# Calculus and Kinematic equations---need help seeing the logic

• I

## Summary:

I am studying Physics on my own. I understand the Kinematic equations and how to use them and understand their derivation using algebra, but am stuck on understanding the logic of deriving these equations use Calculus.

I am especially not sure what Calculus rule allows us to take the integral of both sides of an equation, but there are a few other questions as well.

Details of Question:

ds/dt= v which becomes ds=v dt, where s=displacement, t =time, and v=velocity

Then we can integrate both sides of this equation, and do a little algebra, and turn the above equation into:

s − s0 = v0t + ½at2

My main question is about the integration of both sides of the equation ds=v dt: First off, is this a differential equation? If not, what type of equation is this? Going back, we first multiply dt on both sides of ds/dt= v by dt to get ds= v dt and then we integrate. 1) what math rule allows us to multiply dt to both sides of this equation? 2) What calculus rule allows us to take the integral of both sides of this equation? 3) How can I visualize that these 2 integrals as being equal? 3) Would a one year college Calculus class be enough to understand this? I took 1 year of college Calculus but don't recall learning about this.

s

⌡ ds
s0

equals

t

⌡ (v0 + at) dt
0

Thanks for any help!

PeroK
Homework Helper
Gold Member
2020 Award
Paul's online notes are a good reference for all things calculus:

https://tutorial.math.lamar.edu/

In terms of integrating the equations ##v =\frac{ds}{dt}## and ##a =\frac{dv}{dt}##: first note that if two functions are equal, then their definite integral must be equal. First we have:
$$\int_{t_0}^{t_1} a(t) dt = \int_{t_0}^{t_1} \frac{dv}{dt} dt = v(t_1) - v(t_0)$$
That's by the fundamental theorem of calculus. And in the special case where ##a## is constant we have:
$$\int_{t_0}^{t_1} a dt =a(t_1 - t_0)$$
Putting these together gives:
$$a(t_1 - t_0) = v(t_1) - v(t_0)$$
Which is normally written:$$v(t) = v_0 + a(t-t_0)$$
Similarly we have:
$$\int_{t_0}^{t_1} v(t) dt = \int_{t_0}^{t_1} \frac{ds}{dt} dt = s(t_1) - s(t_0)$$ And again in the special case of constant acceleration $$\int_{t_0}^{t_1} v(t) dt = \int_{t_0}^{t_1} v_0 + a(t-t_0) dt = v_0(t_1 - t_0) + \frac 1 2 a(t_1-t_0)^2$$ Putting these together gives: $$s = s_0 + v_0(t - t_0) + \frac 1 2 a (t-t_0)^2$$ Or, with ##t_0 = 0##: $$s = s_0 + v_0t + \frac 1 2 at^2$$

Lnewqban
sophiecentaur
Gold Member
2020 Award
Summary:: I am studying Physics on my own. I understand the Kinematic equations and how to use them and understand their derivation using algebra, but am stuck on understanding the logic of deriving these equations use Calculus.

I am especially not sure what Calculus rule allows us to take the integral of both sides of an equation, but there are a few other questions as well.

First off, is this a differential equation?
Those basic equations are based on an assumption of constant acceleration. (A problem I, myself, had when this important assumption was not emphasised enough for my grasshopper brain in class).
I assume that you know enough about Calculus to follow the rules for differentiation and basic integration. That relationship "ds=v dt" contains infinitesimals and it is an equation so it has to be a differential equation. If v is not a varying function of time then that can be solved by integrating both sides: s=vt but (of course) the initial value of s needs to be added in, as with all solutions AND will need modification if v is a function of t.
Likewise for dv = a dt, you get v = u + at (but, again, a may be a function of time so the simple Suvat formula again may not apply.

You have to go beyond the simple (graphical) algebraic suvat equations for nearly all practical situations like motion of a mass on a spring, an orbiting planet or even an accelerating motor car.

If your calculus is up to it, then you may or may not need to follow the above post from @PeroK which shows the steps to solve a simple (linear acceleration) case.

Lnewqban
PeroK