# Problem: The maximum acceleration attained on the interval 0&lt;=t&lt;=3 by the particle whose velocity is given by v(t) = t3 - 3t2 + 12t + 4 is?Is the derivate acceleration or the second derivative?

###### FREE Expert Solution

We are given an equation with some powers of t. We recognize that this is a kinematics problem with calculus.

We'll need the chart below showing the relationship between acceleration, velocity, and position equations.

2D&3D Motion with Calculus:

$\overline{){\mathbit{P}}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}\underset{\frac{\mathbit{d}}{\mathbit{d}\mathbit{t}}}{\overset{{\mathbf{\int }}{\mathbit{d}}{\mathbit{t}}}{\mathbit{V}}}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}{\mathbit{A}}}$

Power rule of derivation:

$\overline{)\frac{\mathbit{d}}{\mathbit{d}\mathbit{t}}\mathbf{\left(}\mathbit{a}{\mathbit{t}}^{\mathbit{n}}\mathbf{\right)}{\mathbf{=}}{\mathbit{n}}{\mathbf{·}}{\mathbit{a}}{{\mathbit{t}}}^{\mathbit{n}\mathbf{-}\mathbf{1}}}$

95% (369 ratings) ###### Problem Details

The maximum acceleration attained on the interval 0<=t<=3 by the particle whose velocity is given by v(t) = t3 - 3t2 + 12t + 4 is?

Is the derivate acceleration or the second derivative?