so we have s(t), namely the positional equation for the particle.
now, let's recall that the derivative of s(t) will be v(t), namely the derivative of the positional equation will be the velocity, and the derivative of the velocity will be the acceleration.
s'(t) = v(t) and s''(t) = a'(t) <--- acceleration
so we need the 2nd derivative of s(t) firstly, and then we can check for any critical values and find its extrema.
[tex]\bf s(t)=t^4-rt^3+6t^2-20\implies \stackrel{\textit{velocity}}{s'(t)=4t^3-12t^2+12t} \\\\\\ \stackrel{\textit{acceleration}}{s''(t)=12t^2-24t+12}\implies \stackrel{\textit{derivative of acceleration}}{s'''(t)=24t-24} \\\\[-0.35em] ~\dotfill\\\\ 0=24t-24\implies 24=24t\implies \cfrac{24}{24}=t\implies 1=t[/tex]
now, if we run a first-derivative test on that acceleration derivative, namely s'''(t), let's say we test s'''(0) and that gives us a negative value, and then test s'''(2) and that gives us a positive value, so the test goes like in the picture below, clearly, 1 = t is a minumum, and is between [0, 3].
s'''(0) = -24
s'''(3) = 48
so clearly the maximum happens at the s'''(3) endpoint, when t = 3.
