Please explain part (d) and (e)

Dear Student,
Answer to your question is given below:
$$\begin{gathered} Given, \\ x\left(t\right) = \frac{t^3}{3} - \frac{5t^2}{2} +6t + 4 \\ Now, we know that, velocity, v\left(t\right) = \frac{dx\left(t\right)}{dt} = \frac{3t^2}{3} - \frac{5\times 2t}{2} + 6 = t^2 - 5t+ 6 \\ Now, acceleration, a\left(t\right) = \frac{dv\left(t\right)}{dt} = 2t - 5 \\ Now at the time the particle changes its direction, a\left(t\right)=0 \\ Hence, 2t - 5=0 \Rightarrow t = \frac{5}{2} seconds \Rightarrow t = 2.5 seconds \\ \left(d\right) Time at which particle changes its direction is 2.5 seconds. \end{gathered}$$
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Regards,