13.393 The distance between two points is $d$. Under the influence of certain forces, both points start moving uniformly towards each other. To meet at the midpoint of the path, the first point needs to start moving $t$ units of time earlier than the second. If the points start moving towards each other simultaneously, then after $T$ units of time, the distance between them will be $k$-th part ($k>1$) of the initial distance. Find the speeds of the points.
Solution. Let $x$ and $y$ be the speeds of the points. According to the first condition of the problem, $\frac{d}{2 x}-t=\frac{d}{2 y}$. According to the second condition, $T(x+y)=\frac{(k-1)}{k} d$. From these two equations, we have: $y^{2}-y(a-b)-\frac{a b}{2}=0$, where $a=\frac{(k-1) d}{k T}, b=\frac{d}{t} \Rightarrow$ $\Rightarrow y=\frac{a-b}{2}+\sqrt{\frac{a^{2}+b^{2}}{4}}=\frac{a}{2}-\frac{b}{2}\left(1-\sqrt{1+\frac{a^{2}}{b^{2}}}\right) \Rightarrow x=a-y=\frac{a}{2}+\frac{b}{2}\left(1-\sqrt{1+\frac{a^{2}}{b^{2}}}\right)$. Answer: $\frac{k-1}{2 k T} d \pm \frac{d}{2 t}\left(1-\sqrt{1+\frac{(k-1)^{2}}{k^{2} T^{2}} t^{2}}\right)$.