A particle is moving in a straight line. At time $t$, the distance from its starting point is given by $x = t - 6t^2 + t^3$. Its acceleration will be zero at

✅ Correct Answer: $t = 2$ units

Explanation

Velocity $v = \frac{dx}{dt} = 1 - 12t + 3t^2$. Acceleration $a = \frac{dv}{dt} = -12 + 6t$. Setting $a=0$ gives $t=2$.

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