- Joined
- 10 Dec 06
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On a normal day a Ferry takes 10 minutes to makes its journey across the great RHP river. The speed of the ferry in still water is 12 m/s. The current of the river is 5 m/s. The ferry's path is straight line directly across the river ( shortest possible path length ).
Today the river is flowing much faster that normal - double its normal current. The ferries' path across the river is the same, how long will it take to make the journey today?
- Joined
- 10 Dec 06
- Moves
- 8528
@bigdoggproblem saidThis problem doesn't appear to have generated much interest. It's up to you.
I can show my work, if anyone would like to see it.
@bigdoggproblem saidIt all helps the learning process Bigdogg.
I can show my work, if anyone would like to see it.
As has become obvious, I'm out of my depth with Joe's problems but I still try from time to time.
I might even fluke a right answer once in a while!!
One way to learn is to look at the answer and then try and work out how it is derived!!
One way to sum up force vectors is to create a triangle, using the two vectors, as shown on this site:
https://www.school-for-champions.com/science/force_vectors.htm
So we get a right triangle with a hypotenuse of 12 [boat speed in still water] and one side equal to 5 [river current]. Use the Pythagorean theorem to solve for the remaining side [the actual rate of the boat as it goes straight across, fighting the current somewhat].
5² + R² = 12²
R = sqrt(119) m/s
We were given the time (10 min, which equals 600 s) for crossing the river, so Rate * Time gives a distance of:
600 * sqrt(119) m
Now, we bump the current up to 10 m/s. We have a right triangle, similar to above, but with a longer side. Friend Pythagoras tells us that:
10² + RP² = 12² [where RP is 'rate prime']
RP = sqrt(44) = 2 * sqrt(11) m/s
Since we found the distance across the river above, we can solve for the new crossing time.
2 * sqrt(11) m/s * TP = 600 * sqrt(119) m
TP = 300 * sqrt(119) / sqrt(11) s
= 5 * sqrt(119) / sqrt(11) min
@bigdoggproblem saidThanks.
One way to sum up force vectors is to create a triangle, using the two vectors, as shown on this site:
https://www.school-for-champions.com/science/force_vectors.htm
So we get a right triangle with a hypotenuse of 12 [boat speed in still water] and one side equal to 5 [river current]. Use the Pythagorean theorem to solve for the remaining side [the actual rate of ...[text shortened]... 1) m/s * TP = 600 * sqrt(119) m
TP = 300 * sqrt(119) / sqrt(11) s
= 5 * sqrt(119) / sqrt(11) min