The distance \( |AB| \) traveled by a vehicle with an average speed V in time t is expressed as:

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

\[ |AB| = x = V \times t \]

Here, if \(t_y= 1 \) hour is taken, it becomes \( |AB | = V_A- V_B \). From this, the following conclusion can be drawn: the vehicle behind closes the distance between them by the difference in speeds every 1 hour.

Example:

As shown in the figure, when two moving objects from points A and B with average speeds \( V_a\) and \( V_b \) start moving simultaneously towards each other (in opposite directions), let them meet at point C. Since both moving objects reach point C in the same duration \( (t_k) \), the distances covered by the objects moving from A and B are written as:

\[ |AC| = V_a \cdot t_k \]

\[ |BC| = V_b \cdot t_k \]

Here, if \(t_k= 1 \) hour is taken, it becomes \( |AB | = V_A + V_B \). From this, the following conclusion can be drawn: two moving objects approach each other by the sum of their speeds every 1 hour.

Let’s examine the motion of vehicles A and B, whose speeds are given in the figure, starting simultaneously from points A and B.

1) The distance between the two vehicles is closed after \[\frac{480}{90 + 70 } = 3 \; \text{hours.} \] The two vehicles meet 3 hours later.

2) In 3 hours, vehicle A covers \( 3 \cdot 90 = 270 \) km. That is, |AC| = 270 km.

3) In 3 hours, vehicle B covers \( 3 \cdot 70 = 210 \) km. That is, |BC| = 210 km.

The ratio (division) of the total distance covered by a vehicle between two points to the total time spent completing this distance is the average speed of the vehicle between these two points.

Accordingly, it is calculated as:

\[ V_{ort} = \frac{\text{total distance} }{\text{total time} } \]

.

As shown in the figure above, let a vehicle moving from A cover the path AB with a speed of \(V_1 \) in time \(t_1 \), the path BC with a speed of \(V_2 \) in time \(t_2 \), and the path CD with a speed of \(V_3 \) in time \(t_3 \). The average speed of this vehicle along the entire path AD is:

\[ V_{ort} = \frac{|AD|}{t_1 +t_2 + t_3} \] or \[ V_{ort} = \frac{|AD|}{\frac{|AB|}{V_1} + \frac{|BC|}{V_2} + \frac{|CD|}{V_3} } \] .

If a vehicle covers one-third of its path at a speed of 60 km/h and the remainder at a speed of 80 km/h, let’s find the average speed of this vehicle along this path.

Assuming the total distance covered by the vehicle is \(3x \) km:

\[ V_{ort} = \frac{\text{total distance} }{\text{total time}} = \frac{3x}{\frac{x}{60} + \frac{2x}{80} } = \frac{3x}{ x \cdot \left( \frac{1}{60} + \frac{1}{40} \right)} \]
\[ V_{ort} = 72 \; \text{km/h is found.} \]

\[
\text{A )} 550 \quad
\text{B) } 600 \quad
\text{C) } 675 \quad
\text{D) } 725 \quad
\text{E) } 750
\]

Let the vehicle with a speed of 90 km/h cover the distance |AB| in \( t\) hours. In this case, the vehicle with a speed of 50 km/h covers the distance between the two cities in \( t+ 6\) hours.

\[
\text{A )} 300 \quad
\text{B) } 250 \quad
\text{C) } 200 \quad
\text{D) } 150 \quad
\text{E) } 100
\]

Since the distance AB is 75 km, the vehicle starting from A would normally catch up with the vehicle starting from B after:

\[ \frac{75}{75-50} = 3 \] hours.

Since the vehicle starting from A takes a 1-hour break before catching up with the vehicle in front, the vehicle starting from B increased the distance between them by \(50 \cdot 1 = 50 \) km during this 1 hour. Accordingly, the total distance that the vehicle starting from A needs to close with the vehicle in front is \( 75 + 50 = 125\) km. Therefore, it will close this 125 km distance in:

\[\frac{125}{75-50} = 5 \] hours. In addition, during the 1-hour break taken by the vehicle starting from A, the vehicle starting from B traveled \( 50 \cdot 1 = 50\) km. Accordingly, the total distance covered by the vehicle starting from B is:

\[ |BC| = 50 \cdot 5 + 50 = 300 \; \text{km.} \]

\(\textbf{Answer: A} \)

The ratio of a runner’s speed to the wind speed is \(\Large \frac{5}{2} \). Since this runner can run a distance of 45 km against the wind in 5 hours, what is the runner’s speed per hour in km?

\[
\text{A )} 3 \quad
\text{B) } 9 \quad
\text{C) } 12 \quad
\text{D) } 15 \quad
\text{E) } 18
\]

If the wind speed is called \( 2v \), the runner’s speed becomes \( 5v \). When the runner runs against the wind, their net speed is \[ 5v \; – \; 2v = 3v \]. Since the relationship belonging to the given problem is:

\[ \text{Distance = Speed x Time} \] , we have: \[ 45 = 3v \cdot 5 \Rightarrow v= 3 \; \text{km/h} \]

A bus moving from point A at a speed of 60 km/h increases its speed by V hours after starting its journey, and consequently reaches point B 1 hour before its scheduled arrival time. Given that the distance AB is 240 km, how many km/h is V?

\[
\text{A )} 20 \quad
\text{B) } 25 \quad
\text{C) } 30 \quad
\text{D) } 35 \quad
\text{E) } 40
\]

A vehicle travels an AB distance at a speed of 80 km/h, and on its return, it covers the path at a speed of 60 km/h. Given that the total round-trip time is 14 hours, how many km is the distance AB?

\[
\text{A) } 480 \quad
\text{B) } 460 \quad
\text{C) } 420 \quad
\text{D) } 300 \quad
\text{E) } 350
\]

Solution:

If the departure time of the vehicle is called t, the return time becomes \(14 – t \). Since the vehicle travels at a speed of 80 km/h on departure and 60 km/h on return, the equation belonging to the given problem is written as:

\[ |AB| = 80 \cdot t = 60 \cdot (14 \; – \; t )\]

\[ 14t = 6 \cdot 14\]

\[\Rightarrow t = 6 \; \text{hours is found.} \]

From here, since \[ |AB| = 80 \cdot t \] :

\[ |AB| = 80 \cdot 6 = 480 \; \text{km.} \]

\(\textbf{Answer: A} \)

\[
\text{A) } 420 \quad
\text{B) } 480 \quad
\text{C) } 540 \quad
\text{D) } 680 \quad
\text{E) } 700
\]

Since the two vehicles moving towards each other with speeds \( V_1 \) and \( V_2 \) meet after 3 hours, the distance AB is:

\[ |AB| = 3 \cdot (V_1 + V_2) \Rightarrow |AB| = 3 \cdot (80 + 60 ) = 420 \; \text{km.} \] If the vehicle with speed \( V_1 \) catches up with the vehicle with speed \( V_3 \) in t hours:

\[ t = \frac{|AB|}{V_1 \;- \; V_3} \Rightarrow t = \frac{420}{80 \;-\; 50 } = 14 \; \text{hours.} \]