Rap:
Column is 1 mile long & traveling at constant rate (Rc). In time T, column traveles 1 mile. Therefore, the total time is t=1/Rc
Lead soldier drops off of the front & travels to the rear at constant speed Rs the time requires to get to the rear of the column is t1, the distance traveled is d1=1-Rc*t1=Rs*t1
Rs*t1=1-Rc*t1
Rearranging
t1=1/(Rs+Rc)
Time for soldier to get back to the front
d2=1+Rc*t2=Rs*t2
Rs*t2=1+Rc*t2
Rearranging
t2=1/(Rs-Rc)
Total time is t=t1+t2=1/Rc
or
1/(Rs+Rc)+1(Rs-Rc)=1/Rc
combining and rearranging
Rs^2-2*Rc*Rs-Rc^2=0
Put an arbitrary speed in for Rc since the total times are the same and the distances ratios are functions of rates (I'm easy use 1 for the column rate), the the above quadratic becomes
Rs^2-2*Rs-1=0
Solving for Rs, using the good old quadratic equation
Rs=[(2+/-(2^2-4*1*(-1))^(1/2)]/2=1+/-2^(1/2)
Negative solution doesn't pass the sanity test so
Rs=1+sqrt(2)~2.42
Therefore the soldier travels 2.42 times the distance the column travels.
Good to see ya buddy; you have confirmed my answer
Let r1 = rate of the column = 1 mile/hr.
Let r2 = rate of the messenger.
d = distance covered by messenger.
t = time to move one mile = 1 hour.
The meeting point of the messenger and the person at the rear is a distance of r2/(1+r2) from the starting point of the messenger. This is just the ratio of the messenger's rate divided by the sum of the two rates.
The messenger will have to cover the distance to the meeting point, turn around and cover it again, and then march one more mile. So...
d = 2*r2/(1+r2).
Distance = rate * time, so...
2*r2/(1+r2) = r2*1
2*r2/(1+r2) = r2 -1
2*r2 = r2^2 -1
r2^2 - 2*r2 - 1 = 0
Use the quadratic equation (same one as Rap) to solve for r2...
r2 = (2 + 8^1/2)/2 = 1+2^1/2 =~ 2.4142 miles/hr
Since he is marching for one hour is distance is 2.4142 miles.
Mark:
10 INCH STICK
If I did my triple integral correctly, the answer is 36/1000.
If you did; them I messed up! Check this out
Instead of 10 inches let's give the stick a length of 1. Let x be the point of the original cut. The probability given x that all four pieces are 0.1 long or greater is (x^2-x+0.16)/(x^2-x).
Next take the integral over this probability for the larger initial cut ranging from 0.5 to 0.8. Note that if it were greater than 0.8 the smaller piece could not have two cuts more than 0.1. So we have 2 times the integral from 0.5 to 0.8 of :
(1) x^2/(x^2-x) + ...
(2) -x/(x^2-x) + ...
(3) 0.16/(x^2-x)
Integral (1) works out to 0.3+ln(.4)
Integral (2) works out to -ln(.4)
Integral (3) can also be expressed as 0.16*(1/(x-1) - 1/x). The integral of 1/x from 0.5 to 0.8 is ln(0.8)-ln(0.5).
The integral from 1/(x-1) from 0.5 to 0.8 = ln(x-1) from 0.5 to 0.8 = ln(-0.2)-ln(-0.5) = ln(-1)+ln(.2)-(ln(-1)+ln(.5)) = ln(.2)-ln(.5).
So the initial expression equals 0.16*(ln(0.2)-ln(0.5)-(ln(0.8)-ln(0.5))) = 0.16*(ln(0.2)-ln(0.8)) = 0.16*ln(.2/.8) = 0.16*ln(1/4) = -0.16*ln(4)=-0.32*ln(2).
So the final answer is:
2*(0.3+ln(.4)-ln(.4)-0.32*ln(2)) = 0.6-0.64*ln(2) = (15-16*ln(2))/16 = 0.1564
Or is it?
GLRET
DSOWR
GLETS
