A body of mass 2 kg is moving according to the equation for displacement at seconds as `x(t) = pt^(2) + rt^(3). If p = 3 ms^(-1) , q = 4 ms^(-1)` and `r = 5 ms^(-1)` the force acting after 2 sec is :

Options

A. `136N`
B. 128N
C. 68N
D. 64N

Answer

a

Explanation

Here `x(t)=pt+qt^(2)+rt^(2)`
`(dx)/(dt)=p+2qt+3et^(2)`
`a=(d^(2)x)/(dt^(2))=0+2q+6rt`
Acceleration at `t = 2s = 0 + 2 xx 4 + 6 xx 5 xx 2`
`= 68 ms^(-2)`
`"Force" = ma = (2 xx 68) N`
`= 136 N`
Hence .a. is the right choice