Kinematics physics homework help -- 2 canoes crossing a river

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Homework Statement


Two canoeists, A and B, live on opposite shores of a 300.0 m wide river that flows east at 0.80 m/s. A lives on the north shore and B lives on the south shore. They both set out to visit a mutual friend X who lives on the north shore at a point 200.0 m upstream from A and 200.0 m downstream from B. Both canoeists can propel their canoes at 2.4 m/s through the water. How much time must canoeist A wait after canoeist B sets out so that they both arrive at X at the same time? Both canoeists make their respective trips by the most direct routes.

Homework Equations


delta v = delta d / delta t

The Attempt at a Solution


since the friend lives downstream from b i did 2.4 m/s - 0.80 m/s= 3.2 m/s. Then i rearranged v=d/t to t=d/t so 200 m/ 1.6 m/s = 125 seconds...
im struggling with canoe A since he has to go both down stream and across the river.. i have no idea how to do that. can you help me please?
 

Answers and Replies

  • #2
HallsofIvy
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Start by drawing a diagram: draw two lines representing the banks of the river, the straight line from B to X, the line down the river from B to the point on the bank opposite to X, and the line from that point to X. The gives a right triangle. You can analyze that by taking the speed of the river for the line down the river and the speed of the canoe for the distance across the river.
 
  • #3
CWatters
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This is a vector addition problem. You should have three velocity vectors on your diagram representing....

* The velocity of the river (direction west to east at 0.8m/s)
* The velocity of B (unknown direction at 2.4m/s)
* The resultant of these two (known direction B to X at unknown velocity)

Solve for the unknown velocity. Calculate the time taken.
 
  • #4
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hey thanks for replying!
so i had the right triangle already made and i found the total distance that he needs to travel is 360.5m. But im having a little trouble analayzing the velocity triangle... the the horizontal vector is 0.8 m/s but is the 2.4 m/s the the diagonal vector or the one straight accross? and would i do 2.4 m/s + 0.8 m/s since he is going downstream?
 
  • #5
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So i did what you said CWatters and i found the unknown velocity to be 2.26 m/s? I think thats right but not really sure... but now i just do t=360.5/2.26?
 
  • #6
CWatters
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I think the unknown velocity must be faster than 2.4m/s. Here is my drawing...

River.jpg
 
  • #7
CWatters
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Actually this diagram might be easier to think about... T is the crossing time...

River.jpg
 
  • #8
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hey CWatters I apologize for such a late reply... i did not have access to the internet. I still need your help with this question if you have the time please. I found the answer to the question but im having a hard time understanding how they got velocity of canoeist A to land diagram ... its the same exact diagram you showed me but im having a really hard time figuring out how you drew that diagram with two unknown angles... the math is pretty easy for me I just don't unders
Screen Shot 2016-03-21 at 2.29.54 PM.png
tand how the diagram was drawn... Also for the answer i found should it be referring to canoeist B in the first part and Canoeist A in the second? Any help would be greatly appreciated...
 
  • #9
CWatters
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Perhaps this helps..

Canoe A

The calculation for Canoe A is easy and doesn't need a diagram. It moves at 2.4m/s relative to the water but the water is flowing the other way at 0.8m/s. So the velocity relative to the land is 2.4-0.8 = 1.6m/s. Time = distance/velocity = 200/1.6 = 125 seconds. Happy with that?

Canoe B

This needs a diagram. I'll refer to the one you found. This shows that the resulting velocity of the canoe VAL is the sum of two other vectors VAW and VWL.

Lets say the time it takes canoe B to make the crossing is T.

The problem statement says that the resultant distance VAL * T is the "most direct route"
VWL * T represents the effect of the current West to East. eg the distance canoe B would drift East in time T.
VAW * T represents the heading the canoe has to take in order to account for the current.

If you multiply all the sides of a triangle by the same quantity (eg T) then it just scales and the angles stay the same. In this case this allows you to show velocities and distance vectors on the same diagram and to work out the angles θ and β.

Their working shows how to caulate the time T that canoe B takes.

Then just subtract the two times.
 
  • #10
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ooh okay that makes so much more sense thanks for your help CWatters just one last question if you dont mind... we are given V_aw= 2.4 m/s right? so what does it mean when it says: "The component of https://www.physicsforums.com/file://localhost/Users/alikhan1/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image002.gif [Broken] across the river is: 2.4sin(56° + 16°) = 2.28 m/s."? I thought we were trying to find the resultant vector (V_al) ? Last question and hopefully ill stop bothering you lol
 
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  • #11
CWatters
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we are given V_aw= 2.4 m/s right? so what does it mean when it says: "The component of across the river is: 2.4sin(56° + 16°) = 2.28 m/s."? I thought we were trying to find the resultant vector (V_al) ?
There are actually three ways to solve this part of the problem. They selected a different method.

You will recognise the basic equation...

Time = distance/velocity

so the next step is to find some distances and velocities to use in that equation. There are three possibilities..

a) The resultant vector method..
Time = Distance from B to X / VAL
where
Distance B to X = Sqrt{2002 + 3002}
VAL = Sqrt{(2.4Cos(56.3°+16°)+0.8)2 + (2.4Sin(56.3°+16°))2}
so
Time = Sqrt{2002 + 3002} / Sqrt{(2.4Cos(56.3°+16°)+0.8)2 + (2.4Sin(56.3°+16°))2}
= 360.56 / 2.751
= 131 seconds

b) The component across the river (the method they used)...
Time = Distance across river / component of velocity across the river
= 300 / 2.4sin(56.3° + 16°)
= 300 / 2.28
= 131 seconds

or

c) The component down the river...
Time = Distance down river / component of velocity down river
= 200 / (2.4cos(56.3° + 16°) + 0.8)
= 200 / 1.53
= 131 seconds

At the risk of confusing you...

Have done any projectile motion problems? Typically what you do to work out the range is to calculate the flight time using the vertical component of the launch velocity. You then use that time and the horizontal component of the launch velocity to work out the range. In other words you rely on the fact that the "vertical flight time" is the same as the "horizontal flight time". That's the same trick used in b and c.

PS: The angle is 56.3, if you use 56 you get slightly different answers for each method.
 

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