Relation between acceleration and velocity

In summary, derivatives are limits, and you can often treat them as if they were small numbers. To make this derivation mathematically respectable, you need to (a) regard your first step as applying the chain rule, (b) put integral signs in front of both sides of your last equation, and (c) solve for v using one of the three methods for interpreting derivatives.
  • #1
Clever_name
125
0
Hi all,
so my question is can i carryout normal algebraic operations on derivatives, for example:
v=ds/dt and a=dv/dt then eliminating dt a=(dv/ds) *v then, a *ds= v*dv
is that how you derive the relationship between acceleration, velocity and displacement?
 
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  • #2
The point is that ##ds## and ##dv##are notations that are undefined. So writing ##a ds = v dv## is undefined. In particular, the notation ##\frac{dv}{dt}## is not a fraction. It is something that merely behaves like a fraction.

That said, there are ways to define ##ds## and ##dv##. But this is usually not done in a calculus course, so I won't say anything about that.

In short: unless you ever define ##ds##, what you wrote down is meaningless.
 
  • #3
So then could i define it as a differential change in distance?
 
  • #4
Clever_name said:
So then could i define it as a differential change in distance?

What do you mean with that?? Can you write it down formally?
 
  • #5
Hmmm, why then can a physics textbook throw that equation out at you, and then expect you to integrate it to find one of those kinematic equations describing motion with constant acceleration assumed?
 
  • #6
Clever_name said:
Hmmm, why then can a physics textbook throw that equation out at you, and then expect you to integrate it to find one of those kinematic equations describing motion with constant acceleration assumed.

Physics textbooks aren't always mathematically correct.

What they do actually does work. And you can explain it physically. But it's mathematically incorrect (unless you actually define ##ds##).

I'm going to move this to the physics section to let you get a physicists' perspective on this.
 
  • #7
It works, and it is, as you suggest, one way of deriving the relationship between start velocity, finish velocity, displacement and acceleration for motion with a constant acceleration. I think all you need do to make it mathematically respectable is (a) to regard your first step, not as eliminating dt, but as applying the chain rule: a = dv/dt = (dv/ds)(ds/dt) = (dv/ds)v, (b) to put integral signs in front of both sides of your last equation.
 
  • #8
Clever_name said:
Hi all,
so my question is can i carryout normal algebraic operations on derivatives, for example:
v=ds/dt and a=dv/dt then eliminating dt a=(dv/ds) *v then, a *ds= v*dv
is that how you derive the relationship between acceleration, velocity and displacement?

Derivatives are not ordinary fractions, they are limits. But it is quite common in derivations to treat differentials as if they were small numbers, and manipulate derivatives as if they were ratios of small numbers. But it's a heuristic--it's a way to get a hint as to the form of the answer, but it's not rigorous. Usually, differentials (quantities like ds) are just used as intermediate steps in a nonrigorous derivation. There are three ways that a differential can appear in the final result:

  1. As a part of a derivative.
  2. As an integrand to an integral.
  3. As a finite difference equation.

Here are those three ways of interpreting your result [itex]a \cdot ds= v \cdot dv[/itex]:
  1. As a part of a derivative: [itex]a \cdot \dfrac{ds}{dt} = v \cdot \dfrac{dv}{dt}[/itex]
  2. As an integrand to an integral: [itex]\int a \cdot ds = \int v \cdot dv[/itex]. If we multiply the left side by the mass [itex]m[/itex], then it gives the work done by the force acting on the object: [itex]W = \int F \cdot ds = \int m a \cdot ds[/itex]. On the right side, if we multiply by [itex]m[/itex], then it gives the change in kinetic energy:
    [itex] m \int v \cdot dv = \int d(\frac{1}{2} m v^2)[/itex]. So this says that the work done by the forces acting on an object is equal to the change in the kinetic energy of the object.
  3. As a finite difference equation. It's approximately true that [itex]a \delta s = v \delta v[/itex], where [itex]\delta s[/itex] is the change in position during a small time period [itex]\delta t[/itex], and [itex]\delta v[/itex] is the change in velocity during that time period.
 

1. What is the difference between acceleration and velocity?

Acceleration is the rate of change of an object's velocity over time, while velocity is the rate of change of an object's position over time. In simple terms, acceleration measures how quickly an object's speed is changing, while velocity measures how fast an object is moving and in what direction.

2. How are acceleration and velocity related?

Acceleration and velocity are related because acceleration is the derivative of velocity with respect to time. This means that acceleration is the change in velocity over time, or the rate of change of velocity. In other words, acceleration is the cause of the change in velocity.

3. Can an object have a constant acceleration and changing velocity?

Yes, an object can have a constant acceleration and changing velocity. This can occur if the acceleration is in the same direction as the velocity, causing the object to speed up, or if the acceleration is in the opposite direction as the velocity, causing the object to slow down. In both cases, the object's acceleration remains constant, but its velocity changes.

4. How does the direction of acceleration affect an object's velocity?

The direction of acceleration plays a crucial role in determining an object's velocity. If the acceleration is in the same direction as the velocity, the object will speed up. If the acceleration is in the opposite direction of the velocity, the object will slow down. If the acceleration is perpendicular to the velocity, the object's direction will change, but its speed will remain constant.

5. What is the formula for calculating acceleration from velocity?

The formula for calculating acceleration from velocity is a = (vf - vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken. This formula is derived from the definition of acceleration as the change in velocity over time.

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