SOLUTION: a boat travels 70 miles and then comes back, the current was 20 mph the total trip time was 4 hours and 40 minutes, how fast was the boat traveling
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Question 1077949: a boat travels 70 miles and then comes back, the current was 20 mph the total trip time was 4 hours and 40 minutes, how fast was the boat traveling Found 2 solutions by ankor@dixie-net.com, MathTherapy:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! a boat travels 70 miles and then comes back,
the current was 20 mph the total trip time was 4 hours and 40 minutes,
how fast was the boat traveling
:
let s = speed of the boat in relation to the water
then
(s-20) = effective speed upstream
and
(s+20) = effective speed downstream
:]
Change 4 hrs, 40 min to hrs: 4 + 40/60 = 4.67 hrs
:
Write a time equation; time = dist/speed
Time downstream + time upstream = 4 hr 40 min + = 4.67 hr
multiply equation by (s+20)(s-20), cancel the denominators
70(s-20) + 70(s+20) = 4.67(s-20)(s+20)
distribute
70s - 1400 + 70s + 1400 = 4.67(s^2 - 400)
140s = 4.67s^2 - 1866.667
Arrange as a quadratic equation
0 = 4.67s^2 - 140s - 1866.667
Using the quadratic formula, I got positive solution
s = 39.977 ~ 40 mph is boat speed
;
:
Check this find the actual time each way
70/20 = 3.5
70/60 = 1.167
--------------
total time 4.667 hrs which is about 4 hrs and 40 min
You can put this solution on YOUR website! a boat travels 70 miles and then comes back, the current was 20 mph the total trip time was 4 hours and 40 minutes, how fast was the boat traveling
Let boat's speed be S
Then we get the following TIME equation: _____
70(3)(S + 20) + 70(3)(S - 20) = 14(S - 20)(S + 20) -------- Multiplying by LCD, 3(S - 20)(S + 20)
(S - 40)(S + 10) = 0
S, or boat's speed = OR S = - 10 (ignore)
