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

[IamVector]

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

 

[PeroK]

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.

 

[IamVector]

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 won't it reduce the x component thus reducing drift?

 

[PeroK]

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 won't 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: $\vec w + \vec u$

 

[archaic]

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.

 

[archaic]

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]

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 won't it reduce the x component thus reducing drift?

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

 

[PeroK]

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.

 

[IamVector]

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?

 

[archaic]

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]

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.

 

[IamVector]

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?

 

[archaic]

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]

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 $\alpha$.

 

[archaic]

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.

 

[IamVector]

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

 

[IamVector]

x will be a*tanα

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

 

[IamVector]

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]

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 realized thanks for help. : )

 

[IamVector]

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]

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.

 

[PeroK]

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

That's the answer I get.

 

[IamVector]

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.

 

[IamVector]

That's the answer I get.

got it too :)

 

[IamVector]

I'm not sure about that. Using Snell's law was a clever idea, but maybe you should check with some conventional calculus.

and calculus approach??

 

[PeroK]

and calculus approach??

If you got the answer using Snell's law, then great. That was very good.

 

[IamVector]

I'm not sure about that. Using Snell's law was a clever idea, but maybe you should check with some conventional calculus.

is calculus approach has something to do with properties of right angle and Pythagoras theorem?

 

[IamVector]

If you got the answer using Snell's law, then great. That was very good.

ohk moving on to next ques thanks for help :)

 

[PeroK]

is calculus approach has something to do with properties of right angle and Pythagoras theorem?

Using calculus you need to minimise $(1 + \frac w v) \sec \alpha - \frac u v \tan \alpha$. Which you do by usual differentiation method.