Optimal Path for Honeybee to Reach Honey Drop in Minimum Time - No Gravity

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The discussion revolves around the optimal path for a honeybee to reach a honey drop in the shortest time, assuming no gravity and maximum acceleration. The honeybee's strategy involves moving in a straight line to minimize distance, which logically correlates with minimizing time. Concerns were raised about the assumption of linear acceleration and whether the bee should consider alternative paths. However, it was concluded that linear acceleration maximizes speed, thereby minimizing time effectively. The consensus is that minimizing distance inherently leads to minimizing time when speed is maximized.
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Homework Statement



A honeybee is flying parallel to tabletop at height
c1975b3ac41216fc73cd36b0e4434cf67a5bcf30.png
with velocity
9ad7fa55a37ab127b0ffc739f278c0c65e1bb490.png
.With its wings it can acquire a maximum acceleration of
e207f7dd1f98225ce817d245322a4dce995d73f3.png
. At an instant when honeybee is vertically above honey drop on tabletop and decides to reach the honey drop. Neglect reaction time of honeybee and there is no gravity. Find minimum time in which honeybee can reach honey drop.

2. The attempt at a solution

Let us work in the frame moving with the constant velocity
a9f23bf124b6b2b2a993eb313c72e678664ac74a.png
, such that now the honeybee is simply moving with
c7d457e388298246adb06c587bccd419ea67f7e8.png
and the honey drop moves in the opposite direction (to the initial velocity of the honeybee) with a constant velocity
a9f23bf124b6b2b2a993eb313c72e678664ac74a.png
. The honeybee obviously wants to travel the least distance in order to catch the honey drop, so should compute the point it will catch the honey drop if it travels on a straight line towards that point. Let the honey drop's coordinates at a certain time defined
5a607fd61885c014512c949fee18869690330407.png
be
f9603ca3089464e548fc6f1366bc474e7efef8d9.png
, and the honeybee be at
57acd195dfc44927ca230fbd512bf32cfeba669c.png
at this time. After some time
e0d2bf360290fd61d1c1557e763f2622363b3d35.png
, the coordinates of the honey drop would be
461eaa0c47244a9b94b69aa8936ba22b85973cd6.png
. Now, obviously
19ad84175ba436ee089e05012a8eb9b9e98c8515.png

The solution of this equation would be at
65360f557164b60bd01f1b1122adc6ff987ba117.png
, that is, when the honey drop is finally caught by the honey bee. Using the well-known quadratic formula for the equation written above,
aafa64c95c85226b146a31ab1462091f6a82a19e.png


My question is: is this wrong? The answer matches, but an acquaintance comes and asks me, why did you assume linear acceleration? Why do you think honeybee only wants to travel the least distance? Because the problem says: find the minimum time...

Please help me asap.
 

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PhysicsKT said:
My question is: is this wrong? The answer matches, but an acquaintance comes and asks me, why did you assume linear acceleration? Why do you think honeybee only wants to travel the least distance? Because the problem says: find the minimum time...

Please help me asap.

It's logical that if you start at a point and need to reach a point on a given line, then a straight line will minimise the distance. If you travel further, along a curve say, then you need a greater average speed to get there in the same time. And, linear acceleration maximises your speed, as well. An acceleration component that changes direction does not increase your speed.
 
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PeroK said:
It's logical that if you start at a point and need to reach a point on a given line, then a straight line will minimise the distance. If you travel further, along a curve say, then you need a greater average speed to get there in the same time. And, linear acceleration maximises your speed, as well. An acceleration component that changes direction does not increase your speed.
But what we want is to minimise time, not distance.
 
PhysicsKT said:
But what we want is to minimise time, not distance.

If you minimise distance and maximise speed, then you must minimise time. You can't go further and slower in less time.
 
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Thanks. Got it! :biggrin:
 
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