Show that acceleration varies as cube of the distance given

Thread starter chwala
Start date Oct 27, 2023
Tags

Acceleration Calculus distance

Click For Summary

The discussion presents a mathematical approach to demonstrate that acceleration varies as the cube of the distance. The relationship starts with distance represented as x and velocity as x', leading to the equation x' = kx^2, where k is a constant. The acceleration is derived as x'' = 2k(x)(x'), which simplifies to x'' = 2k^2x^3. This indicates that acceleration is proportional to the cube of the distance, confirming the initial premise. The calculations appear to be correct and consistent with the proposed relationship.

Oct 27, 2023

chwala

Gold Member

Homework Statement: See attached.

Relevant Equations: Rate of change

In my approach i have distance as ##(x)## and velocity as ##(x^{'})##, then,

##(x^{'}) = kx^2##

where ##k## is a constant, then acceleration is given by,

##(x^{''}) = 2k(x) (x^{'})##

##(x^{''}) = 2k(x)(kx^2) ##

##(x^{''}) = 2k^2x^3##.

Correct?

Physics news on Phys.org

Oct 27, 2023

anuttarasammyak

Gold Member

It seems OK.

Thread 'Distance between a Clock's hands when the distance is increasing most rapidly'

Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

View full post »