Question

The distance moved by the particle in time t is given by $x=t^3-12t^2+6t+8$. At the instant when its acceleration is zero, then the velocity is

• A. 42
• B. -42
• C. 48
• D. -48

Answer 1

The correct option is B

-42

We have,
$x=t^3-12t^2+6t+8$
$\Rightarrow \frac{dx}{dt}=3t^2-24t+6and\frac{d^{2}x}{d{t}^2}=6t-24$
Now, Acceleration =0
$\Rightarrow \frac{d^{2}x}{d{t}^2}=0\Rightarrow 6t-24=0\Rightarrow t=4$
Att=4, we have
Velocity =${\left(\frac{dx}{dt}\right)}_{r-4}=3\times 4^2-24\times 4+6=-42$.
Hence (b) is the correct answer.