Two ferries problem
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5 members have voted
November 27th, 2017 at 8:25:53 PM
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Two ferries depart at right angles from opposite shores of a river. Each ferry travels at a consistent speed, although one is faster than the other. The pass each other 700 meters from the nearest shore. Upon reaching the other side, each spends 10 minutes docked. On the return trip they pass each other 400 meters from the opposite shore.
How wide is the river?
Free beer to the first satisfactory answer and solution.
How wide is the river?
Free beer to the first satisfactory answer and solution.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 27th, 2017 at 10:41:23 PM
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Quote: WizardTwo ferries depart at right angles from opposite shores of a river. Each ferry travels at a consistent speed, although one is faster than the other. The pass each other 700 meters from the nearest shore. Upon reaching the other side, each spends 10 minutes docked. On the return trip they pass each other 400 meters from the opposite shore.
How wide is the river?
Free beer to the first satisfactory answer and solution.
1700 meters.
A graph showing the positions of each ferry vs time helps a lot to set up the equations.
Let f = speed of the fast ferry,
s = speed of the slow ferry,
and x = width of the river.
If both ferries leave their shore at t = 0, then they pass at t = 700/s = ( x - 700 ) / f.
They pass again at t = x/f + 10 + ( x - 400 ) / f = x/s + 10 + 400/f.
Solve both equations for the ratios of the ferries speeds:
s/f = 700 / ( x - 700 )
s/f = ( x + 400 ) / ( 2x - 400 )
Equate the two expressions, and solve for x:
700 / ( x - 700 ) = ( x + 400 ) / ( 2x - 400 )
x^2 - 300x - 2800 = 1400x - 2800
x^2 - 1700x = 0
x ( x - 1700 ) = 0
So x = 0 meaning the river has zero width (ignore this answer), or x = 1700 meaning the river has a width of 1700 meters.
A graph showing the positions of each ferry vs time helps a lot to set up the equations.
Let f = speed of the fast ferry,
s = speed of the slow ferry,
and x = width of the river.
If both ferries leave their shore at t = 0, then they pass at t = 700/s = ( x - 700 ) / f.
They pass again at t = x/f + 10 + ( x - 400 ) / f = x/s + 10 + 400/f.
Solve both equations for the ratios of the ferries speeds:
s/f = 700 / ( x - 700 )
s/f = ( x + 400 ) / ( 2x - 400 )
Equate the two expressions, and solve for x:
700 / ( x - 700 ) = ( x + 400 ) / ( 2x - 400 )
x^2 - 300x - 2800 = 1400x - 2800
x^2 - 1700x = 0
x ( x - 1700 ) = 0
So x = 0 meaning the river has zero width (ignore this answer), or x = 1700 meaning the river has a width of 1700 meters.
November 28th, 2017 at 6:33:43 AM
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1,700m wide.
At a given time t1, the positions of the two ferries is the same. So, the ferry on the near shore, travelling at a velocity of v1 goes a distance of 700m (v1·t1 = 700), and the other ferry, travelling at v2, goes a distance of W - 700m, where W is the width of the river (v2·t1 = (W-700)). Now, solve both equations for t1 and combine the two, eliminating t1, and solve for the ratio of the velocities:
v2/v1 = (W-700)/700
Next, we know that at time t2, their positions are once again the same. Using the same methods as above, v1·t2 = W+400 and v2·t2 = 2W-400. I originally included the time at the dock in the equation, but found that it dropped out while solving. So, it doesn't really matter how long they are docked, as long as it is the same for both. Again, solve for t2, combine, and then solve for v2/v1:
v2/v1 = (2W-400)/(W+400)
Now, set the two speed ratio terms equal to each other and solve for W, which turns out to be 1,700m.
At a given time t1, the positions of the two ferries is the same. So, the ferry on the near shore, travelling at a velocity of v1 goes a distance of 700m (v1·t1 = 700), and the other ferry, travelling at v2, goes a distance of W - 700m, where W is the width of the river (v2·t1 = (W-700)). Now, solve both equations for t1 and combine the two, eliminating t1, and solve for the ratio of the velocities:
v2/v1 = (W-700)/700
Next, we know that at time t2, their positions are once again the same. Using the same methods as above, v1·t2 = W+400 and v2·t2 = 2W-400. I originally included the time at the dock in the equation, but found that it dropped out while solving. So, it doesn't really matter how long they are docked, as long as it is the same for both. Again, solve for t2, combine, and then solve for v2/v1:
v2/v1 = (2W-400)/(W+400)
Now, set the two speed ratio terms equal to each other and solve for W, which turns out to be 1,700m.
Edit: Didn't Chris de Burgh sing about AceOfSpades' nemesis at the Aria?
"Dealer has 'rock'... Pay 'paper!'"
November 28th, 2017 at 7:17:24 AM
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Let W be the width of the river in meters, A be the speed of the faster ferry in m/sec, and B the speed of the slower ferry in m/sec
At time 0, they depart
When they first meet, the closest shore is the one where the slower boat started
At time T1, the slower boat has traveled B * T1 = 700m, and the faster boat has traveled A * T1 = W - 700m
A/B = (W - 700) / 700
B = 700 A / (W - 700)
The faster boat takes W / A + 600 + (W - 400) / A = (2W - 400) / A + 600 seconds to reach the second meeting point
The slower boat takes W / B + 600 + 400 / B = (W + 400) / B + 600 seconds to reach the second meeting point
(2W - 400) / A = (W + 400) / B
(2W - 400) / A = (W + 400) (W - 700) / (700 A)
700 (2W - 400) (W + 400) (W - 700)
1400 W - 280,000 = W2 - 300 W - 280,000
W2 = 1700 W
W = 1700m
November 28th, 2017 at 9:02:44 AM
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I applaud the three brilliant guys that have responded so far and I believe that their answers are probably what the Wizard was looking for. However, I have a twist on the answers they've given.
The width of the river is 1,700m + 2x, where x is the length of the ferry boats. This does assume that the ferry boats are of equal length and thus had an equal distance to travel.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
November 28th, 2017 at 10:24:56 AM
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A more careful reading of the problem leads me to this slightly modified answer
W = width of river, m
L = length of each identical ferry, m
"meet" means that the front end of the ferries meet
The 1st ferry travels 700-L to the first meeting. The 2nd ferry travels W-700-L to that first meeting.
The 1st ferry travels (W-L)+ (400-L) or W+400-2L to the 2nd meeting, the 2nd ferry travels (W-L)+(W-400-L), or 2W-400-2L to that 2nd meeting.
That means you must solve the following equation to express W as a function of L:
(2W-2L-400)*(700-L) = (W-700-L)* (W+400-2L)
Let's simplify (hopefully) by defining x= W-L, thus we need to solve this equation for x as a function of L:
(2x-400)*(700-L)= (x-700)*(x+400-L)
1400x -2xL+400L-280000 = x^2-300x-xL+700L-280000
or
xL-300L=x^2-1700x
0=x^2-(1700+L)*x+300L
No simple analytical solution for x =f(L)! Given a ferry boat length, L, = 10 m, I get that an approximate value of x is 1708.24381 and that the width of the river is approximately 1718.24381 meters.
*************
Of course, I have ignored relativistic effects . . . .
W = width of river, m
L = length of each identical ferry, m
"meet" means that the front end of the ferries meet
The 1st ferry travels 700-L to the first meeting. The 2nd ferry travels W-700-L to that first meeting.
The 1st ferry travels (W-L)+ (400-L) or W+400-2L to the 2nd meeting, the 2nd ferry travels (W-L)+(W-400-L), or 2W-400-2L to that 2nd meeting.
That means you must solve the following equation to express W as a function of L:
(2W-2L-400)*(700-L) = (W-700-L)* (W+400-2L)
Let's simplify (hopefully) by defining x= W-L, thus we need to solve this equation for x as a function of L:
(2x-400)*(700-L)= (x-700)*(x+400-L)
1400x -2xL+400L-280000 = x^2-300x-xL+700L-280000
or
xL-300L=x^2-1700x
0=x^2-(1700+L)*x+300L
No simple analytical solution for x =f(L)! Given a ferry boat length, L, = 10 m, I get that an approximate value of x is 1708.24381 and that the width of the river is approximately 1718.24381 meters.
*************
Of course, I have ignored relativistic effects . . . .
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
November 28th, 2017 at 10:33:37 AM
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duplicate post removed
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
November 28th, 2017 at 11:30:45 AM
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Yes, 1700 is the answer was I was looking for. Congratulations to ChesterDog for being first. A beer your way your next Vegas visit, and I think you're overdue for one.
Here is how I worded the solution:
Let's call x the distance between the two meetings points.
We can ignore the 10 minutes each spends turning around and assume the turn around instantly. This is simply obvious.
At the first meeting, the two boats combined would have traveled the width of the river.
At the second meeting, the two boats combined would have traveled three times the width of the river.
Since both boats travel at a consistent speed, both boats would have traveled three times as far at the second meeting as the first.
Let's solve for x looking at the fast boat.
3 × (time traveled until first meeting) = (time traveled until second meeting).
3 × (400+x) = 1800 + 2x.
x = 600.
So the total width of the river is 400+600+700 = 1700 meters.
Here is how I worded the solution:
Let's call x the distance between the two meetings points.
We can ignore the 10 minutes each spends turning around and assume the turn around instantly. This is simply obvious.
At the first meeting, the two boats combined would have traveled the width of the river.
At the second meeting, the two boats combined would have traveled three times the width of the river.
Since both boats travel at a consistent speed, both boats would have traveled three times as far at the second meeting as the first.
Let's solve for x looking at the fast boat.
3 × (time traveled until first meeting) = (time traveled until second meeting).
3 × (400+x) = 1800 + 2x.
x = 600.
So the total width of the river is 400+600+700 = 1700 meters.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 28th, 2017 at 2:53:57 PM
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But are they taking the armoire?
I am a robot.
