# Understanding the kinematic equations

## Main Question or Discussion Point

So I am trying to fully understand how to come about the kinematic equations for uniformly accelerated motion.

So v = dr/dt and a = dv/dt

How does the math work behind this I know the second is the derivative of the first.

dr = v*dt dv = a*dt

So dv is the derivative of dr, a is the derivative of v, then why do we not do anything with dt?

Rusty on my calculus anyone who could explain this to me I would greatly appreciate it. Thanks

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Doc Al
Mentor
So v = dr/dt and a = dv/dt
These are true in general, not just for uniformly accelerated motion.

How does the math work behind this I know the second is the derivative of the first.

dr = v*dt dv = a*dt
OK. Now you just need to integrate a to get v, then integrate again to get r.

So dv is the derivative of dr, a is the derivative of v, then why do we not do anything with dt?
dv is not the derivative of dr, v is the derivative of r with respect to time. When you integrate, you'll get time as a variable.

I'll start you off. We know that a is a constant for uniformly accelerated motion. So what's v? v = ∫a dt. Evaluate that simple integral.

Okay I think I get it so v = a*t + V0 because "a" is a constant

Then

∫dr = ∫v*dt

∫dr = ∫(a*t + v0)dt

r = (1/2)a*t^2 + V0t +r0

Is this all correct?

And then I assume it must be fundamentally wrong but don't know exactly why to start with v=dr/dt and get ∫dr = ∫v*dt and then go something like r = (1/2)v^2 * t? Or is it possible to work the problem out starting from ∫dr = ∫v*dt?

Doc Al
Mentor
Okay I think I get it so v = a*t + V0 because "a" is a constant

Then

∫dr = ∫v*dt

∫dr = ∫(a*t + v0)dt

r = (1/2)a*t^2 + V0t +r0

Is this all correct?
Yes. All good.

And then I assume it must be fundamentally wrong
Why would you assume that?
but don't know exactly why to start with v=dr/dt and get ∫dr = ∫v*dt and then go something like r = (1/2)v^2 * t? Or is it possible to work the problem out starting from ∫dr = ∫v*dt?
You need to start with what you know. All you know is that acceleration is constant. While it's certainly true that r = ∫v*dt (that's just a restatement of the definition of v, v = dr/dt), since you don't know v(t) you cannot do the integral.

And also is calculus the only way to derive these euqations or is it also possible to do it with just algebra or another way?

And also am I right that there are four kinematic equations total with the other two being combinations of the original two?

Doc Al
Mentor
And also is calculus the only way to derive these euqations or is it also possible to do it with just algebra or another way?
You could certainly derive the basic equations using algebra.
And also am I right that there are four kinematic equations total with the other two being combinations of the original two?
The number of kinematic equations depends on how you slice it. Here's one version: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"

Last edited by a moderator:
jtbell
Mentor
The two equations that you derive by integration are all that you really need for constant acceleration:

$$v = v_0 + at$$

$$x = x_0 + v_0 t + \frac{1}{2}at^2$$

provided that you're willing to solve them together with two unknowns for certain problems. The other one-dimensional kinematic equations are derived by solving these two, using different pairs of unknown quantities.

(hmm, looks like LaTeX doesn't quite work yet again after the server move... hopefully it won't take too long to fix this.)

Last edited:
Redbelly98
Staff Emeritus
Homework Helper
Just to add: those other two equations are essentially statements about 2 familiar concepts.

(1) conservation of energy (kinetic + potential):

v2 = v02 +2 a (x-x0)

(Substitute a=-g, then multiply by m/2 to get a more familiar expression for conservation of energy)

and

(2) the average velocity of a uniformly accelerated particle

(v + v0) / 2 = (x-x0) / t

So while they can be derived from the other 2 equations by eliminating either t or a, I remember them by thinking about energy conservation or average velocity.