The Student Room Group

Relative Velocity Help

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).

Scroll to see replies

Anyone?
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).


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 stream-wise 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
v_b\cos\theta = v_r[\latex]

River at 2m/s:
\theta = cos^{-1}\frac{2}{3}[\latex]
\theta = 41^\circ[\latex]

River at 4m/s:
\theta = cos^{-1}\frac{2}{4}[\latex]
\theta = 60^\circ[\latex]
(edited 7 years ago)
(edited 7 years ago)
Reply 4
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).


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.
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.

IMG_20160707_183406.jpg
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.
Reply 6
Original post by tangotangopapa2
IMG_20160707_183406.jpg
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.

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.
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.


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.
Reply 8
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.


I will write down the solution and send it to you.
Original post by oShahpo
I will write down the solution and send it to you.


Thank you. Please do it.
Original post by tangotangopapa2
Thank you. Please do it.


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.
Original post by tangotangopapa2
Thank you. Please do it.


The solution is quite complicated to explain, so give me some time to make it as clear as possible.
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 is not true. The velocity of the boat is not perpendicular to the banks, rather with a bearing > 270.
Original post by tangotangopapa2
Thank you. Please do it.


WIN_20160707_19_32_29_Pro.jpg
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.
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.


Thanks for the reply.
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?
Original post by oShahpo
WIN_20160707_19_32_29_Pro.jpg
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.


Thank you so much. Could you please explain it a bit?
Why is the angle 90 degrees?
Original post by tangotangopapa2
Thank you so much. Could you please explain it a bit?
Why is the angle 90 degrees?

Okay, let me explain it to you in a diagram. Give me a moment.
Original post by tangotangopapa2
Thank you so much. Could you please explain it a bit?
Why is the angle 90 degrees?


WIN_20160707_19_55_11_Pro.jpg
Attachment not found

Attachment not found

ucles sending.jpg
Original post by oShahpo
WIN_20160707_19_55_11_Pro.jpg
Attachment not found

Attachment not found

ucles sending.jpg


Thank you so much. Really appreciate it.
Original post by tangotangopapa2
Thank you so much. Really appreciate it.


Glad I could help :smile:

Quick Reply

Latest