Problema Solution
A motorboat can travel 15 mph in still water. On a particular day, it took 15 minutes longer to travel a distance of 5 miles upstream than it took to travel the same distance downstream. What was the rate of the current in the stream that day?
Answer provided by our tutors
let
v = 15 mph the speed of the boat in still water
c = the current of the stream
d = 5 miles the distance traveled in one direction
t = the time of the travel downstream
t + 15/60 = the time of the travel upstream
the speed upstream is: v - c
the speed downstream is: v + c
since speed = distance/time => time*speed = distance
t(v + c) = d
t(15 + c) = 5
t = 5/(15 + c)
(t + 15/60)(v - c) = d
(t + 1/4)(15 - c) = 5
lets plug t = 5/(15 + c) into the last equation
(5/(15 + c) + 1/4)(15 - c) = 5
by solving we find and consider the positive root
c = 5 mph
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the rate of the current in the stream that day was 5 mph.