A foce acts on a 30.g particle in such a way that the position of the particle as a function of time is given by
`x=3t-4t^(2)+t^(3)`, where x is in metre and t in second. The work done during the first 4s is

Options

A. 576 mJ
B. 450 mJ
C. 490 mJ
D. 530 mJ

Answer

A

Explanation

`x=3t-4t^(2)+t^(3)`
`(dx)/(dt)=3-8t+3t^(2)`
Acceleration `(d^(2)x)/(dt^(2))=-8+6t`
Acceleration after 4 sec =`-8+6xx4=16 ms^(-2)`
displacemetn in 4 sec `=3xx4-4xx4^(2)+4^(3)=12 m`
`=3xx10^(-3)xx16xx12=576 mJ`