2. A point moving in the positive direction of the $O x$ axis has a horizontal coordinate given by $x(t)=5(t+1)^{2}+\frac{a}{(t+1)^{5}}$, where $a$ is a positive constant. Find the minimum value of $a$ that satisfies $x(t) \geqslant 24$ for all $t \geqslant 0$.
2. Using the arithmetic-geometric mean inequality, we have
$x(t)=\underbrace{(t+1)^{2}+\cdots+(t+1)^{2}}_{5}+\frac{a}{2(t+1)^{5}}+\frac{a}{2(t+1)^{5}} \geqslant 7 \sqrt[7]{\frac{a^{2}}{4}}$, with equality if and only if $(t+1)^{2}=$ $\frac{a}{2(t+1)^{5}}$. Therefore, $7 \sqrt[7]{\frac{a^{2}}{4}} \geqslant 24$, which implies $a \geqslant 2 \sqrt{\left(\frac{24}{7}\right)^{7}}$.

On the other hand, when $a=2 \sqrt{\left(\frac{24}{7}\right)^{7}}, t=\sqrt{\frac{24}{7}}-1$, we have $x(t)=24$, thus the minimum value of $a$ is $2 \sqrt{\left(\frac{24}{7}\right)^{7}}$.