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?

I had to make two related assumptions:
1) You can't overshoot B and come back to it.
2) A negative acceleration from A to B is unallowed. (You can't go backwards.)

The slowest time comes from the lowest acceleration for the longest time. Since you can't increase the acceleration, start with the lowest acceleration possible and keep it constant.

If I remember my physics, a graph of D vs. Time will be a parabola. If the initial velocity is zero, the vertex is at the origin.

We want to reach the point (T,D) with velocity (derivative) V.
Subsitute the point into the parabola equation y=ax^2 to get y=(D/T^2)x^2
derivative y'=(2D/T^2)x
at time T, velocity is V, so
V=(2D/T^2)T
solve for T

T = 2D/V

I hope this is what you are looking for.

-Jer

Posted by Jer on 2004-05-06 09:51:54