Question

The distance moved by the particle in time t is given by x = t³ − 12t² + 6t + 8. At the instant when its acceleration is zero, the velocity is
(a) 42
(b) −42
(c) 48
(d) −48

Answer 1

(b) −42

$$\begin{gathered} x=t^3-12t^2+6t+8 \\ \Rightarrow \frac{dx}{dt}=3t^2-24t+6 \\ \Rightarrow \frac{d^{2}x}{d{t}^2}=6t-24 \\ \Rightarrow 6t-24=0\left[\because \mathrm{acceleration}\mathrm{is}\mathrm{zero}\right] \\ \Rightarrow t=4 \\ \mathrm{So}, \\ \mathrm{V}\text{elocity at}\mathit{\text{t}}\text{=4} \\ \Rightarrow \frac{dx}{dt}=3{\left(4\right)}^2-24\times 4+6 \\ \Rightarrow \frac{dx}{dt}=48-96+6 \\ \Rightarrow \frac{dx}{dt}=-42 \end{gathered}$$