The distance moved by the particle in time  is given by `x = t^3-12t^2+6t+8` . At the instant when its acceleration is zero, the velocity is (a) 42       (b) -42          (c) 48           (d) -48

Options

A. 42
B. `- 42`
C. 48
D. `- 48`

Answer

B

Explanation

We have,
`x=t^(3)-12t^(2)+6t+8`
`implies(dx)/(dt)=3t^(2)-24t+6 and (d^(2)x)/(dt^(2))=6t-24`
Now,
Acceleration = `0implies(d^(2)x)/(dt^(2))=0implies6t-24=0impliest=4`
At t=4, we have
Velocity = `((dx)/(dt))_(t=4)=3xx4^(2)-24xx4+6=-42`