Marching Cadets & the Trotting Dog
A square formation of Army cadets, 50 feet on a side, is marching forward at a constant pace. The company mascot, a small terrier, was standing at the center of the rear line of cadets when the march began. Also when the march began, the mascot began running, at a constant pace, along the perimeter of the square identified by the cadets. (For the sake of argument, we will define the dog's path as always directly on the line of the square identified by the cadets... and also that he loses no time in making a turn). When the mascot reaches the point of the formation, where he originally started, the Cadets halt their march.
The formation has moved 50 feet ... how far did the terrier run?
- Zaux
The formation has moved 50 feet ... how far did the terrier run?
- Zaux
Labels: mathemagic, SharedPuzzle
posted by Rajesh Lal at
12:05 AM
22 Comments:
Total distance covered is the four sides of the formation (200 feet) plus the distance the formation has moved (50 feet), i.e. 250 feet.
I'll be surprised if that's the right answer.
Can't look at now, as 'puter is playing up.
Nope ... not 250ft....
I get 209.056 feet. But I'm not happy that I haven't dropped a clanger somewhere.
If I'm right, the exact is expression is very horrible:
d = 50 +
1/2 Sqrt[
40000/3 + 1/3 (23500000000000 - 1500000000000 Sqrt[93])^(1/3) +
5000/3 2^(2/3) (47 + 3 Sqrt[93])^(1/3)] +
1/2 \[Sqrt](80000/3 -
1/3 (23500000000000 - 1500000000000 Sqrt[93])^(1/3) -
5000/3 2^(2/3) (47 + 3 Sqrt[93])^(1/3) + 2000000/Sqrt[
40000/3 +
1/3 (23500000000000 - 1500000000000 Sqrt[93])^(1/3) +
5000/3 2^(2/3) (47 + 3 Sqrt[93])^(1/3)])
This post has been removed by the author.
This post has been removed by the author.
For convenience, let the dog start at the rear left corner and go
round ant-clockwise. Let the square's speed be u and the dog's
speed be v. (v > u > 0).
Let time taken to cross the back edge be t_b. Then distance the
platoon travels is u*t_b. So the dog travels a distance
v*t_b = √(50² + (u*t_b)²) => t_b = 50/√(v²-u²)
The same time and distance apply when the dog crosses the front.
Let t_r be the time taken for the dog to move up the right side,
then the square travels u*t_r and the dog travels 50+u*t_r = v*t_r.
Rearranging => t_r = 50/(v-u).
Let t_l be the time taken for the dog to move down the left side,
then the square travels u*t_l and the dog travels 50-u*t_l = v*t_l.
Rearranging => t_l = 50/(v+u).
The total time taken t = (2*t_b + t_r + t_l)
=> t = 2*50/√(v²-u²) + 50/(v-u) + 50/(v+u)
=> t = (√(50/(v-u)) + √(50/(v+u)))²
=> t = (50/u)(1/√((v/u)-1) + 1/√((v/u)+1))²
The distance the square moves is: 50 = u*t => t = 50/u and the
dog moves D = v*t = (v/u)*50. Substituting into the previous
and square rooting
=> 1 = 1/√((D/50)-1) + 1/√((D/50)+1)
Solving for D => D ≈ 209.056
To do the calculation I used:
http://www.wolframalpha.com/input/?i=NSolve%5B1%2F%E2%88%9A%28%28D%2F50%29-1%29+%2B+1%2F%E2%88%9A%28%28D%2F50%29%2B1%29%3D%3D1%2Cd%5D
http://www.wolframalpha.com/input/?i=Solve%5B1%2F%E2%88%9A%28%28D%2F50%29-1%29+%2B+1%2F%E2%88%9A%28%28D%2F50%29%2B1%29%3D%3D1%2Cd%5D
I may be underthinking this but wouldnt it be a simple rectangle with sides of 50 and 100 giving a total distance of 300?
On second thought... I'm way off.
For it to be a simple rectangle, the start and end point would be the same ... in this case, they are 50ft. apart.
Hi Zaux, did I get the right answer?
Hi Chris ... yep ... nice work. You have a math background or just love it?
Hi Zaux. Thanks for putting me out of my misery - although I was fairly sure that the dog didn't travel much more than the minimum 200 feet. It seemed reasonable that the dog would be about 4 (or 5 worst case) times quicker than the cadets, and so saw that the left and right sides distance path differences would largely cancel, and also thanks to Pythagoras and small angle approximations, that the angled parts of the path would only be a little bigger than 50 feet.
I've a degree in electronics, so learnt plenty of maths getting that. But I also love maths and physics and have learnt a lot of stuff about both informally.
Dog Mascot
Label the following points on the moving marching square.
Call starting point A
Bottom right corner B
Top Right Corner C
Top Left corner D
Bottom left corner E
D-----C
* *
* *
* *
E--A--B
-Dog’s speed is contant
-Cadet’s march forward at a contant speed
-In the time for Dog to make journey AB+BC+CD+DE+EA, the cadet’s have marched forward 50 feet
Call vs the cadet’s forward speed
Call vd the dog’s speed
Distance=velocity * time
Find distance for each side
---------------------------
Distance of each leg is the vector sum of the relative distance from the point to the next point plus the distance travel by the cadets:
d(rside)=50+vs*Trside
d(top)=sqrt(50^2+(vs*Ttop)^2)
d(lside)=50-vs*Tlside
by symmetry d(AB)+d(EA)=d(top)
Find time spent on each side
----------------------------
for clarity
Trside=Time from B to C
Tlside=Time from D to E
Ttop= Time from C to D
for left and right sides solve for when dog’s position matches the corner cadet’s position
Corner C Cadet's position=Dog's position
50+vs*Trside= vd*Trside
50=(vd-vs)*Trside
Trside=50/(vd-vs)
Dog's position =Corner E Cadet's position
50-vd*Tlside=vs*Tlside
50=(vs+vd)*Tlside
Tlside= 50/(vs+vd)
for top use d=v*t
d(top)=sqrt(50^2+(vs*Ttop)^2)= vd*Ttop
vd^2*Ttop^2=50^2+vs^2*Ttop^2
Ttop^2*(vd^2-vs^2)=50^2
Ttop=50/sqrt(vd^2-vs^2)
Total Time relation
-------------------
Ttotal= Tlside + Trside + 2*Ttop
vs*Ttotal =50
vs*( 50/(vs+vd) + 50/(vd-vs) + 2*50/sqrt(vd^2-vs^2)) = 50
vs*( 1/(vs+vd) + 1/(vd-vs) + 2/sqrt(vd^2-vs^2)) = 1
Multiply both sides by (vs+vd)*(vd-vs)
Note: (vs+vd)*(vd-vs)= vs*vd-vs^2+vd^2-vs*vd= vd^2-vs^2
vs*( (vd-vs) + (vs+vd)) + 2*sqrt(vd^2-vs^2)) = vd^2-vs^2
vs*vd-vs^2+vs^2+vs*vd+2*vs*sqrt(vd^2-vs^2)-vd^2+vs^2=0
vs^2+2*vs*vd-vd^2+2*vs*sqrt(vd^2-vs^2)=0
The relation in this form is not useful unless we can bring it down to one variable. Thinking carefully about the problem, one may realize that the actual time and actual speed cannot be solved for, as enough information has not been provided to do so. But what we do know is that the relative position of the dog to the cadets must be maintained and the relative speed of the dog to the cadets must be maintained, like watching the event on a video tape in slow motion, normal speed or fast forward.
We introduce N where N=vd/vs
We then attempt to use this variable to bring the relation to a single variable.
vs^2+2*N*vs*vs-N^2*vs^2+2*vs*sqrt(N^2*vs^2-vs^2)=0
(1+2N-N^2)*vs^2+2*vs*sqrt(vs^2*(N^2-1))=0
(1+2N-N^2)*vs^2+2*vs^2*sqrt(N^2-1)=0
divide both sides by vs^2 to yield the single variable equation
1+2N-N^2+2*sqrt(N^2-1)=0
solve via numerical methods (I use a version of Newton's method, but goal seek in Excel should work fine)
N~=4.1809
Sounds reasonable as the time for the dog to travel 4 sides of a non-moving square in the time the cadets travel 1 side would be 4. It is interesting to note that N is independent of the side length.
vs*Ttotal =50
Ttotal= 50/vs
dtotal=vd*Ttotal=N*vs*50/vs=N*50
dtotal~=4.1809*50
dtotal~=209.0573
Answer:
dtotal~=209.0573
Lots of equations. Hopefully no mistakes above.
Cam
This one was a lot trickier than I initially thought it would be. When I had the equation reduced to terms of vd an vs I was scratching my head for a while as to how to get it down to a single variable.
Chris:
Nice work. Looks like we followed similar paths to the solution.
Zaux:
Nice problem. I enjoyed it.
Cam
Hi Cam. D is nearer to 209.056272264633715027229128 ;) and so N is nearer to 4.18112544529267430054458256.
Try this (for N): http://www.wolframalpha.com/input/?i=1+%2B+2N+-+N%5E2+%2B+2*Sqrt%28N%5E2-1%29+%3D+0
This post has been removed by the author.
Hi again Cam. I was a bit worried that I couldn't easily get it down to one variable either. The expressions looked like they were ready to become quite lengthy. Luckily, I found some good substitutions on the first try and knew that www.wolframalpha.com could do the donkey work.
Thanks for the problem Zaux.
The Answer is 300 Feet. The dog initially has to travel 50 feet because the cadets are moving but to return to its original position (which is moved 50 meters forward) it also has to travel another 50 feet back. Hence the dog has to travel 200f + 100f = 300 feet
Glad it was fun ...
Hi Karlossy. Where do you think Cam and I goofed?
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