Smart Ant

For the ant to reach the minimum time in both cases, of course it needs to use the maximum acceleration, $a$ .

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First, look at the ant's motion from $A$ to $B$ .

When the ant arrives at $B$ , it needs to move to $C$ . However, $A$ , $B$ , and $C$ are not collinear. Therefore, the ant has to change directions when it arrives at $B$ .

However, changing directions immediately to $C$ is not possible if the ant still has some velocity left. This means the ant has to stop at $B$ because no matter how large the acceleration toward $C$ is, if the ant still has some velocity left, the ant won't follow the sides perfectly.

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Therefore, the only way for the ant to stop at $B$ while going as fast as possible is like the graph below (suppose $t$ is the time needed to move from $A$ to $B$ :

If the ant stops before the midpoint, it will have to decelerate really slow to make sure it doesn't stop and still arrives at $B$ .

If the ant stops after the midpoint, its speed will become too large to stop at $B$ even with the maximum deceleration.

From the graph,

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x = \frac{1}{2} (\frac{1}{2}a) t^2 = \frac{1}{4}at^2$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad t = 2 \sqrt{\frac{x}{a}}$

Hence,

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad T_1 = 6t = 12 \sqrt{\frac{x}{a}}$

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For case 2, the ant doesn't have to stop at $A$ . This means the ant can just keep accelerating from $F$ to $A$ .

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad T_2 = 5t + \sqrt{\frac{2x}{a}} = (10+\sqrt{2})\sqrt{\frac{x}{a}}$

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Therefore,

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{T_1}{T_2} = \frac{12}{10+\sqrt{2}} \approx 1.05$