Show that acceleration varies as cube of the distance given

  • #1
chwala
Gold Member
2,697
360
Homework Statement
See attached.
Relevant Equations
Rate of change
1698409888454.png


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
  • #2
It seems OK.
 
  • Like
Likes chwala

Related to Show that acceleration varies as cube of the distance given

What is the basic principle behind acceleration varying as the cube of the distance?

The principle is based on a specific physical relationship where the acceleration of an object is directly proportional to the cube of the distance from a reference point. This relationship can arise in certain theoretical models or specific force fields, although it is not common in everyday physical situations.

Can you provide a mathematical expression for acceleration varying as the cube of the distance?

Yes, the mathematical expression can be written as \( a = k \cdot d^3 \), where \( a \) is the acceleration, \( d \) is the distance, and \( k \) is a constant of proportionality.

What are some examples or scenarios where acceleration varies as the cube of the distance?

While it is rare in classical mechanics, such a relationship might be found in certain theoretical physics models or specific gravitational or electromagnetic field configurations. However, these are generally more complex and less intuitive than the inverse-square law commonly seen in gravity and electromagnetism.

How do you determine the constant of proportionality in the equation \( a = k \cdot d^3 \)?

The constant of proportionality \( k \) can be determined experimentally by measuring the acceleration and distance for different values and then fitting these data points to the equation \( a = k \cdot d^3 \). Alternatively, it can be derived from theoretical considerations based on the specific system being studied.

What are the implications of acceleration varying as the cube of the distance in physical systems?

If acceleration varies as the cube of the distance, it implies that the force causing this acceleration increases very rapidly with distance. This can lead to highly non-linear and potentially unstable systems, depending on the context. Such a relationship would significantly affect the dynamics and behavior of objects within the system.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
624
  • Calculus and Beyond Homework Help
Replies
15
Views
923
  • Calculus and Beyond Homework Help
Replies
7
Views
547
  • Calculus and Beyond Homework Help
Replies
5
Views
507
Replies
12
Views
563
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top