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Particle Acceleration (Posted on 2004-05-06) Difficulty: 2 of 5
A particle is travelling from point A to point B. These two points are separated by distance D. Assume that the initial velocity of the particle is zero.

Given that the particle never increases its acceleration along its journey, and that the particle arrives at point B with speed V, what is the longest time that the particle can take to arrive at B?

No Solution Yet Submitted by SilverKnight    
Rating: 2.3333 (3 votes)

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Solution My uninfluenced solution | Comment 13 of 24 |

I assume that decreasing the acceleration will not maximize the time, that the puzzle is one-dimensional, and both D and V are positive.

Considering the movement of the particle on an x-y graph, it should look like a parabola.  The equation of the parabola is y=Ax/2, where y=distance, A=acceleration, and x=time.  The speed is equal to Ax.  When this parabola intersects y=D, the particle hits point B.

Before maximizing time, I'm just going to figure out the possibilities for A.

The coordinates of the intersection of y=Ax/2 and y=D:
D=Ax/2
x=+sqrt(2D/A)
y=D
(+sqrt(2D/A),D)

The speed of this intersection will be A*sqrt(2D/A).
V=A*sqrt(2D/A)
V=sqrt(2DA)
A=V/2D

It seems that there is only one possibility for A, so the time is already maximized.
x=+sqrt(2D/A)
x=+sqrt(4D/V)
x=2D/V

Now, to prove my first assumption: "that decreasing the acceleration will not maximize the time."  Decreasing the acceleration will make the ending speed slower then it would have been.  Since it must be at speed V, the beginning speed must be faster to compensate.  This makes the time shorter.


  Posted by Tristan on 2004-05-07 23:16:25
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