Two ferries, Alpha and Beta, set off simultaneously towards each other from two harbors that are 240 kilometers apart. Ferry Alpha travels at a speed of 14 km/h in still water, while Ferry Beta travels at 10 km/h in still water. If Alpha moves downstream and Beta moves upstream, after how many hours will they meet?

The correct answer is 10 hours.

Step-by-step explanation:

1. Understand the movement of both ferries:

Let the speed of the river current be $v$ km/h.

Ferry Alpha moves downstream, so the river current increases its speed relative to the riverbank:

$\text{Speed of Ferry Alpha}=14+v\text{km/h}$

Ferry Beta moves upstream, so the river current decreases its speed relative to the riverbank:

$\text{Speed of Ferry Beta}=10-v\text{km/h}$

2. Calculate the relative speed of approach:

Since the two ferries travel towards each other, their relative speed of approach is the sum of their individual speeds relative to the riverbank:

$\text{Relative Speed}=(14+v)+(10-v)$

$\text{Relative Speed}=14+10=24\text{km/h}$

Notice that the current speed $v$ cancels out completely. The effective rate at which the gap between the two ferries decreases depends solely on their speeds in still water.

3. Calculate the time until they meet:

The total distance between the two harbors is 240 kilometers. The time taken to meet is given by the relation:

$\text{Time}=\frac{\text{Distance}}{\text{Relative Speed}}$

Substituting the values:

$\text{Time}=\frac{240}{24}=10\text{hours}$

Therefore, the two ferries will meet after 10 hours.