The Fastest Path

[wblipp]

Originally Posted by a1call

Along a straight downhill path the ball will accelerate constantly (linearly) as a function of g and the angle of decline.

See xilman above. "underspecified" means that your original problem statement doesn't say anything about balls nor gravity. There are MUCH faster paths from A to B if, for example, you strap a rocket engine onto the ball.

[a1call]

Originally Posted by wblipp

See xilman above. "underspecified" means that your original problem statement doesn't say anything about balls nor gravity. There are MUCH faster paths from A to B if, for example, you strap a rocket engine onto the ball.

I thought it was clear.
The power source is gravity of Earth near the surface of the Earth.
When I get the chance I will post the answer to the 1st part.
For the 2nd part, infinite right-left symmetrical cycloids can be drawn through the 2 points A and B. Any of them would yield a faster path than a straight path from A to B.
I only assume that the fastest cycloid path would be the one with it's highest point at A.
Perhaps it would be more complete to state the 2nd question as the following.
What is the fastest time a ball can roll from A to B with initial speed 0 and disregard to friction/drag. The choice of the path construct is up to you.

[science_man_88]

Originally Posted by a1call

The choice of the path construct is up to you.

okay well looking up trig functions the ratio of linear falling and distance along the slope would depend on the angle at b theta which depends on if you mean the 1 degree to the horizontal to be at b or at a ( and yes it does matter as a one degree down slope at a is the equivalent of a 89 degree upslope from b) but the angle made theta would allow the function of the linear falling distance to be equated to the sloped fall, for them both to equal along the slope you would have d=csc(theta)*(vi*t+1/2*a*t^2) under classical newtonian mechanics using the fact that vi=0 and the distributive law of multiplication over addition we get d= 1/2*csc(theta)*a*t^2.

[a1call]

Originally Posted by science_man_88

okay well looking up trig functions the ratio of linear falling and distance along the slope would depend on the angle at b theta which depends on if you mean the 1 degree to the horizontal to be at b or at a ( and yes it does matter as a one degree down slope at a is the equivalent of a 89 degree upslope from b) but the angle made theta would allow the function of the linear falling distance to be equated to the sloped fall, for them both to equal along the slope you would have d=csc(theta)*(vi*t+1/2*a*t^2) under classical newtonian mechanics using the fact that vi=0 and the distributive law of multiplication over addition we get d= 1/2*csc(theta)*a*t^2.

I do not understand what you are saying. AB makes a 1° angle with the horizontal. The only way this would make a difference if measured at A or B is if you assume Horizontal being arced along the surface. at 4.2 km the curvature of the Earth is negligible and would make a marginal difference from a flat horizontal assumption.
But fair enough: Please base your calculations based on a simplified flat horizontal.

[science_man_88]

Originally Posted by a1call

I do not understand what you are saying. AB makes a 1° angle with the horizontal. The only way this would make a difference if measured at A or B is if you assume Horizontal being arced along the surface. at 4.2 km the curvature of the Earth is negligible and would make a marginal difference from a flat horizontal assumption.
But fair enough: Please base your calculations based on a simplified flat horizontal.

__A
;;/
B/_

the sum of the angles of a right angle triangle sum to 180 one is 90 I'm asking which one if a triangle is formed is the 1 degree it could be the interior angle b or the exterior angle a.

[a1call]

Originally Posted by science_man_88

__A
;;/
B/_

the sum of the angles of a right angle triangle sum to 180 one is 90 I'm asking which one if a triangle is formed is the 1 degree it could be the interior angle b or the exterior angle a.

The constraints are complete. You are unnecessarily confusing yourself.
Take a piece of paper:

* Draw a horizontal line
* Draw line AB at an angle of 1 degree related to the previous line.

There are only 2 orientations/directions possible for AB. One descending to right and the other descending to left. Both are equivalent. The relative positions of points A and B relative to the original horizontal line are irrelevant.

[a1call]

Here is what I have for 1st part of the problem (i.e. a straight path from A to B):

AB
a = g sin 1° = 0.171149641
v = v₀+gt sin 1°
x = v₀t+(g(t²) sin 1°)/2

t = √(2x/(g sin 1°) = 220.91 s ≈ 3.68 min.

[ET_]

Originally Posted by a1call

Here is what I have for 1st part of the problem (i.e. a straight path from A to B):

AB
a = g sin 1° = 0.171149641
v = v₀+gt sin 1°
x = v₀t+(g(t²) sin 1°)/2

t = √(2x/(g sin 1°) = 220.91 s ≈ 3.68 min.

3.68 min = 4.08 min?

[science_man_88]

Originally Posted by ET_

3.68 min = 4.08 min?

if the 220 s is correct then the 3.68 estimate is correct 4 minutes would be 240 s or more.

[ET_]

Originally Posted by science_man_88

if the 220 s is correct then the 3.68 estimate is correct 4 minutes would be 240 s or more.

So 3'40"910 = 3.68 min.

[science_man_88]

Originally Posted by ET_

So 3'40"910 = 3.68 min.

(10:45) gp > 220.91/60.
%1 = 3.6818333333333333333333333333333333333