Orson rides his power boat up and down a canal. the water in the canal flows at 6 miles per hour. orson takes 5 hours longer to travel 360 miles against the current than he does to travel 360 miles with the current. what is the speed of orson's boat?

To solve this problem, let us say that:

v1 = the speed or velocity of Orson in travelling against the current

t1 = the time taken in travelling against the current

v2 = the speed or velocity of Orson in travelling with the current

t2 = the time taken in travelling with the current

It was stated in the problem that the time difference in travelling against the current and travelling with the current is 5 hrs, therefore:

t1 – t2 = 5

We know that:

t = d / v

Therefore:

(360 / v1) – (360 / v2) = 5

However we also know that:

v1 = v – 6

v2 = v + 6

where v is the velocity of the boat alone

[360 / (v – 6)] – [360 / (v + 6)] = 5

Multiplying everything by (v – 6) * (v + 6):

360 (v + 6) – 360 (v – 6) = 5 (v – 6) (v + 6)

360 v + 2160 – 360 v + 2160 = 5 v^2 – 180

4320 = 5 v^2 – 180

5 v^2 = 4500

v^2 = 900

v = 30

Therefore the speed of Orson’s boat is 30 miles per hour