The Army Advances
An army 4 miles long advances steadily for 4 miles along a straight road. A dispatch rider leaves from the extreme rear and rides to the front to deliver important information to the commanding general. Although different speeds, the army's and the rider's speed remain constant. After delivering the info, the rider gallops back to the rear.
How far has the rider traveled?
How far has the rider traveled?
posted by Zaux at
4:16 PM
28 Comments:
let x=speed of army
let y=speed of rider
let pi=exactly 3
rider going to the front will travel 4 miles + (y-x)
headed for the rear,he will travel 4 miles - (y-x)
answer 8 miles?
I can only see an algebraic answer. The distance is at least 8 miles. But there is no upper limit. This is independent of the value of pi, LOL.
Is there some info missing?
Sorry guys ... must have been tired when I posted this one ... I am editing it with additional info
hi there Chris...i see your point.
BUT can you give an answer in a way thats...i don't know...something like...rider traveled for 8 miles + 2(y\x)?
just made that up,but you know what i mean.
sorry ....missing info is now posted
Hi Ragknot. That's why I said I could only give an algebraic answer. I didn't bother as I was sure that it was a mistake in the question.
I see that Zaux has added the distance moved by the army, which is the exact info I was expecting. Thanks Zaux.
Stunningly fiddly, but I get 4(√2+1) ≈ 9.657 miles.
Ooops. I meant Knightmare, not Ragknot.
Hi Chris ...
The rider did, in fact, travel 9.657 miles.
The Army Advances
Assuming the rider has a speed faster than the army.....
Let a=speed of army
let r=speed of rider
let z=r-a
let N=r/a, a=r/N
let L=length of army at rest
If the rider travels at infinite speed, he will travel 4+4=8
1st stage:
for the rider to catch up with the front
r*t1=L+a*t1
(r-a)*t1=L
t1=L/(r-a)
d1=r*t1= distance of first leg
d1=L*r/(r-a)=L*r/(r-r/N)=L*r/(r*(1-1/N))=L/(1-1/N)=L/((N-1)/N)=L*N/(N-1)
d1= L*N/(N-1)
Clearly as N approaches 1 d1 approaches infinity
2nd stage:
d1-r*t2=0+a*t1+a*t2
d1-a*t1=(r+a)*t2
but r*t1=L+a*t1 OR a*t1=r*t1-L=d1-L
d1-(d1-L)=(r+r/N)*t2
L=r*(1+1/N)*t2
L/r*1/(1+1/N)=t2
L/r*1/((N+1)/N)=t2
L/r*N/(N+1)=t2
d2=r*t2
d2=r*L/r*N/(N+1)
d2=L*N/(N+1)
d1+d2= L*N/(N-1)+ L*N/(N+1)
d1+d2=L*N*(1/(N-1)+1/(N+1))
d1+d2= L*N*((N+1+N-1)/((N-1)(N+1)))
d1+d2= 2*L*N^2/(N^2-1)
As N -> infinity, d1+d2 -> 2*L
As N-> 1 , d1+d2 -> infinity
Answer:
So in this case where L=8, the minimum distance is 8, and d1+d2 can be expressed by
d1+d2= 2*L*N^2/(N^2-1)
OR
d1+d2= 2*8*N^2/(N^2-1)
where N is the ratio of the speed of the rider to the army
Cam
Let a and r be the army's and the rider's speed. r > a > 0
Outward journey, d1 = r*t1 = 4 + a*t1 => t1 = 4/(r-a)
Return journey, d2 = r*t2 = 4 - a*t2 => t2 = 4/(r+a)
Also have d = d1 + d2 = r(t1 + t2) = 4r(1/(r-a) + 1/(r+a))
End up with, d = 8r²/(r²-a²)
Let x = r/a. Then d = 8x²/(x²-1)
During all this, the army has moved 4 miles, so 4 = a(t1 + t2)
So d = r(t1+t2) = 4r/a = 4x
Comparing the two expressions => 4x = 8x²/(x²-1)
Assuming x <> 0 => x² - 2x - 1 = 0
=> x = (2 +/- Sqrt(4+4))/2 = 1 +/- Sqrt(2
As x > 0, x = 1 + √2
So d = 4(1 + √2) ≈ 9.657 miles
That was messy.
Meh, I solved it for when the distance the soldiers travelled wasn't given..... I see the problem has now been changed.
Cam
Hi Cam, I guess you missed the updated info. But I see that you have it right, as expected ;)
Posts crossing. I delayed doing it (unlike you) until I was sure what the precise form of the missing info was going to be. I knew it was going to be fiddly.
That didn't sound right. I meant I was too lazy to give a general answer.
definitely its 8 miles.. but if asked how many miles going to the gen comm and then back, then definitely an info is lacking.. the ditanc traveled going to the gen comm should be greater due to the moving line which increases the distace traveled from 4 miles.. then the distance traveled going back would be deacreasing due to the line traveling in a direction against to the post man..
definitely its 8 miles.. but if asked how many miles going to the gen comm and then back, then definitely an info is lacking.. the ditanc traveled going to the gen comm should be greater due to the moving line which increases the distace traveled from 4 miles.. then the distance traveled going back would be deacreasing due to the line traveling in a direction against to the post man..
definitely its 8 miles.. but if asked how many miles going to the gen comm and then back, then definitely an info is lacking.. the ditanc traveled going to the gen comm should be greater due to the moving line which increases the distace traveled from 4 miles.. then the distance traveled going back would be deacreasing due to the line traveling in a direction against to the post man..
Sorry Dadot ...
definitely its not 8 miles
Right - It's definitely 4 miles, just like everyone else in the column...
-h
noooooo... it's 4(1 + √2) ≈ 9.657 miles.
If you read it as a trick question, then I'll concede that 4 miles is the answer.
When the rider gets to the front, he will then have travelled
4(1 + √2)/√2 ≈ 6.828 miles
Chris is right ... and it's not a trick question ... actually the trick is to find the solution.
He travels 4 miles plus 1 step. He rides from the rear to the front, a distance of 4 miles. (regardless of speed the line is 4 miles long). He then takes one very slow step towards the rear and the end of line passes him. Thus he travels 4 miles plus 1 step. Albeit, his one step is very slow.
Pamela...that's the single smartest thing i've ever heard anybody say about anything.i mean,why put in the work to travel to a place,when you can sit there and wait for that place to come to you!...me likes ;)
Hi Pamela. I'm not sure if you're being serious-ish.
By the time the rider gets to the front of the army, the army has walked approx 2.828 miles, you can't just dismiss that.
Also the question says that the rider and the army move at constant speed. The rider isn't allowed to just wait for the army to catch up with him.
Coming to this question lat, I see most people have come up with an answer.
The alternative is radio communications, with the rider covering no miles.
Post a Comment
Links to this post:
Create a Link
<< Home