# Calculus and Kinematic equations--- seeing the logic

• LouisL
In summary, the equations in this post are based on an assumption of constant acceleration, and if this is not the case then the equations will need to be changed.
LouisL
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!

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
LouisL said:
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
sophiecentaur said:
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.

And also the same steps can be followed where acceleration if a function of time (##a(t)##).

Lnewqban and sophiecentaur

## What is calculus and how is it related to kinematic equations?

Calculus is a branch of mathematics that deals with the study of change and motion. It is used to find the rate of change of a function and is closely related to kinematic equations, which are used to describe the motion of objects.

## What is the difference between derivatives and integrals in calculus?

Derivatives are used to find the rate of change of a function at a specific point, while integrals are used to find the total change of a function over a given interval. In other words, derivatives are used to find instantaneous rates of change, while integrals are used to find cumulative rates of change.

## How are kinematic equations derived using calculus?

Kinematic equations are derived using the principles of calculus, specifically derivatives and integrals. These equations describe the motion of objects in terms of displacement, velocity, and acceleration, and can be derived from the fundamental equations of calculus.

## Can calculus be used to solve real-world problems involving motion?

Yes, calculus is commonly used to solve real-world problems involving motion, such as finding the velocity of a moving object at a specific time or determining the distance traveled by an object over a given time period. It is a powerful tool for analyzing and predicting the behavior of physical systems.

## What is the importance of understanding the logic behind calculus and kinematic equations?

Understanding the logic behind calculus and kinematic equations allows us to apply these concepts to a wide range of problems and scenarios. It also helps us to develop critical thinking skills and a deeper understanding of the fundamental principles of mathematics and physics.

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