From `04 s`, at `t=0, a=5 ms^(-2)`
.
From 4 to 8 s, `a=-5 ms^(-2)`
`v_(2)=v_(1)+ast`
`rArr v_(2)-5xx4=0 ms^(-1)`
Maximum velocity is `20 ms^(-1)`
at `t=4 s`.
.
Desplacement form `2` to `6s` is equal to area from`2` to `6s`, i.e.
`10xx4+(1//2)xx4xx10=60 m`
Since the motiontakes place along the same direction only
distance=displacement`=60 m`.