Aboat takes 3.6 h to travel downstream from its dock to a fishing hole. if the water were still, it would have taken 4.5 h to make the trip. let x represent the speed of the boat and let y represent the speed of the water. (a) write an expression for the distance traveled in moving water, using 3.6 h for the time. then write an expression for the distance traveled as if in still water, using 4.5 h for the time. (b) set the expressions in part (a) equal to each other. then solve the equation for y. show your work. (c) what percent of the boat’s speed is the water current?

Given:

t1 = 3.6 h
t2 = 4.5 h
x = speed of boat
y = speed of water

Required:
a) Expression of distance traveled with moving water with 3.6h
     Expression of distance traveled with moving water with 4.5h
b) Solve for y
c) Percent of boat's speed is the water current

Solution:

Working formula: distance = velocity*time

a) For travelling downstream, we get the equation

d = (x +y)*3.6

For travelling upstream, we get the equation

d = (x-y)*4.5

b) Setting the distance as equal for travelling upstream or downstream, we arrive at the equation of

(x+y)*3.6 = (x-y)*4.5
3.6x + 3.6y = 4.5x - 4.5y
8.1y =0.9x
y = x/9

c) percentage = 1/9*100% = 11.1%

ANSWERS: a) d = (x+y)*36; d = (x-y)*4.5
                     b) y = x/9
                     c) 11.1% 

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