Carcass wrote:
The Lirr River runs from Rosedale in the west to Oceanside in the east with the current moving at an average of 10 miles per hour. Sasha is traveling by motorboat from Oceanside to Rosedale and back. If the water were not moving, Sasha’s motorboat would travel at an average speed of 20 miles per hour. Given the current, what was Sasha’s actual average speed, in miles per hour, for the round trip?
Considering the current:
Speed downstream (D) = speed of boat + Speed of current = 20 + 10 = 30 miles/hr
Speed upstream (U) = speed of boat - Speed of current = 20 - 10 = 10 miles/hr
Since in the round-trip, distance traveled downstream = distance traveled upstream, we have:
Average speed = \(\frac{2*D*U}{(D + U)} = \frac{2*30*10}{(30 + 10)}\) =
15 miles/hrAlternate approach: Let distance = D miles
Time taken to travel downstream = D/30 hr
Time taken to travel upstream = D/10 hr
Total time = (D/30 + D/10) = 4D/30 hr
Total distance = D + D = 2D miles
=> Average speed = \(\frac{2D}{(4D/30)}\) = 15 miles/hr
[We could also have chosen some suitable value of D (say 300 miles) and performed the above calculation]