The acceleration - time graph of a particle is as shown in figure Initial velocity of particle is `10 m//s` Displacement of the particle (in m) at the end of 8 sec is ___ Round off to nearest integer.

Answer

`0197`

Explanation

`v=u+at`
`v_(1)=10+4xx4=26 m//s`
`S_(1)=10xx4+1/2xx4xx16=40+32=72` m
Now `a=8-t`
`(dr)/(dt)=8-t`
`underset26 overset rint dr = underset4overset t int(8-t)=dt`
`v-26=[8t-t^(2)/2]_(4)^(t)`
`v=26+(8t-t^(2)/2)-(32-16/2)`
`v=26+8t-t^(2)/2 24`
`v=2+8t-t^(2)/2`
`(dx)/(dt)=2+8t-t^(2)/2`
`x=[2t+4t^(2)-t^(3)/6]_(4)^(8)=(16+64xx4-8^(3)/6)-(8+64-64/6)=8+64xx3-74.6`
`=200-74.6=125.33`
Total distance `=72+125.33=197.33 approx 197]`