Question

A boat downstream from A to B takes 7 hours, upstream from B to 9 hours. How much time does a log take from A to B? (Driftwood along the water velocity)

Answer 1

The log will make the trip from A to B in 63 hrs.

Explanation:

Let s be the distance from A to B, the velocity of the river be `v_r`, and the velocity of the boat in still water be `v_b`.

Analyse downstream travel:

The distance `s = (v_b + v_r)*7 hrs`

Analyse downstream travel:

The distance `s = (v_b - v_r)*9 hrs`

We have 2 expressions for s, so we can say that those 2 expressions are equal to each other.

`(v_b + v_r)*7 hrs = (v_b - v_r)*9 hrs`

Let's simplify that and solve for `v_b`.

`7 hrs*v_b + 7 hrs*v_r = 9 hrs*v_b - 9 hrs*v_r`

`2 hrs*v_b = 16 hrs*v_r`

`v_b = (16 hrs)/(2 hrs) * v_r = 8 * v_r`

The floating log will travel at the velocity `v_r`. Now let's plug `8 * v_r` into the first equation in place of `v_b`.

`s = (8*v_r + v_r)*7 hrs = 9*v_r*7 hrs`

`s = v_r*63 hrs`

That relationship can be rewritten `v_r = s/63 hrs`

That says that `v_r` is such that the log will go the distance s in 63 hrs.

I hope this helps,
Steve