The acceleration versus time graph of a particle moving along a straight line is shown in the figure. Draw the respective velocity-time graph Given `v=0` at `t=0.`

Explanation

Area of acceleration -time graph = change in velocity
For `0 le t le 2`, area of `(a - t)` graph `= 2 xx 2 = 4 m//s`
For `2 le t le 4`, area of `(a - t)` graph `= 0`
For ` 4 le t le 6`, area of `(a - t)` graph `= (-4) xx 2 = - 8 m//s`
So, velocity -time graph will be as shown
Velocity after `2 s rArr v_(2) = v_(0) +` change in velocity in 2 s `= 1 + 4 = 5 m//s`
Velocity after 4 s
`rArr v_(4) = v_(2) +` change in velocity from `t = 2 s " to " t = 4s = 5 + 0 = 5 m//s`

Velocity after `6 s rArr v_(6) = v_(4) +` change in velocity from `t = 4 s " to " t = 6s`
`= 5 - 8 = - 3 m//s`
Area of velocity -time graph = change in position
`=` displacement
`:. " For " 0 lt t lt 2 s`, area of `(v - t)` graph `= (1)/(2) xx (5 + 1) xx 2 = 6 m`
`2 lt t lt 4 s`, area of `(v - t)` graph `= 5 xx 2 = 10 m`
`4 lt t lt 6 s`, area of `(v - t)` graph `= A_(3) + A_(4)`
To find `A_(3)` and `A_(4)` we must find `t_(1)` the time when velocity becomes zero from `5 m//s` with an acceleration `= - 4 m//s^(2)`
`v = v_(0) + at`
`0 = 5 - 4 (Delta t)`
`:. Delta t = (5)/(4) = 1.25 s`
`:. t_(1) = 4 + 1.25 = 5.25 s`
`:. A_(3) = (1)/(2) xx 1.25 xx 5 = 3.125 m`
`A_(4) = (1)/(2) xx 0.75 xx (-3) = - 1.125 m`
Therefore area of `v - t` graph for `4 le t le 6 s = A_(3) + A_(4) = 2m`
So, position after `2 s, bar(r)_(2) = r_(0) +` change in position
`= bar(r)_(0) + Delta bar(r)`
`= 1 + 6 = 7 m`

Position after 4 s, `bar(r)_(4) = bar(r)_(2) + Delta bar(r) = 7 + 10 = 17 m`
Position after `5.25 s, bar(r)_(5.25) = bar(r)_(4) + Delta bar(r)`
`= 17 + 3.125 = 20.125 m`
Position after `6 s, bar(r)_(6) = bar(r)_(5.25) + Delta bar(r)`
`= 20.125 - 1.125 = 19 m`