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Vectors - crossing a river w/ given flow

  • Thread starter Sparky_
  • Start date
  • #1
220
4

Homework Statement



a student want to cross a river flowing east at a rate of 2.0km/hr in a boat on the south bank and arrive at point L on the north (north and west) side. Point L is 0.5km to the left or west from the starting point (point K - point K is on the south side). Point L is 0.5km (to the left) from a perpindicular line drawn from the starting point (K)

The student can row at 5.0km/hr.

How long will it take to paddle north-west up stream and arrive at point L?

Homework Equations



the picture

L ----------0.5km ------------M
........................................|
........................................|
....................................0.25km -> flow = 2km / hr
........................................|
........................................|
(K)







The Attempt at a Solution



It has been a very long time since I have worked these types of problems and just for fun I tried my hand at this - no luck

I said the student can row at 5km/hr north and (5-2) km/hr westward

He will need to paddle the length of the hypotenuse or SQRT((0.25)^2 + (0.5)^2) = 0.559 km

The students velocity vector is SQRT((5)^2 + (3)^2) = 5.83 km/hr.

I see this is wrong - his velocity is faster than he can paddle.

My direction was to acquire his velocity vector and divide the hypotenuse distance (0.5590km) by the new velocity vector.

I know this is wrong - I have the answer - 0.17858 hr.

Am I wrong to think in terms of a velocity vector?

Is this to be solved with simple position equations?

Help??

Thanks
Sparky_
 

Answers and Replies

  • #2
881
40
The velocity with which the student can paddle is 5km/h. So having his velocity as 5km/h northwards AND 3 km/h westwards will obviously mean he is exceeding his capacity.

So instead, solve it this way. You know the student's final velocity has to lie on the line joining K and L. Call the velocity vector in this direction v. Now, the vector sum of the the velocity of stream and the boat's velocity has to be equal to this v. Find v, and then find the time taken.
 
  • #3
220
4
I am not caught up with you.

I see that if the direction was purely west - then the boats net velocity is (5-2)km/hr

I "thought" that the northward component is 5km/hr since the river flow does not affect it. That's how I got the vector sum of 5 north and 3 in westward direction.

I don't see what my vector sum should be.

you mention the velocity V (the hypotenuse) on line K-L is the vector sum of the boat and river flow.

I see the flow is 2km/hr.

Thankls
Sparky_
 
  • #4
881
40
I am not caught up with you.

I see that if the direction was purely west - then the boats net velocity is (5-2)km/hr

I "thought" that the northward component is 5km/hr since the river flow does not affect it. That's how I got the vector sum of 5 north and 3 in westward direction.
You're right that the northward component isn't effected by the river flow. But you have to remember, that the maximum velocity the boy can manage on his own, in still water is 5km/h. If he was going purely west in still water, his maximum velocity would be 5km/h which due to the stream flow is cut down to 3km/h. He can't go ANY faster than this towards the west(or move in any other direction along with this 3km/h) So, him having a northward velocity with the westward 3km/h is not possible.

Hope that makes sense.
 
  • #5
220
4
My next attempt - (still wrong answer though)

I first solved for the vector components assuming no flow in the river.

I found the angle (using the distances 0.25km and 0.5km) between the hypotenuse and the southern shore = 26.57 and a distance of 0.5590km

Next I said the velocity along the hypotenuse is 5km/hr (with no river flow) - the x-component is 4.494km/hr and the y-component = 2.2365 km/hr.

Next I subtract the 2km/hr from the x-component and get a new x-component = 2.494 km/hr (westward)

With 26.57 degrees, I get a new velocity hypotenuse pointed to the end location of 2.7885 km/hr. (the y-component is 1.2361 km/hr northward)

Traveling at 2.7885 km/hr along the hypotenuse of 0.5590km gives a time of 0.2023 hrs.

I have re-worked this keeping all decimals to see if it is rounding - the answer is 0.17858 hr.

Where am I wrong (again)?

Thanks
Sparky_
 
  • #6
881
40
This diagram will help.

I found this method will be much easier, and you don't even need to find Vb. You know Vm(=5km/h) and Vs(2km/h), and the angle x can be found out, as you did above, just find it from the vertical in this case. Now use equations of motion for the respective x and y axes velocities to find out time. You will have two equations in time and angle.
 

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  • #7
220
4
Thanks!!!
 

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