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I've not clearly understood what it's meant by the text of this problem. To me, it seems that I have to show why an object travels AB in less time than ACB. However, if an object travels from A to C, why doesn't it stop at C?

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closed as off-topic by Bob D, John Rennie, Bill N, Kyle Kanos, tpg2114 May 11 at 7:03

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    $\begingroup$ Would help if I fully understood which is B and C... $\endgroup$ – Rick May 5 at 13:06
  • $\begingroup$ The book doesn't show. If we consider the triangle in the figure, i think A is the top-left vertex, B is the top-right vertex and C the remaining vertex $\endgroup$ – AleQuercia May 5 at 13:18
  • $\begingroup$ You have to prove why an object travels faster along ACB than along AB according to the text, which says, "Prove that wherever pt. C is chosen on the arc AB, an object will always get from A to B faster along the slopes ACB than along the original slope AB." If an object travels from A to C, it wouldn't stop at C because of the inertia of motion. $\endgroup$ – Tapi May 5 at 13:24
  • $\begingroup$ Okay thanks. So is the inertia what i didn't understood. How can i understand deeply why an object at C doesn't stop ? $\endgroup$ – AleQuercia May 5 at 13:41
  • $\begingroup$ No change in speed means there must be an elastic collision when it hits C. $\endgroup$ – Cuspy Code May 5 at 13:54
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You have to prove why an object travels faster along ACB than along AB according to the text, which says, "Prove that wherever pt. C is chosen on the arc AB, an object will always get from A to B faster along the slopes ACB than along the original slope AB."

If an object travels from A to C, it wouldn't stop at C because of the inertia of motion. Moreover, it is a perfectly elastic collision when it hits C because the direction changes but the speed doesn't. So, the object will always move with the same speed through CB as it moved with along AC.

The speed when traveling from A to C is more than the speed any object would have when traveling from A to B (without external force). This is because the path from A to C is steeper than the path from A to B.

Mathematically, if you take the acceleration of the object to be 'a', and take its vertical component, it turns out to be $a cos\theta$, $\theta$ being the angle AB/AC makes with the normal (parallel to the side of the container). Now, $cos\theta$ is more for AC than for AB, because $cos \theta$ decreases as $\theta$ increases. Hence, $a cos\theta$ is more for AC, than for AB. Thus, acceleration is more for AC, making the journey through ACB faster than through AB.

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