At `t=0`, a particle starts from reat and moves along a straight line, whose acceleration-time graph is shown in .
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Convert this graph into velocity-time, From the velocity-time graph, find the maximum velocity attained by the particle. Also find from `v-t` graph, the sisplacement and distance travelled by the particle from `2` to `6 s`,

Explanation

From `04 s`, at `t=0, a=5 ms^(-2)`
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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`.
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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`.