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:

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

Power rule of derivation:

$\overline{)\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathbf{\left(}\mathit{a}{\mathit{t}}^{\mathit{n}}\mathbf{\right)}{\mathbf{=}}{\mathit{n}}{\mathbf{\xb7}}{\mathit{a}}{{\mathit{t}}}^{\mathit{n}\mathbf{-}\mathbf{1}}}$

The maximum acceleration attained on the interval 0<=t<=3 by the particle whose velocity is given by v(t) = t^{3} - 3t^{2} + 12t + 4 is?

Is the derivate acceleration or the second derivative?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Kinematics in 2D concept. You can view video lessons to learn Kinematics in 2D. Or if you need more Kinematics in 2D practice, you can also practice Kinematics in 2D practice problems.