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

विकल्प

• 42

• −42

•  48

• −48

Answer 1

−42

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