3.18 The distance between points $A$ and $B$ is 270 m. A body moves uniformly from $A$ to $B$; upon reaching $B$, it immediately returns with the same speed. A second body, which leaves $B$ for $A$ 11 s after the first body leaves $A$, moves uniformly but more slowly. On its way from $\boldsymbol{B}$ to $\boldsymbol{A}$, it meets the first body twice: 10 and 40 s after its departure from $B$. Find the speed of each body.
3.18 A convenient model of the problem is a graph of uniform motion in the coordinate system "path" ($s$ - in meters), "time" ($t$ in seconds). Let $AC$ (Fig. 3.3) be the graph of the motion from $A$ to $B$ of the first body with speed $v_{1}=\operatorname{tg} \alpha$ (time axis $At$); $CD$ be the graph of the motion from $B$ to $A$ of the same body with the same speed $v_{1}=\operatorname{tg} \alpha$ (time axis $Bt$); $\boldsymbol{EF}$ be the graph of the motion of the second body from $B$ to $\boldsymbol{A}$ with speed $v_{2}=\operatorname{tg} \beta, \beta<\alpha$ (time axis $Bt$).

The time interval $BE=11$, the time interval to the first meeting $EH=10$, between the first and second meetings $HK=30$; then $NM=21 v_{1}, HM=10 v_{2}$, $KF=40 v_{2}, NM+HM=AB=270$, i.e.,

$21 v_{1}+10 v_{2}=270$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-099.jpg?height=441&width=757&top_left_y=95&top_left_x=229)

Fig. 3.3

The time interval $HC=\frac{HM}{v_{1}}=\frac{10 v_{2}}{v_{1}}$; the time interval $CK=\frac{KF}{v_{1}}=\frac{40 v_{2}}{v_{1}}$. Since $HC+CK=30$, then

$\frac{10 v_{2}}{v_{1}}+\frac{40 v_{2}}{v_{1}}=30$, hence

$5 v_{2}=3 v_{1}$.

Solving equations (1) and (2) simultaneously, we find $v_{1}=10 \mathrm{m} / \mathrm{s}$, $v_{2}=6 \mathrm{m} / \mathrm{s}$.

Answer: 10 and $6 \mathrm{~m} / \mathrm{s}$.