goi gia toc doan tac la abt chuyen dong cua toa tau dau tien la
$S=\frac{at_1^2}2=>t_1=\sqrt{\frac{2S}a}$
bt chuyen dong cua toa tau o toa thu $n-1$ va $n$
$t_2=\sqrt{\frac{2(n-1)S}a}$ $t_3=\sqrt{\frac{2nS}a}$ =>$\Delta t=\sqrt{\frac{2S}a}(\sqrt n-\sqrt{n-1})=t_1(\sqrt n-\sqrt{n-1})$