Introduction to kinematics
Kinematics is the study of how to describe the motion of objects using mathematics. Galileo long ago thought about the parabolic motion of cannonballs. This is one of the earliest applications of kinematics that I'm aware of and this will be our starting point. The study of the motion of objects near Earth's surface is a branch of kinematics called projectile motion. We're going to start off just looking at one-dimensional projectile motion and, later on, we'll also delve into two-dimensional and three-dimensional projectile motion. Now, if your cannonball gets too fast (meaning its speed is a considerable fraction of the speed of light), then we'll have to talk about four-dimensional projectile motion—but for now, let's not go there.
We're going to start by learning from examples and saving the abstract generalizations for later since this, to me, seems like the most pedagogic way of teaching this material. It is also very pedagogic to give a glimpse of topics later to come and present them in a way which captures interest, even though it might be a new concept which the student doesn't yet know—but it is one's not fully knowing which largely makes it interesting to begin with. We will mention Newton's second law and the concept of force throughout these lessons and hopefully we'll have a friendly first encounter with these ideas and hopefully they'll spark some interest. But we'll save a more thorough treatment of these ideas for lessons which will be covered after we're done with kinematics.
Position vectors and displacement
Figure 1 (click to enlarge)
We're going to start off with something very general but we'll see that the results are very important for analyzing one-dimensional projectile motion. In physics, we use something called a position vector (written as \(\vec{R}\)) to specify the location of an object at each instant of time. You could imagine a set of \(x\), \(y\), and \(z\) Cartesian axes sitting stationary on the ground (say, to make things easier to visualize, on Earth's surface) as in Figure 1. These Cartesian coordinate axes should be thought of as imaginary rulers (that are infinitely long) with a bunch of pink tick marks on them that specify how far away the object is from the origin. Indeed, if you were standing at the origin of this coordinate system, then these rulers would measures how far away the object is in the left-right direction, the up-down direction, and the in-out direction.
The position vector of the object is defined as
$$\vec{R}(t)≡x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}.$$
Notice that the position vector is a parametric function of time: at each moment into time \(t\), it specifies the location of the object. So, for example, in Figure 1 you can see that at \(t=0\), if you were standing at the origin of the coordinate system, then the ball would start off right next to you since \(\vec{R}(0)=0\). After a time \(t_1\) has gone by, the object will have moved to a position specified by \(\vec{R}(t_1)\). After waiting an additional time of \(Δt=t_2-t_1\), the object will have moved to a position specified by \(\vec{R}(t_2)\). Now, if you were to take a tape measure and run it down the entire arclength of the red line in Figure 1, you would measure some amount of course; and that amount is what we call the distance the object traveled. Distance is a useful concept in everyday life but it, to a large extant, takes a backseat in most discussions of kinematics. The purpose of our entire discussion on position vectors was to develope a far more useful concept called displacement. Let me explain what that is. Let the position vectors \(\vec{R}(t)\) and \(\vec{R}(t_0)\) represent any two arbitrary positions of the object at any arbitary instants of time. The displacement (represented as \(Δ\vec{R}\)) of the object as it moved from \(\vec{R}(t_0)\) to \(\vec{R}(t)\) is given by the difference of these two vectors:
$$Δ\vec{R}≡\vec{R}(t)-\vec{R}(t_0).\tag{1}$$
First off, I'd like to start by saying that there is a big difference between distance and displacement and the two should nevver be confused. To see this distinction, let's consider the displacement of the object as it moves from \(\vec{R}(t_1)\) to \(\vec{R}(t_2)\). From Equation (1), we know that this displacement is given by \(Δ\vec{R}=\vec{R}(t_2)-\vec{R}(t_1)\). What does this look like graphically? Notice that in order to get \(Δ\vec{R}\), we had to add the two vectors \(\vec{R}(t_2)\) and \(-\vec{R}(t_1)\). Since whenever you add any two vectors you always get a new vector, it follows that \(Δ\vec{R}\) must also be a vector. But what does the vector \(Δ\vec{R}\) look like on the graph? Well, we already know what the vectors \(\vec{R}(t_1)\) to \(\vec{R}(t_2)\) look like since they are already drawn for us in Figure 1. If we wanted to find out what the vector \(\vec{R}(t_1)+\vec{R}(t_2)\) looked like, we'd just have to put the 'tail' of \(\vec{R}(t_2)\) next to the 'head' of \(\vec{R}(t_1)\) to get out new vector. The same procedure is, of course, true for adding any two vectors. But before we can find out what \(\vec{R}(t_1)-\vec{R}(t_2)\) looks like, we have to first figure out what \(-\vec{R}(t_2)\) is. To get \(-\vec{R}(t_2)\), all you need to do is rotate \(\vec{R}(t_2)\) by 180 degrees. (The reason why this is is explained very well in the Linear Algebra section of the Khan Academy.) If we then put the tail of \(-\vec{R}(t_2)\) next to the head of \(\vec{R}(t_1)\), then we'll get the vector \(Δ\vec{R}\) shown in Figures 1 and 2.
Figure 2
From Figure 1, you can immediately see that there are two big differences between distance and displacement. First of all, distance is a scalar (meaning it has only a magnitude) whereas displacement is a vector (meaning it has both magnitude and direction). The second big difference is that the magnitude of the displacement is not the same as the distance. As you can see graphically, the length of the vector \(Δ\vec{S}\) is shorter than the distance from \(\vec{R}(t_1)\) to \(\vec{R}(t_2)\) (as a reminder, the distance is the arclength of the red curve from the point at \(\vec{R}(t_1)\) to the point at \(\vec{R}(t_2)\)). An extreme example is often given in most textbooks to make this distinction clear. If you watch a runner run for 1 mile on a circular racetrack and return to his starting point, the total distance he would have run would be 1 mile. But his total displacement \(Δ\vec{S}\) would be zero. The reason is because the position vector \(\vec{R}(0)\) specifying his location at the beginning of the run points at the same location as the position vector \(\vec{R}(t)\) specifying his location at the end of the run (which is, of course, the same location). Since the two position vectors are the same, it follows that there difference \(Δ\vec{S}=\vec{R}(t_1)-\vec{R}(t_2)\) must be zero.
Instantaneous velocity
Here's where things start to get interesting. The quotient \(\frac{Δ\vec{S}}{Δt}\) gives us the average velocity of the object as it moves during the time interval \(Δt\). It isn't exact though since the object might be accelerating or deceleration during that time interval. It is oftentimes best to start with the extreme examples to get the basic point across of when \(\frac{Δ\vec{S}}{Δt}\) can be a good estimate and when it can be totally off. If the runner Usain Bolt runs a total distance of 1 mile in a circle and returns back to his starting point, his average velocity \(\frac{Δ\vec{S}}{Δt}\) over the time interval of that entire run would be zero since \(Δ\vec{S}=0\). The reason why this calculation gave us such a poor estimate of how fst he was running is becuse we did the calculation over a time interval \(Δt\) that was so big that his running speed varied wildly over that time. But if we considered a very small time interval, his speed would remain roughly constant and the calculation of \(\frac{Δ\vec{S}}{Δt}\) would give us a good estimate.
A long time ago, Newton encountered this same kind of problem. When an apple falls, it acellerates and its speed keeps changing. Newton, of course, realized that if you calculated \(\frac{Δ\vec{S}}{Δt}\) over a very small time interval, the object wouldn't acellerate much and its speed would be almost constant. For a small value of \(Δt\), the quotient \(\frac{Δ\vec{S}}{Δt}\) would give you a pretty good estimate of how fast the apple was falling. Now, if you keep choosing values of \(Δt\) that get smaller and smaller, the displacement would keep getting smaller and smaller (each of the values would approach zero). But (assuming the apple is in motion) the ratio of the two would be approaching some finite number. This number is called the instantaneous velocity. We can define the instantaneous velocity mathematically as
$$\vec{v(t)}≡\lim_{Δt→0}\frac{Δ\vec{S}}{Δt}=\frac{d\vec{S}}{dt}.\tag{2}$$
Equation (2) can be used to tell you how fast the apple is moving right at the time \(t\).
Instantaneous acelleration
Another very important idea is how quickly is the velocity changing? By taking the time derivative of the position \(\vec{R}\), the velocity \(\vec{v}\) is able to capture how quickly the position \(\vec{R}\) is changing. To find out how quickly the velocity \(\vec{v}\) is changing, we need to take its time derivative as well; this is called acelleration and is defined as
$$\vec{a}(t)≡\lim_{Δt→0}\frac{Δ\vec{v}}{Δt}=\frac{d\vec{v}}{dt}.\tag{3}$$
Displacement, and instantaneous velocity and acelleration, will be the three fundamental quantities which will be used to describe kinematics: the motion of objects.
This article is licensed under a CC BY-NC-SA 4.0 license.