1.

A particle moves along x-axis and its displacement at any time is given by x(t) = 2t^(3) -3t^(2) + 4t in SI units. The velocity of the particle when its acceleration is zero is

Answer»

`2.5 ms^(-1) `
`3.5 ms^(-1)`
`4.5 ms^(-1)`
`8.5 ms^(-1)`

Solution :Given `x(t) = 2t^(3)-3t^(2)+4t`
Differentiating with respect to t we get
`(dx)/(dt) = 6t^(2)-6t+4 or nu= 6t^(2)-6t+4 ( because(dx)/(dt)=nu)` …(i)
Again differentiating with respect to t.
`(d^(2)x)/(dt^(2))=12T -6`
a=12t-6 `( because (d^(2)x)/(dt^(2))=a)`
According to question 0=12t -6 , `t = (6)/(12)=(1)/(2)s`
Putting the value of t in equation (i) we get
`nu= 6xx((1)/(2))^(2)-6xx((1)/(2))+4 `
`= 6xx(1)/(4) -6xx(1)/(2)+4=(5)/(2)=2.5 ms^(-1)`


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