SOLUTION: A cruise ship going across a lake 15km wide travels a distance of 6km at a certain speed and then finishes the trip at a speed 1 kph slower than its original speed. The boat arri

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Question 1170177: A cruise ship going across a lake 15km wide travels a distance of 6km at a certain
speed and then finishes the trip at a speed 1 kph slower than its original speed. The
boat arrives 20 minutes later than if the original speed is maintained throughout the
trip. What is the original speed of the cruise ship?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A cruise ship going across a lake 15km wide travels a distance of 6km at a certain speed and then finishes the trip at a speed 1 kph slower than its original speed.
The boat arrives 20 minutes later than if the original speed is maintained throughout the trip.
What is the original speed of the cruise ship?
:
let s = original speed
then
(s-1) = slower speed
:
Change 20 min to 1%2F3 hr
:
Total dist was 15 km therefore he traveled 6 km at s and 9 km at (s-1)
Find the actual time take for the trip
6%2Fs + 9%2F%28%28s-1%29%29 = %286%28s-1%29%2B9s%29%2F%28s%28s-1%29%29 = %286s-6%2B9s%29%2F%28s%28s-1%29%29 = %2815s-6%29%2F%28s%28s-1%29%29
then
actual time - normal time = 1/3 hr
%2815s-6%29%2F%28s%28s-1%29%29 - 15%2Fs = 1%2F3
multiply by 3s(s-1), get rid of the denominators
3(15s-6) - 15(3(s-1)) = s(s-1)
45s - 18 - 45s + 45 = s^2 - s
27 = s^2 - s
A quadratic equation
s^2 - s - 27 = 0
using the quadratic formula, the positive solution
s = 5.72 km/h is the normal speed
:
:
seems like we deserve to get an integer here, check in time equation with calc
(6%2F5.72 + 9%2F4.72) - 15%2F5.72 =
1.049 + 1.907 - 2.622 = .334 which is about 1/3 hr or 20 min
seems to check out, if you find that I screwed this up, let me know. ankor@att.net