A particle moves along a straight line in such a way that its acceleration is increasing at the rate of `2 m//s^(3)`. Its initial acceleration and velocity were zero. Then, the distance which it will cover in the `3^(rd)` second `(t=2 " to "t =3 " sec")` is :

Options

A. `19//3 m`
B. `12//5 m`
C. `17//5 m`
D. `19//4 m`

Answer

A

Explanation

`(da)/(dt)=2impliesunderset(o)overset(a)intda=underset(o)overset(t)int2dtimplies a=2t`
`a=(dv)/(dt)=2timpliesunderset(o)overset(v)intdv=underset(D)overset(t)int2tdtimpliesv=t^(2)`
`v=(dx)/(dt)=t^(2)impliesunderset(o)overset(x)intdx=underset(2)overset(3)intt^(2)dt`
`impliesx=(t^(3))/(3)|_(2)^(3)=(27)/(3)-(8)/(3)=(19)/(3)m`