Two cars, $A$ and $B$, start from one city to another. In the first 5 minutes, they travel the same distance. At this point, due to a mechanical issue, $B$ has to reduce its speed to 2/5 of its original speed, and as a result, it arrives 15 minutes after the steadily traveling $A$. If the malfunction had occurred 4 km further, $B$ would have arrived only 10 minutes after $A$. How far apart are the two cities?
2. Task. Two cars, "A" and $B$, start from one city to another. In the first 5 minutes, they cover the same distance. Then, due to a motor failure, $B$ has to reduce its speed to $2/5$ of its original speed, and thus arrives 15 minutes after the still steadily moving "A". If the failure had occurred 4 km further, $B$ would have arrived only 10 minutes after "A". How far apart are the two cities?

I. Solution. Let the starting point of the cars be $I$, the destination $C$, the point of failure $H$, and the hypothetical point of failure $H^{*}$. Let the speed of $A$ be $v \mathrm{~km}$ per minute, and the total time required for $A$ to travel from $I$ to $C$ be $x$ minutes. Then, $B$, after the first 5 minutes, needs $(5x-25)/2$ minutes to cover the remaining distance, which is 15 minutes more than the time $A$ takes to reach $C$, so

\[
\frac{5 x-25}{2}=x-5+15
\]

From this, the travel time of $A$ is

\[
x=15 \text{ minutes}
\]

and the total distance is $15v \mathrm{~km}$.

If $B$'s motor failure occurs 4 km further, then the distance covered until the failure is $I H^{*}=5v+4 \mathrm{~km}$. The remaining distance $H^{*} C=15v-(5v+4)=10v-4 \mathrm{~km}$. $A$ covers this distance in $(10v-4)/v$ minutes, while $B$ covers it in

\[
\frac{10 v-4}{2 v / 5}=\frac{25 v-10}{v}
\]

minutes. The latter is 10 minutes longer than the former, so

\[
\frac{25 v-10}{v}=\frac{10 v-4}{v}+10
\]

from which $v=1.2 \mathrm{~km} /$ minute.

Now, the distance between the two cities is $v \cdot x=1.2 \cdot 15=18 \mathrm{~km}$.

According to this, $B$'s motor failure occurred after $I H=1.2 \cdot 5=6 \mathrm{~km}$, and there was still $H C=12 \mathrm{~km}$ left to cover. $B$'s new speed became $0.48 \mathrm{~km} /$ minute, and with this speed, the $H C$ segment was covered in $12:0.48=25$ minutes. The total travel time was $5+25=30$ minutes, which is indeed 15 minutes later than $A$. - If the failure had occurred only $6+4=10 \mathrm{~km}$ later, then $B$'s total travel time would similarly be

\[
\frac{10}{1.2}+\frac{8}{0.48}=\frac{25}{3}+\frac{50}{3}=25 \text{ minutes }
\]

and this is indeed only 10 minutes longer than $A$'s time.

II. Solution. It is sufficient to talk only about car $B$. This car would have completed the entire journey 15 minutes earlier if it had maintained its initial speed throughout, instead of reducing its speed to $2/5$ of its original speed after 5 minutes, - and 10 minutes earlier if this speed reduction had occurred 4 km further. According to this, it takes 5 minutes longer to cover 4 km at the reduced speed than at the original speed. Thus, the 15-minute delay occurs over a 12 km distance, so after the first 5 minutes of driving, 12 km still needed to be covered.

It is easy to calculate the time required for the distance covered after the first 5 minutes. This distance was covered in $5/2$ times the time it would have taken at the original speed, so the delay is $3/2$ times (one and a half times) the originally required time. At the original speed, this distance would have been covered in 10 minutes.

Since we have determined that this distance is 12 km, in the first 5 minutes $B$ covered 6 km; thus, the total distance was 18 km.

![](https://cdn.mathpix.com/cropped/2024_05_02_3ad3f3495d7d6f33ee4eg-1.jpg?height=574&width=748&top_left_y=2139&top_left_x=686)

Remark. Knowing the above results, we can draw the graph of the movements. The movement of $A$ is represented by the straight line segment $O D$, the movement of $B$ by the broken line segment $O T E$, and the hypothetical variant by the broken line segment $O T^{*} E^{*}$. However, we can also arrive at this diagram independently of the previous results, only the determination of the unit on the distance axis must be postponed, or rather, this will be the task of the problem.

The movement of $A$ is represented by a slanted line $a$ starting from $O$ (at any inclination). The point above the 5-minute mark $T^{\prime}$ on the time axis is $T$. By constructing the point $U$ on the segment $T^{\prime} T$ such that $T^{\prime} U: T^{\prime} T = 2:5$, we get the direction of $B$'s reduced speed in the direction of $O U$; if $B$ had started with the reduced speed immediately, its movement would have been represented by the line $O U$. Thus, $b$ is given by the line parallel to $O U$ passing through $T$.

Now imagine a third car $A^{\prime}$ moving identically to $A$ but starting later from $I$ and catching up with $B$ exactly at $C$. Since $B$ is 15 minutes late at $C$ compared to $A$, this is also true for $A^{\prime}$, so the graph $a^{\prime}$ of $A^{\prime}$'s movement starts from the 15-minute mark $V$ on the time axis and is parallel to $a$. Thus, the point $E$ is determined by the intersection of $b$ and $a^{\prime}$, and the point $C$ on the distance axis is determined by the line parallel to the time axis passing through $E$. (Of course, $D$ is also on this line.) - Similarly, we get the graph $a^{\prime \prime}$ of the hypothetical fourth car $A^{\prime \prime}$ moving identically to $A$ and arriving at $C$ with $B$ if the motor failure had occurred later: this is parallel to $a$ and starts from $W$, where $O W$ represents the 10-minute delay due to the later failure. Thus, $E^{*}$ is given by the intersection of $a^{\prime \prime}$ and the line $E C$, and from $E^{*}$, we can draw the graph $b^{*}=E^{*} T^{*}$ of the movement $B$ would have completed from the later failure to the destination ( $T^{*}$ is where $b^{*}$ intersects $a$).

Since the distance between the projections $H$ and $H^{*}$ of $T$ and $T^{*}$ on the distance axis is 4 km, constructing one quarter of $H H^{*}$ gives us the unit of the distance axis (1 km). Measuring the segment $O C=I C$ with this unit, we get the desired distance.

The diagram also shows the visual equivalents of the calculations in the second solution. $\triangle D T E \sim \triangle D T^{*} E^{*}$, so $D T^{*} : D T = D E^{*} : D E = 10 : 15$, thus $D T^{*} = 2 D T / 3$, $T^{*} T = D T / 3$; furthermore, $\triangle I H T \sim \triangle I H^{*} T^{*} \sim \triangle I C D$, so $C H : H^{*} H = D T : T^{*} T$, $C H = 3 \cdot 4 = 12 \mathrm{~km}$. Additionally, if the projection of $T$ on $E C$ is $T^{\prime \prime}$, then $T^{\prime \prime} E = 5 T^{\prime \prime} D / 2$, $D E = 3 T^{\prime \prime} D / 2 = 15$ minutes, so $T^{\prime \prime} D = 10$ minutes.