Posted by Alison on Saturday, October 20, 2012 at 8:51pm.
Geesh, I wish authors would stick to the traditional wording of the "cable through a river" problem, instead of trying to be "current"
anyway...
Make a sketch labeling his position B and the point on the road 32 km away as A. let the point 16 km down from A be C
Suppose he aims for a point on the road between A and C, call that P, let AP=x
So his path is BP +PC
then BP^2 = x^2 + 32^2
BP = (x^2 + 1024)^(1/2)
time to go BP = (x^2 + 1024)^(1/2)/48
time to go PC = (16-x)/80
T = (x^2+1024)^(1/2) /48+ 16/80 - x/80
dT/dx = (1/96)(x^2 + 1024)^(-1/2) (2x) - 1/80
= 0 for a min of T
2x/(96√(x^2+1024)) = 1/80
96√(x^2+1024) = 160x
3√(x^2+1024) = 5x
square both sides
9(x^2 + 1024) = 25x^2
9216 = 16x^2
x^2 = 576
x = √576 = 24
But we clearly expected x to be between 0 and 16, so let's investigate the two direct routes.
1. directly to A, then AP
T = 32/48 + 16/80 = .86666 hrs
2. directly to C
distance = √(16^2 + 32^2) = √1280
T = √1280/48 = .7454
3. If we "blindly" sub in our x = 24 we get a total time
of T = √1600/48 + (-8/80) = 11/15 = .73333
3. consider a point between A and C, say x = 4
distance through sand = √(32^2+4^2) = √1040
time through sand = √1040/48 = .67185 hrs
time along road = 12/80 = .15
total time = .82185
I am puzzles by these results, and I can only guess that I made some arithmetic error somewhere. Just can't seem to find it, perhaps somebody else can find it.
anyway , he has 50 minutes to get there.
my 3 answers are :
.866666.. hrs = 52 minutes , go boom!
.7454 hrs = 44.7 minutes, that would do it
.7333333.. hrs = 44 minutes, so would that.
let me take a whack at it.
If the jeep arrives at the road a distance x from the bomb, then
the distance on road is x
distance on sand is √(32^2 + (16-x)^2)
travel time is thus
t = x/80 + √(32^2 + (16-x)^2)/48
dt/dx = 1/80 - (16-x)/48√(x^2-32x+1280)
dt/dx = 0 when x = -8
Also puzzling. And the graph of t(x) appears not to have a minimum for x>0.