Asked by Alison
It is noon. A secret agent is driving a jeep through the desert in the tiny principality of Alta Loma. He is 32km from the nearest point on a straight, paved road. Down the road 16km is a power plant in which a band of international terrorists has placed a time bomb set to explode at 12:50pm. The jeep can travel at 48km/hr in the sand at 80 km/hr on a paved road. If he arrives at the power plant in the shortest possible time, how long will our hero have to diffuse the bomb?
Answers
Answered by
Reiny
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.
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.
Answered by
Steve
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.