In this video, we’re going to learn about kinematics, the study of objects in motion. The topic of kinematics covers all kinds of motion, whether it be a ball flying through the air, a water balloon we’re dropping on an unsuspecting friend, or a hockey puck sliding along the ice. Often, but not always, kinematics involves the study of projectile motion. Projectiles are objects which move only under the influence of one force, the force of gravity, as in our first two examples.
Knowing kinematics gives us the ability to understand all sorts of object motion scenarios. Using this framework, we can solve questions such as: at what angle should we throw a ball so that it lands a certain distance, 𝑑, away from our original release point? Or, how much time, Δ𝑡, should we allow for our water balloon to fall from our hand onto our friend’s head. Or even, how fast do we have to hit a hockey puck so that it goes to our teammate before it’s intercepted by a skating opponent? Kinematics is an extremely practical topic for understanding object motion.
Before we get into the equations of kinematics, let’s consider some principles of this subject. The first principle to learn is that kinematics describes object motion without reference to forces.
Let’s consider an example of this. Let’s say we have a ball that’s free-falling straight toward earth. Clearly, this object is under the influence of a force, the force of gravity, accelerating it toward earth. But, within the kinematic framework, we wouldn’t describe the motion of this ball using that force or any other force. Instead, our description of the object motion is constrained to a set of variables which include velocity, acceleration, displacement, and a few others we’ll get to later. So that’s the first principle of kinematic motion. Even when forces are involved, we don’t reference them in our equations.
The second principle of this motion has to do with one of the variables we just mentioned, acceleration. In order to fit into a kinematic framework, the acceleration in any given scenario must be constant. Note that constant means just that: constant. It doesn’t have to be zero. It can be a positive or negative value. It’s just that that value can’t change over time.
So for example, when we have a ball flying through the air, we know that ball is being accelerated by the force of gravity and that acceleration is 9.8 meters per second squared. It’s not zero, but it is constant. And that means that the kinematic framework for solving problems still applies here.
Another helpful principle of kinematics can be shown using this ball in flight. Say that we put a pair of coordinate axes with its origin at the place where the ball was launched. We’ll say motion in the horizontal direction is 𝑥-displacement and motion in the vertical direction is 𝑦-displacement. We can see that for much of the ball’s flight, it exhibits motion in these two different directions. That is, at any particular point in its path, it has some 𝑥-displacement and a different, separate 𝑦-displacement.
In the world of kinematics, motion in these two perpendicular directions, the 𝑥- and 𝑦-directions, is completely independent one from another. If we wanted to solve for some motion of the ball in the vertical or 𝑦-direction, we can completely neglect its motion in the horizontal direction. That’s what we mean when we say they’re independent of one another. This is a helpful point because in some problems we’ll encounter, we’ll find that we need to understand motion in one dimension in order to let us solve for the motion in the other dimension.
So here is how can we put this principle in words. What we’ve seen is that kinematic motion in one direction does not depend on or impact motion in another perpendicular direction. So kinematics describes object motion without referring to the forces that act on those objects. For the kinematics solution framework to apply, the acceleration of the objects we’re considering needs to be constant. If it changes over time, this approach won’t work. And then, motion in any one direction is independent of motion in another perpendicular direction.
This, as we’ll see, leads to the importance of using subscripts on variables when solving kinematic problems. For example, if an object has a velocity in the 𝑥-direction, we call it 𝑉 sub 𝑥, then we need to be careful to keep that separate from any velocity it has in the 𝑦-direction, what we might call 𝑉 sub 𝑦. Knowing all this, let’s now move on to the kinematic equations of motion, what we’ll actually use to solve problems.
As introduction to the kinematic equations, let’s first consider the kinematic variables. That is, the variables that appear in those equations. There are five of them. They are displacement, initial velocity, final velocity, acceleration, and time. Sometimes these variables are referred to as 𝑑, 𝑉 sub zero, 𝑉 sub f, 𝑎, and 𝑡. However, depending on where you learn about these variables, they may go by another name. You may have learned about these equations by the name: the SUVAT equations.
Here, each one of the letters in this name refers to one of the five variables in these equations. S is displacement. U is initial velocity. V is final velocity. And then A and T, acceleration and time, are the same as we had before. Whichever set of variable names you see being used, just know that the equations they apply to are mathematically equivalent. Speaking of those equations, let’s look at them now.
Here are the four main kinematic equations. And we see that indeed they involve the variables we wrote off to the left. Looking at these equations, the first thing we might wonder is: where did they come from? For example, just looking at the first kinematic equation, we might say: how do we know that the final velocity of an object is equal to its initial velocity plus its acceleration times the time over which it’s moving? It turns out that the kinematic equations can be derived from relationships between quantities of motion, such as displacement, velocity, and acceleration.
For example, consider that acceleration is equal to a change in velocity divided by a change in time. That’s the definition of acceleration. Well, what if we rewrite Δ𝑉 in terms of its initial and final velocities. That is, what if we replace it with 𝑉 sub f, the final velocity of the object, minus 𝑉 sub i, its initial velocity? And then, as a next step, let’s say we multiply both sides of the equation by Δ𝑡, which is the time passed. We see that on the right-hand side, this means that Δ𝑡 cancels from numerator and denominator. And then, lastly, let’s say that we add 𝑉 sub i, the initial velocity, to both sides of the equation.
That gives us an equation that says: 𝑉 sub i plus 𝑎 times Δ𝑡 is equal to 𝑉 sub f. And notice that this equation, which we arrived at simply by rearranging an expression for acceleration, is the first of the four kinematic equations. In the kinematic equation, we’ve called the passage of time 𝑡, whereas in our equation, we’ve called it Δ𝑡. But it’s the same quantity. And it’s the same equation. And the remaining kinematic equations are like this too.
As complicated as they may look, they come from simple origins, relationships between velocity, acceleration, and displacement. Even though these equations are derivable from other simpler mathematical equations, there’s no escaping the fact that it’s just more straightforward, from our problem-solving perspective, to have these four equations memorized. When we have these four equations at our fingertips, we’ll be much more effective at solving kinematics problems. Let’s get some practice doing that now. We’ll slide our four equations over to the side of our screen to clear some space.
Say that we have a swan which, starting from rest, begins to flap its wings and run across the surface of the lake until it gets up enough speed to go airborne. Imagine further that the swan is accelerating at a constant value of 0.44 meters per second squared. And that once it achieves a final velocity, we’ll call it 𝑉 sub f, of 5.25 meters per second, then it will be able to take off. Knowing this, let’s say we want to solve for this swan’s displacement from the time it starts moving until the time it lifts off. We can call that value lowercase 𝑑. How would we solve for that distance?
Among other things we know about this situation, we know that the acceleration of the object in motion, the swan, is constant. That is the critical sign that the kinematic equations of motion apply to the swan’s motion. Without that constant acceleration, the equations wouldn’t apply. But since it is constant, they do. Now before we start to look through our list of kinematic equations to see which one might be most helpful to us, let’s consider again what we know about this scenario.
We know the swan’s acceleration. We know its final velocity right before taking off. And we know one other thing about it. We know that the swan started from rest. That tells us that 𝑉 sub zero, the initial velocity of the swan, is 0.0 meters per second. So then, we know 𝑎, we know 𝑉 sub f, we know 𝑉 sub zero, and we want to solve for displacement 𝑑. Knowing all that, which of the four kinematic equations on our list would be the best place to start to solve for 𝑑?
It might seem confusing which one will work best for us. But, here’s a general point for picking the right kinematic equation from the list of the four. If we look at the variable we want to solve for, in this case 𝑑, the right kinematic equation will certainly have that. That means that the last three kinematic equations are still in contention here. The first one is out because it doesn’t involve displacement 𝑑. And that’s what we want to solve for.
A next step in helping us pick the right kinematic equation is to ask the question: which, if any, of these equations involve variables which we don’t currently know. In our case, an unknown variable is the time 𝑡. That’s the time it takes for the swan to go from rest to its takeoff speed, 𝑉 sub f. And we see that both of the third and the fourth kinematic equations involve that unknown variable. That quickly suggests that these two equations might not be the ones we want to work with.
Finally then, looking at the second equation, we see that this equation is a perfect match for what we know and what we want to solve for. It has 𝑉 sub f which we know, 𝑉 sub zero which we also know, 𝑎 which we know, and 𝑑 which we don’t know but want to solve for. This then is the equation we’ll use to solve for 𝑑. And we see that to choose the right kinematic equation, a great way to do it is to look at what we want to solve for and what we know and then find an equation which contains those variables.
So then, here we have our equation. And as we recall, we want to solve this for the displacement 𝑑. That means a bit of algebraic rearrangement. Which having done, we see that the displacement 𝑑 is equal to the swan’s final velocity squared minus its initial velocity squared all divided by two times its acceleration. We now simply need to plug in for our values of 𝑉 sub f, 𝑉 sub zero, and 𝑎. And we’re on our way solving for 𝑑.
Plugging these values in and then calculating this expression, to two significant figures, we find a result of 31 meters. That’s how far the swan will travel across the surface of the water before it’s able to take off. Now knowing that, let’s consider another question about this takeoff scenario.
How much time, we wonder, does it take from when the swan starts to when it’s able to lift off? To figure this out, we see that, once again, we’ll have to go to our list of kinematic equations and pick the one that most easily lets us solve for 𝑡. In this case, unlike before, we have a few choices. Notice that the first equation involves variables we know — 𝑉 sub f, 𝑉 sub zero, and 𝑎 — and the one we want to solve for, 𝑡. So we could use that one.
But then, we also see that the third equation is like that. We know 𝑑, the displacement, we know 𝑉 sub zero, and we know 𝑎. And we want to solve for the time 𝑡. And even the fourth equation will work as well. We know 𝑑, we know 𝑉 zero, and we know 𝑉 sub f and want to solve for 𝑡. This is a good example of how, when it comes to kinematic equations, there is often more than one way that it can work. In this case, it’s just a matter of choosing which of these three kinematic equations will let us solve for 𝑡 most easily.
Well, it’s a close competition between the first and the fourth equations. But let’s choose the first equation to solve for 𝑡. It says that the swan’s final velocity is equal to its initial velocity plus its acceleration times the time elapsed. Rearranging to solve for 𝑡, we see it’s equal to 𝑉 sub f minus 𝑉 sub zero all over 𝑎. Entering in these values and solving for 𝑡, we find it rounds to 12 seconds. So according to our kinematic equations, this graceful creature takes about 12 seconds of time to start from rest and get up to takeoff speed.
To recap this exercise, our important points were that we realized acceleration is constant. And therefore, our kinematic equations do apply. And we learned an approach for figuring out which of the four kinematic equations fits best for a given scenario.
Okay, let’s summaries this topic of kinematics. Our first main point is that kinematics describes object motion without reference to the forces acting on the object. As we mentioned, an example of this would be to describe the motion of a ball flying through the air without referring to the force of gravity acting on the ball. Or, for that matter, any other force such as air resistance.
We also learned the four kinematic equations of motion. We learned that these equations can be derived from relationships between object motion parameters such as acceleration, velocity, and displacement. As challenging as it can be to memorize all four of these equations, it typically ends up being very worthwhile to help with solving problems. And we also saw that the kinematic equations of motion only apply when acceleration 𝑎 is constant. Knowing all this will give us a good start in being able to solve problems involving kinematic motion.