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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 :A. `19//3 m`B. `12//5 m`C. `17//5 m`D. `19//4 m` |
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Answer» Correct Answer - A `(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` |
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