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Old 17th Jan 2017, 23:26   #1
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Maths Problem - stumper

I've been knocking this one around for a little while now but can't seem to reduce it to something simple. It's a relative velocity problem:

_________________________________

A river flows at 5ms-1 from west to east between parallel banks which are at a distance 300m apart. A man rows a boat at speed of 3ms-1 in still water.

Find the direction in which the boat must be steered in order to cross the river from the southern bank to the northern bank by the shortest possible route. Find the time taken and the actual distance covered by the boat for this crossing.

_________________________________

Treating it as a vector problem where the 3ms-1 velocity is at x degrees to the north (upstream) and splitting it into components I came up with a function which I could graph giving an optimal solution of x=36.9 degrees but I need to find a graphical/algebraic solution.

Any ideas?
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Old 18th Jan 2017, 00:02   #2
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http://www.physicsclassroom.com/clas...rboat-Problems
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Old 18th Jan 2017, 00:32   #3
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Thanks for the link. Unfortunately all the problems and examples assume a perpendicular velocity to the bank. In the problem I've posed, I think both the boat's velocity and it's velocity relative to the river are at angles to the bank and that's what makes this so tough to pick apart.
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Old 18th Jan 2017, 00:44   #4
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The two vectors 5ms and 3ms go togethor into a right angle triangle, the shortest route is the hypotenuse of that triangle.

angle is x
opposite = 3ms
adjacent = 5ms

tan(x) = 3/5
tan^-1(3/4) = x
x = 31 degrees


I would assume by direction it means heading, so that would be 360 - 31 = 329 degrees (headings are always 0 degrees is north and rotate clockwise, so 90 degrees is east). realised thats wrong, it would be 31 degrees, as he'd be pointing the boat downstream to the North/East.

You could then use either sin or cos to work out the hypotenuse to get that velocity, then using the 300m and that velocity work out the time taken.
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Old 18th Jan 2017, 01:01   #5
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Originally Posted by Cerberus90 View Post
The two vectors 5ms and 3ms go togethor into a right angle triangle, the shortest route is the hypotenuse of that triangle.

angle is x
opposite = 3ms
adjacent = 5ms

tan(x) = 3/5
tan^-1(3/4) = x
x = 31 degrees


I would assume by direction it means heading, so that would be 360 - 31 = 329 degrees (headings are always 0 degrees is north and rotate clockwise, so 90 degrees is east). realised thats wrong, it would be 31 degrees, as he'd be pointing the boat downstream to the North/East.

You could then use either sin or cos to work out the hypotenuse to get that velocity, then using the 300m and that velocity work out the time taken.
That was my first thought, but as the river is flowing to the East at 5ms-1 and the boat can only travel at 3ms-1, the resultant cannot be perpendicular to the bank.

I've gone back to my component formula:

If the boat's velocity, relative to the river, makes an angle of x degrees with the perpendicular to the bank upstream then the components of velocity are 3sin(x) upstream and 3cos(x) perpendicular to the bank.

The time taken to cross the river is given by t=300/(3cos(x))=100/cos(x).

The drift is given by (5-3sin(x)) multiplied by time. The resultant distance squared (by Pythagoras' Theorem) is 300^2+(((5-3sin(x))*100)/(cos(x)))^2

I'm trying a differentiation of the distance squared with respect to x and then solving equal to 0 in order to find the optimal angle between 0 and 90 degrees. Fingers crossed.
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Old 18th Jan 2017, 01:49   #6
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It worked! Optimal angle is 36.9 degrees upstream of the perpendicular to the bank (53.1 degrees above the upstream bank). Total time is 125s and total distance is 500m.
Solved using quotient rule and chain rule to differentiate the distance squared. It produced a horrific mess of trigonometric terms but by factorising I obtained a slightly less messy fraction which I could then set equal to zero to determine the minimum point for the domain of 0<x<90. Setting the numerator equal to zero helped me discard some solutions which were not feasible (sin x>1) and find the only feasible solution in the domain.
A victory cup of tea is in order.
How the heck will I explain it tomorrow as it looks hideous. Must be a simpler method but I can't spot anything obvious with the geometry.

Simpler solution still up for grabs vs my somewhat inelegant brute force approach.
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Old 18th Jan 2017, 09:51   #7
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I think the optimal angle is just 90 degrees. Which will give some some shallower angle of inverse tan of 3/5 due to the water.

Since the vertical distance is 300 metres you get an equivalent tiangle with horizontal component of 500 metres and a total distance travelled of 583.09 metres using pythagoras.
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Old 18th Jan 2017, 11:38   #8
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It worked! Optimal angle is 36.9 degrees upstream of the perpendicular to the bank (53.1 degrees above the upstream bank). Total time is 125s and total distance is 500m.
Solved using quotient rule and chain rule to differentiate the distance squared. It produced a horrific mess of trigonometric terms but by factorising I obtained a slightly less messy fraction which I could then set equal to zero to determine the minimum point for the domain of 0<x<90. Setting the numerator equal to zero helped me discard some solutions which were not feasible (sin x>1) and find the only feasible solution in the domain.
A victory cup of tea is in order.
How the heck will I explain it tomorrow as it looks hideous. Must be a simpler method but I can't spot anything obvious with the geometry.

Simpler solution still up for grabs vs my somewhat inelegant brute force approach.
If the boat is pointing upstream then it wouldn't be rowing at 3ms-1, as it would have the 5ms pushing against it. The boat can travel at more than 3ms, it's just that the rower can only row at 3ms, so to get the fastest/shortest way across, you use the 3ms and 5ms, anything other than that is not using the optimal speeds.

If it's pointing more downstream, then the river would push you further than required, the boat would end up going faster, but it would take longer to go 300m north. Anything pointing upstream would get you to a point more perpendicular to where you started, but would take longer as the 5ms would cancel out some of the 3ms speed of the boat.



*EDIT*

Getting stuck now trying to work it all the way through.

*EDIT 2*

So changed tack,

Vector addition on 3ms and 5ms using pythagoras gives 5.8ms-1, still at 31 degrees NorthEast. Using Sin then comes up with 576m for the total distance, and 99 seconds at 5.8ms.





Here's another way to think about it.

That 3ms that the boat can be rowed at, all of that speed must be put towards going north. If you row upstream, some of that 3ms is used in fighting the current, so you won't get across as quick, downstream, you're just propelling yourself further downstream.
I think the time taken can actually be worked out right from the start, just by using 3ms and 300m, which comes out to 100 seconds, which is the same as my 99 seconds above accounting for rounding errors.
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Old 18th Jan 2017, 15:01   #9
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Old 18th Jan 2017, 17:04   #10
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Unless I'm misreading the question, surely you just row towards the bank?

You have a horizontal component of whatever the river is flowing at, but that will never make any difference to the vertical speed of the rower.

I mean, imagine it's the banks that are moving at 5m/s horizontally instead, but then they might as well be staying still because the rower doesn't need to get to any particular point on the bank, just to the bank. it makes a difference for the distance, but then you just do a simple pythagoras calc.
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Old 18th Jan 2017, 17:09   #11
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Unless I'm misreading the question, surely you just row towards the bank?

You have a horizontal component of whatever the river is flowing at, but that will never make any difference to the vertical speed of the rower.

I mean, imagine it's the banks that are moving at 5m/s horizontally instead, but then they might as well be staying still because the rower doesn't need to get to any particular point on the bank, just to the bank. it makes a difference for the distance, but then you just do a simple pythagoras calc.
That's what I came to eventually, I started doing what I usually do and make it more complicated than it needs to be.
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Old 18th Jan 2017, 17:15   #12
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Unless I'm misreading the question, surely you just row towards the bank?

You have a horizontal component of whatever the river is flowing at, but that will never make any difference to the vertical speed of the rower.

I mean, imagine it's the banks that are moving at 5m/s horizontally instead, but then they might as well be staying still because the rower doesn't need to get to any particular point on the bank, just to the bank. it makes a difference for the distance, but then you just do a simple pythagoras calc.
If it was the quickest or most efficient that may be true, but you want the "shortest", which means accounting for the flow to keep the movement perpendicular to the bank.
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Old 18th Jan 2017, 17:39   #13
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If that was the case you would travel at as near as possible to zero degrees( the opposite direction to the water). That would minimise the effect of the water while still retaining a tiny vertical motion component to actually move you to the other side.

That doesn't really fit in as a solution for this kind of question.
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Old 18th Jan 2017, 17:54   #14
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I think he means that the shortest possible route is in reference to time, rather than distance. Also, the current is able to add momentum to the boat, which can be steered, mid-trip, to balance momentum gained from the current with angle of traversal to optimise travel time.
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Old 18th Jan 2017, 21:12   #15
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Surely its a curve, as you row out perpendicular from the bank you use the rivers speed to assist you.

Or you could get a motor on the damn thing and not care.
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Old 19th Jan 2017, 00:12   #16
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If it was the quickest or most efficient that may be true, but you want the "shortest", which means accounting for the flow to keep the movement perpendicular to the bank.
Oh yeah, balls!

But wait, if you can't row faster than the speed of the river then there's no point rowing against the current is there?


Quote:
Originally Posted by theshadow2001
If that was the case you would travel at as near as possible to zero degrees( the opposite direction to the water). That would minimise the effect of the water while still retaining a tiny vertical motion component to actually move you to the other side.

That doesn't really fit in as a solution for this kind of question.
If you did that you'd have a huge distance and time to get to the other side, as for every second you'd be pushed downstream 5-2.9m and only 0.1m across.


I still think it's 3ms perpendicular to the bank, then for every second, you get pushed 5ms downstream, but you'd move 3m across, with a total of 576m distance.


Quote:
Originally Posted by Awoken View Post
It worked! Optimal angle is 36.9 degrees upstream of the perpendicular to the bank (53.1 degrees above the upstream bank). Total time is 125s and total distance is 500m.
Solved using quotient rule and chain rule to differentiate the distance squared. It produced a horrific mess of trigonometric terms but by factorising I obtained a slightly less messy fraction which I could then set equal to zero to determine the minimum point for the domain of 0<x<90. Setting the numerator equal to zero helped me discard some solutions which were not feasible (sin x>1) and find the only feasible solution in the domain.
A victory cup of tea is in order.
How the heck will I explain it tomorrow as it looks hideous. Must be a simpler method but I can't spot anything obvious with the geometry.

Simpler solution still up for grabs vs my somewhat inelegant brute force approach.

Using that time and distance, that means that a speed of 4ms is required, which when pointing upstream against the current of 5ms I don't think is possible if the boat can only be rowed at 3ms, if you pointed the boat perpendicular to the flow of water you could only travel at 2ms, so any direction pointing into the current cannot be faster than 2ms.
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Old 19th Jan 2017, 01:55   #17
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thanks to this thread, I am going back through maths to get back to a decent state of affairs, thanks for the jolt (and nice space pix's).
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Old 19th Jan 2017, 15:26   #18
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Nah, you can do it in 500m at roughly 40 degrees.

EDIT: Basically what awoken said, except I used differentiation by Wolfram.
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Old 19th Jan 2017, 20:43   #19
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So, I asked my dad what he thought (he's a chartered civil engineer), he came up with the same solution as me, row straight across.
300m at 3ms which takes 100s, that gives a drift downstream of 500m, which using pythagoras, gives a total distance of 583m.

So the direction the boat must be steered is 0 degrees, which makes the resultant vector of the boat 5.83ms at an angle of 30.97 degrees to horizontal, which would be ~59 degree heading.



I worked out that at 45 degrees upstream, you would have 1.5ms north component and 3.5ms downstream component (5ms-1.5ms), giving a travel time of 200s, which makes 300m north and 700m drift downstream distances, which works out to be 684m total distance at 26 degrees from horitzontal, which is a 64 degree heading. Which is longer than the 583m distance that we worked out.



What level/grade of maths is this by the way?



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Originally Posted by Awoken View Post
That was my first thought, but as the river is flowing to the East at 5ms-1 and the boat can only travel at 3ms-1, the resultant cannot be perpendicular to the bank.

I've gone back to my component formula:

If the boat's velocity, relative to the river, makes an angle of x degrees with the perpendicular to the bank upstream then the components of velocity are 3sin(x) upstream and 3cos(x) perpendicular to the bank.

The time taken to cross the river is given by t=300/(3cos(x))=100/cos(x).

The drift is given by (5-3sin(x)) multiplied by time. The resultant distance squared (by Pythagoras' Theorem) is 300^2+(((5-3sin(x))*100)/(cos(x)))^2

I'm trying a differentiation of the distance squared with respect to x and then solving equal to 0 in order to find the optimal angle between 0 and 90 degrees. Fingers crossed.

Using all of that and plugging it back in to get components, accounting for the river flowing east at 5ms, the boat would have to travel at 7.35ms in the upstream/east velocity component to make that work. As with a 36 degree angle, and the distance of 500m at 125s, which gives a resultant of 4ms, that gives a North velocity component of 3.2ms and an East velocity component of 2.35ms, but because of the river at 5ms, the east velocity must be 7.35ms to work, which isn't possible.





Make sure you get what the correct answer is and how it's worked out and post it up here, I need to know otherwise it'll be bugging me for ages,
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Old 21st Jan 2017, 01:34   #20
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Hi All,
I worked through my solution the following morning and at the end saw two numbers staring me in the face: 5 and 3
My initial values...s**t.
After a little fiddling with the vectors it became obvious. In order to obtain the shortest possible distance, the boats velocity and it's velocity relative to the river had to be perpendicular. Point the relative velocity upsteam (at an angle to the upstream bank) and the actual velocity, relative to its starting position, would point downstream (perpendicular to the velocity of the boat relative to the velocity). From there it just took a couple of minutes to realise that this whole thing could be reduced to a little bit of trig (cos x = 3/5) and some suvat calculations. With the optimal angle being 53.1 degrees (to 3sf) upstream (measured from the river bank), you can calculate the velocity of the boat, the minimum total distance travelled is therefore 500m and the time is 125s.

Thanks to everyone for giving this a shot, there were some interesting ideas that led me to a better understanding of this. The work on my original, blunt & complex solution was not wasted as it turned up some interesting results such as the plot of the function I created for the distance dependent upon the angle.

For those interested, this is Mechanics 3 from the AQA A-Level Further Maths syllabus. I've still not mastered this topic of Relative Motion but I'm getting closer every day. This has been, without a doubt, the toughest of all the topics so far (I'm looking at you Vertical Circular Motion from M2 and nth roots of unity from FP2).

The M3 textbook is a free download online. M1 and M2 are relatively inexpensive but I'd recommend that you were confident with double angle formulae and your differentiation/integration rules before you progress on to M2 and later M3.
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Last edited by Awoken; 21st Jan 2017 at 01:41.
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