A body moves along the Ox axis according to x = (t³)/3

A body moves along the $Ox$ axis according to $x = \dfrac{t^{3}}{3}-\dfrac{3}{2}t^{2}-4t$. Find the velocity and acceleration, and determine the moments at which the body changes direction.

Step 1: Velocity and acceleration

\[ v(t) = \frac{dx}{dt} = t^{2}-3t-4, \qquad w(t) = \frac{dv}{dt} = 2t-3. \]

Step 2: Where the velocity vanishes

A reversal requires $v = 0$ with a change of sign: \[t^{2}-3t-4 = 0 \;\Longrightarrow\; \bigl(t-4\bigr)\bigl(t+1\bigr) = 0 \;\Longrightarrow\; t = -1 \text{ or } t = 4.\]

Only $t = 4$ is admissible, since $t\ge0$.

Step 3: Sign of the velocity

\[ \begin{array}{c|c|c|c} & 0\le t\lt 4 & t = 4 & t>4 \\ \hline t+1 & + & + & + \\ t-4 & - & 0 & + \\ \hline v & - & 0 & + \end{array} \]

The body first moves in the negative direction, halts at $t = 4$, and then moves in the positive direction. That is a genuine reversal, since the sign actually changes.

Step 4: Verification

\[v(0) = -4,\quad v(3) = -4,\quad v(4) = 0,\quad v(5) = 6 \;\;\checkmark\]

\[x(0) = 0,\quad x(4) = \frac{64}{3}-24-16 = -\frac{56}{3} \approx -18.67 \text{ (the leftmost point)} \;\;\checkmark\]

\[w(4) = 5>0, \text{ confirming that } t = 4 \text{ is a minimum of } x \;\;\checkmark\]