A boat can travel at a speed of 3 m/s on still water. A boatman wants to cross a river whilst covering the shortest possible distance. In what direction should he row with respect to the bank if the speed of water is (i) 2 m/s (ii) 4 m/s ? Assume that the speed of the water is the same everywhere.
Part (i) seems to be trivial but I couldn't solve part (ii). Please help (Speed of water is greater than the speed the boat can travel, so resultant velocity is not perpendicular).
Relative Velocity Help
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 06072016 20:51

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 07072016 10:44
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 07072016 17:44
(Original post by tangotangopapa2)
A boat can travel at a speed of 3 m/s on still water. A boatman wants to cross a river whilst covering the shortest possible distance. In what direction should he row with respect to the bank if the speed of water is (i) 2 m/s (ii) 4 m/s ? Assume that the speed of the water is the same everywhere.
Part (i) seems to be trivial but I couldn't solve part (ii). Please help (Speed of water is greater than the speed the boat can travel, so resultant velocity is not perpendicular).
If the angle is measured from the bank, then
[latex]v_b\cos\theta = v_r[\latex]
River at 2m/s:
[latex]\theta = cos^{1}\frac{2}{3}[\latex]
[latex]\theta = 41^\circ[\latex]
River at 4m/s:
[latex]\theta = cos^{1}\frac{2}{4}[\latex]
[latex]\theta = 60^\circ[\latex]Last edited by ActualPhysTchr; 07072016 at 17:45. Reason: Seems like LaTeX support doesn't work? 
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 07072016 18:21
(Original post by ActualPhysTchr)
To cover the shortest possible distance, the boatman wants to travel directly across the river. This means that his resultant velocity needs to be straight across the river, which means that the streamwise component of his velocity needs to be equal to the speed of the river, but in the opposite direction.
If the angle is measured from the bank, then
Tried changing Latex syntax to Tex syntax, it seems to be working.Last edited by tangotangopapa2; 07072016 at 19:03. 
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 07072016 18:26
(Original post by tangotangopapa2)
A boat can travel at a speed of 3 m/s on still water. A boatman wants to cross a river whilst covering the shortest possible distance. In what direction should he row with respect to the bank if the speed of water is (i) 2 m/s (ii) 4 m/s ? Assume that the speed of the water is the same everywhere.
Part (i) seems to be trivial but I couldn't solve part (ii). Please help (Speed of water is greater than the speed the boat can travel, so resultant velocity is not perpendicular).
So Velocity(boat) = Velocity(boat relative to water) + Velocity(water)
The velocity of the boat should be perpendicular to banks if the boat were to cover a minimal distance, the velocity of water is parallel to the banks and thus perpendicular to the velocity of the boat.
Draw a vector diagram and find your angles. 
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 07072016 18:39
(Original post by oShahpo)
Well, velocity of (boat relative to water) = velocity(boat)  Velocity(water) *using vectors*.
So Velocity(boat) = Velocity(boat relative to water) + Velocity(water)
The velocity of the boat should be perpendicular to banks if the boat were to cover a minimal distance, the velocity of water is parallel to the banks and thus perpendicular to the velocity of the boat.
Draw a vector diagram and find your angles.
In part 2, if you attempt to draw vector triangle with relative velocity perpendicular, you end up getting the right triangle with 3 as the hypotenuse while other side 4. 
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 07072016 18:43
(Original post by tangotangopapa2)
In part 2, if you attempt to draw vector triangle with relative velocity perpendicular, you end up getting the right triangle with 3 as the hypotenuse while other side 4.
So you have to draw a different triangle (non right), then use the sine law.
To figure out the shape of the triangle, think about how the boat would move if the speed of the water was 100m/s, i.e. very very fast. 
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 07072016 18:55
(Original post by oShahpo)
Yea it makes sense, there is no way a boat, moving slower than the water, can cancel the velocity of the water in the x direction, while still having some velocity in the y direction to allow him to move across the river. In other words, there is no way the boat can move in a straight direction across the river when the water is faster than the speed of the boat relative to it.
So you have to draw a different triangle (non right), then use the sine law.
To figure out the shape of the triangle, think about how the boat would move if the speed of the water was 100m/s, i.e. very very fast.
Let's take the case presented in the problem. We know the velocity of water (4 m/s) and the velocity of boat (3 m/s). For different direction of the boat, there are different directions of the relative velocity i.e. there are infinite such cases. We need to find the direction the boat moves, such a way that the angle made by the resultant velocity with the perpendicular distance to be minimum.
I couldn't figure out where to go following this argument. 
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 07072016 18:56
(Original post by tangotangopapa2)
Well, if you consider the speed of water to be 100 m/s, then you will get triangle with one side 100 units long and other side 3 units long, which is really difficult to scale.
Let's take the case presented in the problem. We know the velocity of water (4 m/s) and the velocity of boat (3 m/s). For different direction of the boat, there are different directions of the relative velocity i.e. there are infinite such cases. We need to find the direction the boat moves, such a way that the angle made by the resultant velocity with the perpendicular distance to be minimum.
I couldn't figure out where to go following this argument. 
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 07072016 18:59
(Original post by oShahpo)
I will write down the solution and send it to you. 
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 07072016 19:20
(Original post by tangotangopapa2)
Thank you. Please do it.
For part b I believe it is a bearing of 053.1 degrees. This is because the river is flowing faster, therefore it is not possible for the resultant velocity to be perpendicular to the flow of the river while the speed is less than 4. The only solution here is that the boat sets out to travel perpendicular to the banks. This gives the bearing to be arctan(4/3) for the shortest distance. Anything other than that will delay it. 
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 07072016 19:26
(Original post by tangotangopapa2)
Thank you. Please do it. 
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 07072016 19:28
(Original post by RDKGames)
For part a it's a bearing of 041.8 degrees (considering river flowing horizontally to the right)
For part b I believe it is a bearing of 053.1 degrees. This is because the river is flowing faster, therefore it is not possible for the resultant velocity to be perpendicular to the flow of the river while the speed is less than 4. The only solution here is that the boat sets out to travel perpendicular to the banks. This gives the bearing to be arctan(4/3) for the shortest distance. Anything other than that will delay it. 
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 07072016 19:33
(Original post by tangotangopapa2)
Thank you. Please do it.
alpha = 41.409622109271 so the bearing = 41.409622109271 + 270.
If you'd like an explanation, give me some time to come up with a simple one. 
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 07072016 19:33
(Original post by RDKGames)
For part a it's a bearing of 041.8 degrees (considering river flowing horizontally to the right)
For part b I believe it is a bearing of 053.1 degrees. This is because the river is flowing faster, therefore it is not possible for the resultant velocity to be perpendicular to the flow of the river while the speed is less than 4. The only solution here is that the boat sets out to travel perpendicular to the banks. This gives the bearing to be arctan(4/3) for the shortest distance. Anything other than that will delay it.
That gives the solution to the question: " Which direction to travel in order to reach the next edge as fast as possible?" and anything other than that delays the time taken to reach the next end. Your solution assumes that the fastest route is the shortest route.
But why does the shortest route have to be the fastest route? 
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 07072016 19:42
(Original post by oShahpo)
alpha = 41.409622109271 so the bearing = 41.409622109271 + 270.
If you'd like an explanation, give me some time to come up with a simple one.
Why is the angle 90 degrees? 
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 07072016 19:46
(Original post by tangotangopapa2)
Thank you so much. Could you please explain it a bit?
Why is the angle 90 degrees? 
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 07072016 20:01
(Original post by tangotangopapa2)
Thank you so much. Could you please explain it a bit?
Why is the angle 90 degrees?
Attachment 559598559602
Attachment 559598559602559600
[attach]5.595985596025596e+23[/attach] 
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 07072016 20:05
(Original post by oShahpo)
Attachment 559598559602
Attachment 559598559602559600
[attach]5.595985596025596e+23[/attach] 
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 07072016 20:05
(Original post by tangotangopapa2)
Thank you so much. Really appreciate it.
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Updated: July 7, 2016
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