straight line of -ve slope.

(b)
`Area (1)=20, Area (2)=-10, Area (3)=30`
`t=0` to `t=2 s`, Displacement`=20 m`
`t=0` to `t=4 s`, Displacement`=10 m`
`t=0` to `t=6 s`, Displacement`=40 m`

(c )
`Area(1) =25, Area(2)=-9`
`t=0` to `t=5 s`, Displacement `=25 m`
`t=0` to `t=8 s`, Displacement =16 m
`t=0` to `8 s`, Acceleration=Slope of v-t graph=-2 `m//s^(2)`
From `t=0` to `t=5 s`,
(i) v gt 0, i.e. slope of the s-t slowing down, slope of the s-t graph is decreasing
From `t=5` to `t=8 s`,
(i) v lt 0, i.e. slope of the s-t graph is -ve
(ii) v lt 0, a lt 0, particle is speeding up, slope of the s-t graph is increasing
At t=5 s
v=0, slope of the s-t graph is 0, direction of motion changes

(d)
Area (1)=10, Area (2) =10
`t=0` to `t=2s`, Displacement`=10m`
`t=0` to `t=4s`, Displacement`=20m`
`t=0` to `t=2s`, `a=5 m//s^(2)`
(i) v gt 0, slope of the s-t graph is +ve
(ii) v gt 0, a gt 0, particle is speeding up, slope of the s-t graph is increasing
`t=2` to `t=4s`, `a=-5 m//s^(2)`
(i) v gt 0, slope of the s-t graph is +ve
v gt 0, a lt 0, particle is slowing down, slope of the s-t graph is decreasing