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Problem of physics. Help!!?
An ant and honey are on a perfect sphere miraculously suspended in mid-air. The radius of the sphere is 5cm.
The drop of honey is at the north pole of the sphere, moving towards the south pole at a constant speed of 1.5 cm/s (for no good reason ... gravity would neither account for the drop leaving its cosy northern spot, nor for the constant speed) and stops as soon as it arrives there (unless eaten prior to that).
The ant sits at the equator, unfortunately on the opposite side of the drop's path . It decides for the shortest intercept route and sets off immediately.
When and at which latitude will the ant catch the drop?
I forgot to mention the ants speed: 2.5 cm/ second. My apologies!
1 Answer
- falzoonLv 79 years agoFavorite Answer
The interesting thing here, is that the ant's shortest path can only be in either one of
two directions -
(1) If the honey lands in the northern hemisphere, then the ant must follow the great
circle that goes over the north pole along the same meridian as the honey, that is,
the ant is on the same line of longitude, trying to catch up to the honey.
(2) If the honey lands in the southern hemisphere, then the ant must follow the great
circle that goes over the south pole along the same meridian as the honey, that is,
the ant is on the same line of longitude and meets the honey head on from the
opposite direction.
CASE (1):
Equation for honey : d = sT, that is, d = 1.5T. Therefore, T = d/1.5.
Assume the ant goes over the north pole. Then it has to travel an extra 1/4 of the
circumference of the sphere, which is 2π(5)/4 = 5π/2 cm.
Equation for ant : D = ST, that is, 5π/2 + d = 2.5T. Therefore, T = (5π/2 + d)/2.5.
Equating both T's gives :
d/1.5 = (5π/2 + d)/2.5
Solving gives : d = 15π/4 cm.
1/4 of the circumference of the sphere = 5π/2 cm ≈ 7.9 cm.
1/2 of the circumference ≈ 15.8 cm.
Now, d = 15π/4 ≈ 11.8 cm, which puts the honey between those two results,
and thus, the ant must reach the honey in the southern hemisphere.
The ant has now covered a bit more than half a great circle, so it looks like the
second option is the quickest.
CASE (2):
Equation for honey : d = sT, that is, d = 1.5T. Therefore, T = d/1.5.
If the honey's distance is d, then the ant's distance must be C - d - C/4, where
C is the circumference of the sphere. Simplifying, C - d - C/4 = 15π/2 - d.
Equation for ant : D = ST, that is, 15π/2 - d = 2.5T. Therefore, T = (15π - 2d)/5.
Equating both T's gives :
d/1.5 = (15π - 2d)/5
Solving gives : d = 45π/16 cm (≈ 8.8 cm),
which puts the honey just under the equator.
T = d/1.5, so ant catches the drop when T = (45π/16)/1.5 = 15π/8 sec. ≈ 5.9 sec.
1/2 circumference = 5π cm = 180º. Therefore, 45π/16 cm = (45π/16)*180/(5π)º,
which is 101.25º. Subtracting 90º gives the latitude of the meeting as 11.25º S.
It may have been the long way to go about it, but I sure hope it's correct.
