解；设甲用时间为$T$，乙用时间为$2t$，步行速度为$a$，跑步速度为$b$，距离为$s$，则$T=\dfrac{s}{2a}+\dfrac{s}{2b}=\dfrac{sa+sb}{2ab}$，$ta+tb=s$，$\therefore 2t=\dfrac{2s}{a+b}$，
$\therefore T-2t=\dfrac{sa+sb}{2ab}-\dfrac{2s}{a+b}=s\times \left(\dfrac{a+b}{2ab}-\dfrac{2}{a+b}\right)=s\cdot \dfrac{\left(a-b\right)^{2}}{2ab\left(a+b\right)} \gt 0$，
$\therefore $乙先到达五四广场.
