Understanding Derivatives Through Kinematics: A Detailed Exploration of Motion Scenarios

This post delves into the concept of derivatives, explaining them as the "rate of change" or the slope of a graph. It begins with an intuitive explanation of derivatives, contrasting the calculation of slopes in linear and non-linear graphs. The post then explores three kinematic scenarios—stationary objects, uniform linear motion, and non-uniform linear motion—demonstrating how derivatives are used to analyze distance, velocity, and acceleration. Through these examples, readers learn how derivatives provide valuable insights into the behavior of moving objects and the fundamental connections between distance, velocity, and acceleration in kinematics.

1. First of all we need to define intuitively the concept of derivative

Derivative:

If we recall the concept of derivative, we remember that is the “rate of change” which, as we have seen in the past article “Interpretation of Kinematics: Unravelingthe Relationship Between Kinematic Variables with Graphical Analysis”, this rate of change stand for the slope, therefore, the derivative is nothing more than another way of calling the state of change or slope of a graph. The only difference, is that when we calculate the slope of a linear graph, we can use this simple formula:

Selecting two random points in the graph that were a straight line we were able to find this slope, but when the shape is different than a line, like for example, when is quadratic. We cannot have a good approximation of the shape with a piece of a straight line.

As we can see, it looks apparently good, but when we do a close-up, we see that the line we draw does not perfectly fit the curve. So with the usual equation for a slope, we cannot calculate a good slope for a quadratic graph, unless we make that piece of a straight line smaller and smaller, where its length tends to zero, and then we add them up to approximate the value of the slope in that time interval. Well, how do we do this? Intuitively we use this same formula:

But, where the distance between d2 to d1 tends to zero, in other words, where the two selected points are the closest without being at the same point. So we can rewrite this expression conveniently in the following way using limits:

Where if we analyze this expression, we can interpret the following:  

if hà0, then we can deduce that f(t+h) is super close to f(t), and they are divided in a time interval close to 0 also. So here we can see the similarities of both expressions. Both calculate slope, but one makes the distance between one point to another tend to be zero. Therefore, when we talk about the derivative of something, we still just talk about finding the rate of change of a graph, the slope.

2. Explaining Three simple kinematics scenarios from a derivative perspective

Stationary Object:

When we have a stationary object, from the last review we recognize the following graphs: the difference between acceleration, velocity, and distance graphs. How can we analyze these scenarios with derivatives? We have a distance versus time graph, the function is f(t) = 5, a constant value. Therefore, if we want to calculate the slope of this movement, as we did before, we will calculate the “rate of change of that distance” or "velocity." But because it is a stationary object, we already can think that this velocity will be zero. But how do we express this mathematically?

now we calculate the slope, by selecting two points, and calculating the rate of change, but let’s use a derivative approach (which is the same as I explained before)

This means the derivative of the function f(t)=5 is equal zero, this first derivative is zero, the first derivative = velocity, is equal zero. Now, the derivative of zero?

It is also zero, is there is no velocity, then there is no acceleration either, so the second derivative is zero too. So in summary an stationary object derivatives are the following:

Uniform Linear Motion:

Now in this scenario, where we have a uniform linear motion, we can also recognize the graphs, in the first graph of distance versus time we have a linear behavior, the distance is increasing linearly. In the second graph, the velocity versus time has a constant value, where this means we have a constant velocity. And in the third graph of acceleration versus time we have a value of zero; because our velocity is a constant, therefore, the acceleration is equal to zero. Let's now define the distance function, in the first graph, we know it is lineal, and we start from an initial position of zero (just to keep it simple), so we can say that our function is f(t) = t

So, again if we want to calculate the rate of change of the distance graph, we want to find the slope of the graph, we can take the derivative, where we know the derivative of a variable is equal to 1.

This means that this object has a constant velocity, specifically of 1 m/s. Now, what if we want to calculate the slope of the velocity graph? if we calculate the acceleration?

We know that the object has a constant velocity, so it has no acceleration, so it is pretty intuitive to think that this derivative will be zero, the derivative of 1 is zero. Therefore, the first derivative is the velocity, where the value is 1, and the second derivative is the acceleration, which is zero. In summary, for a Uniform linear motion we have the following derivatives and its meaning:

Non-uniform Linear Motion:

In this third scenario, we have a non-uniform linear motion, if we analyze the distance versus time graph, we can deduce that because the velocity changes over time, the distance it displaces over each time-interval will never be the same. Therefore, because its displacement is not constant or linear over time, the velocity on each interval is also changing. If the velocity is changing linearly, this means that now we have a constant value of acceleration, so when we have a constant acceleration value, our distance versus time graph behaves like a quadratic function.

So, in the distance versus time graph, we have a quadratic function like this f(t)=t^2. Now, if we want to calculate the slope (or the rate of change) of the distance graph, we apply the derivative and we obtain the following:

Where now we see that this is the rate at which velocity is changing, and it is not constant anymore, but a function, a lineal function, so our velocity is also increasing over time, so the slope in the distance graph is f′(t)=2t.

Now, what if we try to calculate the slope of the velocity graph? We can clearly see that the velocity has a lineal behavior of v(t)=2t, so finding its slope, applying the derivative, the acceleration will be a constant value of 2m/s^2.  However, mathematically is expressed as:

In summary, in a Non-uniform Linear motion we have the following derivatives and its meaning:

So, in summary we can see that when we take the derivative of a function we are calculating its slope, in kinematic context, when we have a function that represent distance, velocity or distance, taking its slope will give us different information. For example, beginning from a distance function we obtain:

So we can say that d'(t) = v(t). Now, I can say:

or

I can say for example that acceleration is the second derivative of d(t), but also I can say that is the first derivative of v(t). So, this is really useful because with little information we can obtain a lot of data about the object’s behavior.

Derivative of distance à velocity
derivative of velocity à acceleration

Now, you may be wondering, is the way to obtain the opposite? Like from a v(t) obtain d(t)? or to change a(t) to v(t)?, there is of course a way, and it’s called integration.