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The dispacement of a body is given by 4s...

The dispacement of a body is given by `4s=M +2Nt^(4)`, where `M` and `N` are constants.
The velocity of the body at the end of `1 s` from the start is .

A

`2N`

B

`(M+2N)/(4)`

C

`2(M_N)`

D

`(2M+N)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
`s=(M+2Nt^(4))/(4) rArr v=(ds)/(dt) =2Nt^(3) `
Putting `1 s`, we get `v=2 N` .
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