How do acceleration, velocity, and displacement affect/relate to eachother?

Question

I have been wondering this since learning about position, velocity, and acceleration vs time graphs but can't put numbers/equations to it.

I know that acceleration acts to change velocity, shown by the affect gravity has on an object when thrown upward. Similarly, when driving in a car, you speed changes/impacts your displacement. But how does acceleration, then, impact displacement? How would you show this in some sort of quantitative way? How is your average velocity and instantaneous velocity related to your displacement (same idea with the relationship for acceleration)?

For acceleration, obviously, as you speed up, your displacement will rapidly increase as well, right? So, would you write it something like $ s^{2} \propto a $ or is there some other intuitive relationship that can be represented with numbers?

Is it as simple as something like $v = \frac{\Delta s}{\Delta t}$ so is $ v\Delta t \propto \Delta s $?

Similarity, $a = \frac{\Delta v}{\Delta t}$ so $ a\Delta t \propto \Delta v $?

But how would you relate accleration and displacement together? I know they must be linked, but I'm not sure how exactly.

Could I do something like $\Delta t = \frac{\Delta s}{v}$ and $\Delta t = \frac{\Delta v}{a}$ so $v\frac{\Delta v}{a} \propto \Delta s$? But this doesn't provide much insight at a glance...

Another idea, is it possible to relate $\int_0^T \Delta v\cdot dt = s(T)$ (From the 3Blue1Brown integration video) to acceleration? Or even $a = \frac{d^2s}{d^2t} $ maybe? Simply put, how are these three concepts related and linked to each other as well as how do they impact and affect each other?

Answer 1

The answer is that they are, literally, defined as the differentials/integrands of each other.

Velocity is defined as the rate of change of displacement over time. Acceleration is defined as the rate of change of velocity over time. That is literally, how they are related.

In situations where there is constant acceleration (or none), this gives rise to a set of well known formulae connecting them. But these formulae only apply for constant acceleration. Anything else, and you're usually back to the definitions based on differentials/integrands.

To answer some of your questions:

I know that acceleration acts to change velocity, shown by the affect gravity has on an object when thrown upward. Similarly, when driving in a car, you speed changes/impacts your displacement. But how does acceleration, then, impact displacement?

Literally - acceleration changes your velocity (by definition) and the different velocity means you travel a different distance/displacement per amount of time now vs at another time (also by definition). That's it. There is pretty much no more direct way to say it. We define these terms because they are useful that way. That's all there is.

How is your average velocity and instantaneous velocity related to your displacement.

As before. Instantaneous velocity is a way of asking how your displacement changes over shorter and short time periods. Its a limit, in the sense that all differentiation is a limit as ∆x or ∆t goes to zero. Average velocity asks how your displacement changes when averaged over a non zero time interval - what the equivalent constant velocity would have been., to travel the same total displacement in the same total time.

For acceleration, obviously, as you speed up, your displacement will rapidly increase as well, right?

Not necessarily. Suppose you are accelerating at +10 m/s² from a starting velocity of -100 m/s to +0 m/s, thats acceleration. But your absolute displacement sizes will slow down. Or suppose your acceleration is in a circle or inward spiral. In both cases, you wont see displacements per unit time getting "rapidly larger". Thats because acceleration literally is, rate of change of velocity, nothing more. Even traveling in a circle at a constant speed, is acceleration, because although your speed is constant, your velocity is not constant. (*) So acceleration doesn't mean displacements "getting bigger" (rapidly or any other way) or any other meaning at all. Its just the rate of change of velocity, nothing more.

(*) Because velocity is a stated speed in a stated direction, so changing either speed or direction, changes velocity, and a changing velocity means it's being accelerated.

But how would you relate accleration and displacement together? I know they must be linked, but I'm not sure how exactly.

They are the 2nd derivative or 2nd integral with respect to time. That, and nothing more......

Answer 2

This is how I like to think about position, velocity (ie. change of position) and acceleration (ie. change of velocity):

Suppose that at t=0 (in seconds or any other unit),

Position = [0, 0, 0] Velocity = [10, 20, 30] Acceleration = [1, 0, 0]

Then, at t=1, if the acceleration is constant, the object's motion will be fully described by:

Position = [10, 20, 30] Velocity = [11, 20, 30] Acceleration = [1, 0, 0]

At t=2,

Position = [21, 40, 60] Velocity = [12, 20, 30] Acceleration = [1, 0, 0]

At t=3,

Position = [33, 60, 80] Velocity = [13, 20, 30] Acceleration = [1, 0, 0]

We can think of calculus as just the what happens when the time is very small, for example, when the tick is not 1 second, but 0.0000001 second. The relationships are the same between position, velocity and acceleration.

Hope it helps!

Answer 3

But how would you relate acceleration and displacement together ?

It's easy to derive relationship :

$$ \begin{align} a &= \frac {d }{dt} \left( v \right) \\&= \frac {d }{dt} \left( \frac {dr}{dt} \right) \\&= \frac {d^2 r}{dt^2} \end{align} $$

or simply $$a = \dot v = \ddot r$$

Second form is more cleaner and is more preferred in science, because when it comes to n-th order of differentiation with respect to some variable, it's very messy to write it as $y = \frac {d^nx}{dt^n}$.