Compare travel times for two paths (one longer but lower than the other)

[Aleoa]

Homework Statement

Consider a trough of a semicircular cross-section, and an inclined plane in it that leads from a point A to a point B lying lower than A. Prove that wherever point 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. The change of direction at C does not involve a change in speed. The effects of friction are negligible.

Relevant Equations

(There are no relevant equations)

I'm really sorry, but i don't understand how to solve this problem. Can you give me some help ?

This is the picture of the problem:

 

[haruspex]

The question does not make it clear, but I would assume this is supposed to be under energy conservation.
Does that help, or had you already assumed that?

 

[kuruman]

I'm really sorry, but i don't understand how to solve this problem. Can you give me some help ?

Energy conservation ensures that the change in speed from A to B will be the same regardless of whether the object goes along the incline from A to B directly or from A to C and then back up from C to B.

You are asked to show that the time required to go directly from A to B along a single incline is always longer than the combined time along two inclines that meet at an arbitrary point C on the trough. One can compare the two times by separately calculating $t_{AB}$ and $t_{ACB}=t_{AC}+t_{CB}$ using the standard kinematic equations (they are truly relevant here) under constant acceleration for each inclined plane segment.

 

[Aleoa]

Energy conservation ensures that the change in speed from A to B will be the same regardless of whether the object goes along the incline from A to B directly or from A to C and then back up from C to B.

You are asked to show that the time required to go directly from A to B along a single incline is always longer than the combined time along two inclines that meet at an arbitrary point C on the trough. One can compare the two times by separately calculating $t_{AB}$ and $t_{ACB}=t_{AC}+t_{CB}$ using the standard kinematic equations (they are truly relevant here) under constant acceleration for each inclined plane segment.

Yes, however, I don't understand what is the property of the triangle inscribed in a semicircle that allows me to prove that $t_{AB} < t_{ACB}$

 

[kuruman]

Yes, however, I don't understand what is the property of the triangle inscribed in a semicircle that allows me to prove that $t_{AB} < t_{ACB}$

My guess (without having proved it) is that the sum of the arcs $AC + CB$ is less than a semicircle. I am guessing that the inequality will not hold for something like the configuration shown below. If that is the case, then there must be an intermediate configuration for which the two times are equal. Maybe if $AB$ is a diameter? I have to think about this.

 

[haruspex]

am guessing that the inequality will not hold for something like the configuration shown below

No, it is even a much more general result. As long as the lower path keeps going in the positive x direction and energy is conserved it will be faster. At least, I have seen a neat proof of that for the case where A and B are on the same level.
But in the present case I think the student is only expected to apply the standard SUVAT equations.