39. Ship $K_{1}$ notices ship $K_{2}$, which is at a distance $d$ from $K_{1}$ and is moving with a speed $v$ in a direction,

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-024.jpg?height=388&width=488&top_left_y=1063&top_left_x=131)

Fig. 8. perpendicular to the line $K_{1} K_{2}$ (Fig. 8). Ship $K_{1}$ needs help and therefore wants to draw the attention of ship $K_{2}$, which, however, does not notice $K_{1}$ and continues on the same course. How close can ship $K_{1}$ get to ship $K_{2}$ if

a) the speed of $K_{2}$ is also $v$ and the captain of ship $K_{1}$ at each moment in time directs it straight towards $K_{1} K_{2}$ (which, of course, changes continuously during the movement of both ships);

b) the speed of ship $K_{1}$ is $u < v$, and the captain of $K_{1}$ immediately assumes that the direction of movement of ship $K_{2}$ will not change, and chooses the course of his ship accordingly?
39. a) Let's denote by $\alpha$ the angle between the course of ship $K_{1}$ and (constant) course of ship $K_{2}$ at some moment in time $t$; it is clear that the angle $\alpha$ will change over time. At the moment $t$, ship $K_{1}$ is approaching $K_{2}$ at a speed $v$; as for ship $K_{2}$, it is moving away from $K_{1}$ in the direction $K_{1} K_{2}$ at a speed $v \cos \alpha$ (this is the projection of the speed $\overline{K_{2} T_{2}}$ of ship $K_{2}$ onto the direction $K_{1} K_{2}$; see Fig. 101); thus, the approach of the ships at the moment $t$ occurs at a speed $v - v \cos \alpha = v(1 - \cos \alpha)$, and as time goes on, as the angle $\alpha$ decreases, the speed of approach of the ships becomes smaller. Note now that the projection $P$ of ship $K_{1}$ onto the path $i$ of ship $K_{2}$ "chases" ship $K_{2}$ at a speed $v \cos \alpha$ (this is the projection of the speed $\overline{K_{1} T_{1}}$ of ship $K_{1}$ onto the direction of the line $l$); ship $K_{2}$, on the other hand, moves away from point $P$ at a speed $K_{2} T_{2} = \nu$. Thus, the segment $P K_{2}$ increases at the same speed $v - v \cos \alpha = v(1 - \cos \alpha)$ at which the distance between the ships decreases, from which it follows that the sum $K_{1} K_{2} + P K_{2}$ remains constant at all times - it is equal to $d$, since $d$ is equal to this sum at the initial moment (Fig. 8). The closer ship $K_{1}$ approaches the line $l$, the closer in magnitude the segments $K_{1} K_{2}$ and $P K_{2}$ become (with $K_{1} K_{2}$ always being greater than $P K_{2}$). Thus, the distance $K_{1} K_{2}$ always remains greater than $d/2$; however, with the increase of $t$, it approaches $d/2$ indefinitely.

b) Let $K_{1}^{\prime}$ and $K_{2}^{\prime}$ be the positions of the ships at the moment when the distance between them reaches the smallest possible value (Fig. $102, a$). It is clear that ship $K_{1}$ has arrived at point $K_{1}^{\prime}$ along a straight path $K_{1} K_{1}^{\prime}$: otherwise, it could have shortened the travel time by moving along the segment $K_{1} K_{1}^{\prime}$. Moreover, it is natural to assume that ship $K_{1}$ arrives at point $K_{1}^{\prime}$ along the line $K_{1}^{\prime} K_{2}^{\prime}$, since if this were not the case, the captain of ship $K_{1}$ could have, by choosing the route $K_{1} K_{2}^{\prime}$ from the beginning, approached $K_{2}^{\prime}$ closer than it actually happened (see Fig. $102, \sigma$, where $K_{1}^{0} K_{1}^{\prime \prime} = K_{1}^{0} K_{1}^{\prime}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-128.jpg?height=396&width=441&top_left_y=922&top_left_x=604)

Fig. 101. and point $K_{1}^{\prime \prime}$ is closer to $K_{2}^{\prime}$ than point $K_{1}^{\prime}$).

Let the angle between the lines $K_{1} K_{1}^{\prime}$ and $K_{2} K_{2}^{\prime} = l$ be denoted by $\alpha$; if the angle between the lines $K_{1} K_{2}$ and $l$ is $\alpha^{\prime}$, where $K_{1}$ and $K_{2}$ are the positions of the ships at some moment in time $t$, then the speed $u \cos (\alpha^{\prime} - \alpha)$, at which ship $K_{1}$ approaches the second ship at the moment $t$, must be greater than the speed $v \cos \alpha^{\prime}$, at which $K_{2}$ moves away from $K_{1}$ at the same moment, and this must be true at any moment in time before the moment $t_{0}$ when the ships arrive at points $K_{1}^{\prime}$ and $K_{2}^{\prime}$ (the ships are approaching); it must be equal to the speed $v \cos \alpha^{\prime}$ at the moment $t_{0}$, when $\alpha^{\prime} = \alpha$ (the ships stop approaching); finally, the speed $u \cos (\alpha - \alpha^{\prime})$ must be less than the speed $v \cos \alpha^{\prime}$ at the moment $t$ following $t_{0}$ (the ships start moving away from each other).

From this we obtain (Fig. $102, a$):

$$
u = v \cos \alpha, \quad \text{i.e.} \quad \cos \alpha = \frac{u}{v}
$$

which determines the course that the captain of ship $K_{1}$ should choose.

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-129.jpg?height=44&width=990&top_left_y=271&top_left_x=211)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-129.jpg?height=360&width=456&top_left_y=329&top_left_x=223)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-129.jpg?height=373&width=486&top_left_y=330&top_left_x=700)
b)

Fig. 102.

If the ships $K_{1}$ and $K_{2}$ have traveled the same amount of time, then

$$
\frac{K_{1}^{0} K_{1}^{\prime}}{u} = \frac{K_{2}^{0} K_{2}^{\prime}}{v}, \text{ i.e. } K_{1}^{0} K_{1}^{\prime} = K_{2}^{0} K_{2}^{\prime} \cdot \frac{u}{v} = K_{2}^{0} K_{2}^{\prime} \cos \alpha
$$

And since, obviously, $K_{1}^{0} K_{2}^{\prime} = \frac{d}{\sin \alpha}$ and $K_{2}^{0} K_{2}^{\prime} = d \operatorname{ctg} \alpha$, we finally obtain

$$
\begin{aligned}
& K_{1}^{\prime} K_{2}^{\prime} = K_{1}^{0} K_{2}^{\prime} - K_{1}^{0} K_{1}^{\prime} = \frac{d}{\sin \alpha} - d \operatorname{ctg} \alpha \cdot \cos \alpha = \frac{d}{\sin \alpha} (1 - \cos^2 \alpha) = \\
& = d \sin \alpha = d \sqrt{1 - \cos^2 \alpha} = d \sqrt{1 - \left(\frac{u}{v}\right)^2} = \frac{d \sqrt{v^2 - u^2}}{v}
\end{aligned}
$$