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For the velocity time graph shown in fig...

For the velocity time graph shown in figure, in a time interval from `t=0` to `t=6s` match the following

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The correct Answer is:
A, B, C, D
`upsilon_(i)=+10 ms^(-1)` and `upsilon_(f)=0`
`therefore Delta upsilon = upsilon_(f)-upsilon_(i)=-10 ms^(-1)`
`a_(av)=(Delta upsilon)/(Delta t)=(-10)/(6)=(-5)/(3)ms^(-2)`
Total displacement = area under `upsilon-t` graph (with sign)
`=(1)/(2)xx10xx2-(1)/(2)xx2xx10`
`=-10m`
And acceleration = slope of `upsilon-t` graph
`=-(10)/(2)=-5 ms^(-2)`
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