Combine swimming and running to get to the final point the quickest

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  • #1
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Homework Statement:

A boy is situated at point A in a river, at distance
a from the riverbank. He can swim at speed u or run at speed
v > u on the shore; water flows in the river at velocity w > u.
The boy wants to reach the point C upstream on the riverbank
with minimal time. At what distance x from point B aligned
with point A should he get out of the water?

Relevant Equations:

I think the diagram provided is wrong the question states that he want to reach A with minimum time so he should be aligning himself on the x distance left of AB
Capture3.PNG
 
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  • #2
PeroK
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Homework Statement:: A boy is situated at point A in a river, at distance
a from the riverbank. He can swim at speed u or run at speed
v > u on the shore; water flows in the river at velocity w > u.
The boy wants to reach the point C upstream on the riverbank
with minimal time. At what distance x from point B aligned
with point A should he get out of the water?
Relevant Equations:: I think the diagram provided is wrong the question states that he want to reach A with minimum time so he should be aligning himself on the x distance left of AB

View attachment 258766
As ##w > u##, he can't swim upstream.
 
  • #3
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As ##w > u##, he can't swim upstream.
but If he will align himself x direction to right x component of his velocity will increase but if he will align himself to left wont it reduce the x component thus reducing drift?
 
  • #4
PeroK
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but If he will align himself x direction to right x component of his velocity will increase but if he will align himself to left wont it reduce the x component thus reducing drift?
The question is where he gets out of the water; not where he "aligns" himself. The diagram is not great because ##\vec u## is relative to the water; not to the bank. The dotted arrow in the diagram should be the resultant velocity relative to the bank: ##\vec w + \vec u##
 
  • #5
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The question is where he gets out of the water; not where he "aligns" himself. The diagram is not great because ##\vec u## is relative to the water; not to the bank. The dotted arrow in the diagram should be the resultant velocity relative to the bank: ##\vec w + \vec u##
I think we should interpret ##u## as his absolute speed, i.e with respect to earth, since otherwise it wouldn't make sense to talk about orientation to minimise time.
The statement "he can swim at speed ##u##" is then confusing since if he moves to the right he'd obviously swim with a different speed than if he were to move to the left.
 
  • #6
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I think we should interpret ##u## as his absolute speed, i.e with respect to earth, since otherwise it wouldn't make sense to talk about orientation to minimise time.
The statement "he can swim at speed ##u##" is then confusing since if he moves to the right he'd obviously swim with a different speed than if he were to move to the left.
I mean that one should define a vector for the velocity in water after introducing an orientation angle from the normal to the riverbank and then use some calculus.
 
  • #7
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but If he will align himself x direction to right x component of his velocity will increase but if he will align himself to left wont it reduce the x component thus reducing drift?
Are you sure it's not ##w < u##?
 
  • #8
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I think we should interpret ##u## as his absolute speed, i.e with respect to earth, since otherwise it wouldn't make sense to talk about orientation to minimise time.
The statement "he can swim at speed ##u##" is then confusing since if he moves to the right he'd obviously swim with a different speed than if he were to move to the left.
That can't be right.
 
  • #9
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That can't be right.
I took water frame of reference so his land speed is now (v+w) and time taken to cover x is let say 't' x = (v+w)t
now time it took to cross come out of river be 'T' so hyp = uT now angle be α so sinα = uT/(v+w)t is there a possible relation between T and t?
 
  • #10
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That can't be right.
Why not? It doesn't make sense for ##u## to be with respect to water and say that ##w>u##, since if he swims in the direction of the flow, his speed would be ##u+w>w##.
 
  • #11
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Why not? It doesn't make sense for ##u## to be with respect to water and say that ##w>u##, since if he swims in the direction of the flow, his speed would be ##u+w>w##.
That's precisely what you've done in your solution above. Taken ##u## relative to the water. Although I see that solution has rightly been deleted now.
 
  • #12
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Capture4.PNG
well tried a different approach I studied earlier we have according to water frame of respect velocities v+w and u so we are assuming that the boy didn't took any break when going from water to land so I applied snells law and got sinα = (u/v+w) so alignment angle α = arcsin(u/v+w) is it correct approach?? and what to do next?
 
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  • #13
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That's precisely what you've done in your solution above. Taken ##u## relative to the water. Although I see that solution has rightly been deleted now.
I have deleted it because I haven't included one term in my time expression and don't really want to recompute the derivative.
 
  • #14
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View attachment 258768 well tried a different approach I studied earlier we have according to water frame of respect velocities v+w and u so we are assuming that the boy didn't took any break when going from water to land so I applied snells law and got sinα = (u/v+w) so alignment angle α = arcsin(u/v+w) is it correct approach?? and what to do next?
That looks like the correct equation for ##\alpha##. That was an imaginative approach!

You should be able to calculate ##x## now you know ##\alpha##.
 
  • #15
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That's precisely what you've done in your solution above. Taken ##u## relative to the water. Although I see that solution has rightly been deleted now.
Look at this problem statement:
(IIT JEE 1983) A river is flowing from west to east at a speed of 5 metre/minute. A man on the south bank of the river, capable of swimming at 10 metre/minute in still water, wants to swim across the river in the shortest time. He should swim in a direction?
I think the speed in this problem is of the same spirit of the above.
 
  • #16
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That looks like the correct equation for ##\alpha##. That was an imaginative approach!

You should be able to calculate ##x## now you know ##\alpha##.
x will be a*tanα
 
  • #17
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x will be a*tanα
but when i checked the answer its
x = a( w/(u cos α) − tan α)
 
  • #18
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Look at this problem statement:

I think the speed in this problem is of the same spirit of the above.
I think he should swim straight ahead from s to n
 
  • #19
PeroK
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but when i checked the answer its
x = a( w/(u cos α) − tan α)
You need to be careful, since you used frame moving with respect to the bank.
 
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  • #20
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You need to be careful, since you used frame moving with respect to the bank.
ohk just realised thanks for help. : )
 
  • #21
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You need to be careful, since you used frame moving with respect to the bank.
is there any approach to solve it by using Huygens principle
 
  • #22
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is there any approach to solve it by using Huygens principle
I'm not sure about that. Using Snell's law was a clever idea, but maybe you should check with some conventional calculus.
 
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  • #23
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but when i checked the answer its
x = a( w/(u cos α) − tan α)
That's the answer I get.
 
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  • #24
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That's the answer I get.
here is the approach I saw in hint given below also it says that using huygens principle is considered the shortest way to solve such problem
approach :
once the front meets the point A, the front forms
a cathetus of a right triangle AP Q, where Q is the point
where the front meets the riverbank, and P is the position
of that Huygens source on the riverbank which creates the
circular wave meeting the point A. Notice that the point
P is the point where the boy needs to start swimming in
the water’s frame of reference, and is displaced by wT from
the corresponding point in the lab frame.
 

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