A particle is moving in a straight line along the xaxis. Its velocity-time graph is as shown. Sketch the displacement-time and distance-time grapgs.

Explanation

For displacement,
`Area(1)=2, Area(2)=8, Area(3)=2, Area(4)=-8, Area(5)=-4`
For distance,
`Area(4)=8, Area(5)=4`
`{:(,"Displacement(m)","Distance(m)"),(t=0 "to" 1s,2,2),(t=0 "to" 3s,10,10),(t=0 "to" 4s,12,12),(t=0 "to" 6s,4,20),(t=0 "to" 7s,0,24):}`
From `t=0` to `t=1 s`, `a=4 m//s^(2), v gt 0`, slope of the s-t graph is increasing.
From `t=1` to `t=3 s`, `a=0, v` is constant, slope of the s-t graph is constant.
From `t=3` to `6 s`, `a=-4 m//s^(2)`.
From `t=3` to `4 s, v gt 0, a lt 0` the particle is slowing down, slope of the s-t graph is decreasing.
From `t=0` to `t=4 s`, the graph for diplacement and distance is the same.
From `t=4` to `6 s, v lt 0, a lt 0`, the particle is speeding up.
1. `v lt 0`, s is decreasing, slope of the s-t graph is -ve and decreasing.
2. `|(bs)/(dt)|gt0`, Distance is increasing, slope of the distance-time graph is increasing
From `t=6` to `7 s, a=8 m//s^(2), v lt 0, a gt 0`, the particle is slowing down.
`v lt 0`, s is decreasing, slope of the s-t graph is -ve and increasing.
4. `|(ds)/(dt)| gt 0`, distance is increasing, slope of the distance-time is +ve and decreasing.