A particle, initially at rest, starts moving in a straight line with an acceleration `a=6t+4 m//s^(2)`. The distance covered by it in 3 s is

Options

A. 15 m
B. 30 m
C. 45 m
D. 60 m

Answer

C

Explanation

`a=(dv)/(dt)=6t+4`
`int_0^4dv=int_0^t(16t+4)dt`
`v=(6t^(2))/(2)+4t`
`int_0^s dx = int_0^3 ((6t^(2))/(2)+4t) dt `
`s=(t^(3)+2t^(2))0^(3)`
`s=(3)^(3)+2(3)^(2)=45m`