Question

A particle moves along the x-axis so that at time t ≥ 0 its position is given by x(t) = -t^3 + 6t^2 + 63t. Determine all intervals when the acceleration of the particle is negative.

Answer 1

A particle moves along the x-axis so that at time t ≥ 0 its position is given by x(t) = -t^3 + 6t^2 + 63t. Determine all intervals when the acceleration of the particle is negative.

Solution :

$x(t) =-t^3 + 6t^2 +63t, \quad t\geq0 \\ \text{Velocity} \\ v(t) = \frac{dx}{dt} = -3t^2+12t+63 \\ \text{Acceleration} \\ a = \frac{dv}{dt} = -6t+12 \\ a<0 \quad when \quad -6t+12>0 \rightarrow t>2 \\ \text{Answer : acceleration is negative in } (2,+\infty)$