A certain ferry moves up and down a river between Town A and B. It takes the ferry two hours to travel to Town A and only an hour and thirty minutes to return to Town B. If the current is 5mph how far apart are the two cities?

The current of the river is towards city B; hence, the reason the ferry moves faster from town A to B than when coming back from town B to A. The distance between both towns is 60m.

Let

[tex]s \to[/tex] speed in still water

[tex]d \to[/tex] distance between both cities

The given parameters are:

[tex]t_{AB} =2[/tex] -- from A to B

[tex]t_{BA} =1.5[/tex] -- from B to A

[tex]s_c = 5mph[/tex] -- speed of the current

Speed is calculated as:

[tex]Speed = \frac{Distance}{Time}[/tex]

The speed (s) from town A to town B is:

[tex]s = s_c + \frac{d}{t_{AB}}[/tex] --- i.e. speed of the current + speed in still water from A to B

[tex]s = 5 + \frac{d}{2}[/tex]

Multiply through by 2

[tex]2s = 10 + d[/tex]

Make d the subject

[tex]d = 2s - 10[/tex]

The speed (s) from town B to town Ais:

[tex]s = -s_c + \frac{d}{t_{BA}}[/tex] --- i.e speed in still water from B to A - . speed of the current

[tex]s = -5 + \frac{d}{1.5}[/tex]

Multiply through by 1.5

[tex]1.5s = -7.5 + d[/tex]

Make d the subject

[tex]d = 1.5s + 7.5[/tex]

So, we have:

[tex]d = 1.5s + 7.5[/tex] and [tex]d = 2s - 10[/tex]

Equate both values of d

[tex]2s - 10 = 1.5s + 7.5[/tex]

Collect like terms

[tex]2s - 1.5s= 10 + 7.5[/tex]

[tex]0.5s= 17.5[/tex]

Divide both sides by 0.5

[tex]s = 35[/tex]

Substitute [tex]s = 35[/tex] in [tex]d = 1.5s + 7.5[/tex] to calculate distance (d)

[tex]d =1.5 *35 + 7.5[/tex]

[tex]d =52.5 + 7.5[/tex]

[tex]d =60[/tex]

The distance between town A and B is 60m.