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

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

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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.

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?

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##

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.

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.

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?
Are you sure it's not ##w < u##?

PeroK
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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.

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?

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##.

PeroK
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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.

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?

PeroK
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.

PeroK
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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##.

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.

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

x will be a*tanα
but when i checked the answer its
x = a( w/(u cos α) − tan α)

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

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.

IamVector
You need to be careful, since you used frame moving with respect to the bank.
ohk just realised thanks for help. : )

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

PeroK
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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.

IamVector
PeroK
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but when i checked the answer its
x = a( w/(u cos α) − tan α)

IamVector
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.