The distance x covered by a particle in time | vedclass

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(B) Given the displacement equation: $x = t^3 - 12t^2 + 6t + 8$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac

Vedclass · Accepted Answer

$-42$

Vedclass · Answer

(B) Given the displacement equation: $x = t^3 - 12t^2 + 6t + 8$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = 3t^2 - 24t + 6$. Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 24$. At the instant when acceleration is zero: $a = 0 \Rightarrow 6t - 24 = 0 \Rightarrow t = 4 	ext{ s}$. Now,substitute $t = 4$ into the velocity equation: $v = 3(4)^2 - 24(4) + 6$. $v = 3(16) - 96 + 6 = 48 - 96 + 6 = -42 	ext{ units/s}$.

The distance $x$ covered by a particle in time $t$ is given by $x = t^3 - 12t^2 + 6t + 8$. What is the velocity of the particle at the instant when its acceleration is zero?

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