The velocity - time graph of an object moving along a straight line is as shown.
Calculate distance covered by object between t = 0 to t = 3 and t = 3 to t = 6.

Explanation

t = 0 to t = 3,
velocity at t = 0,u = 0
velocity at t = 3 s, `v=30ms^(-1)`
So, from velocity `v=u+at`
`"Acceleration a"=(u)/(t)=(30)/(3)=10ms^(-2)`
So distance convered between 0 to 3 s
`s=ut+(1)/(2)at^(2)`
initial velocity u =0 and t = 3
`s=ut+(1)/(2)at^(2)=0xx3+(1)/(2)xx10(3)^(2)`
`=0+(1)/(2)xx10xx9=(1)/(2)xx90`
`s=45m`
From 3 to 6 sec, velocity is same `30ms^(-1)`
So distance travelled, `=30xx3=60m`
Total distance covered between 0 to 6 s.
`=30+60=90m`
(ii) At t = 3s, `u=30ms^(-1),` at t = 6 s, `v=0ms^(-1)`
So from, `v=u+at , 0 = 30+axx3`
`a=-10ms^(-2)`
So, distance covered from t = 3 to t = 6 sec is,
`v^(2)=u^(2)+2as`
`(0)^(2)=(30)^(2)-2xx10xxs`
`s=(900)/(20)=45m`
So total distance covered `=90+45=135m`