# Lesson Video: Kinematic Equations Mathematics

In this video, we will learn how to apply the laws of the uniform acceleration motion of a particle in a straight line.

15:38

### Video Transcript

In this video, we’ll learn how to apply the laws of the uniform acceleration motion of a particle in a straight line. Up until now, we’ve used the speed–distance–time formula. Speed is distance divided by time. That formula is applicable when the acceleration of the object is equal to zero. We might also have considered acceleration in terms of the gradient or the slope of the line on a velocity–time graph. Now, we look to introduce some new equations. These equations are sometimes called the equations of constant or uniform acceleration, so called because they’re applied when working with uniform acceleration, that is, acceleration that doesn’t change. So, let’s see where these equations come from.

We begin by considering a velocity–time graph. 𝑣 naught or 𝑣 sub zero is the initial velocity or the original velocity of our object. 𝑣, then, is the velocity after a time 𝑡. We let 𝑎 be equal to acceleration. And Δ𝑥 or change in 𝑥 is the displacement of the object, that sometimes given as 𝑠. Now, we know that the acceleration is given by the slope of the line on a velocity–time graph. Since the acceleration is constant, we have a straight line. And to find the slope or gradient of a straight line, we use the formula rise over run or change in 𝑦 over change in 𝑥.

In our diagram, change in 𝑦 is this length here. It’s the difference between the initial velocity and the velocity after 𝑡 time units. That’s 𝑣 minus 𝑣 naught. Change in 𝑥 is the length of this line. That’s 𝑡 minus zero or just 𝑡. And so, the slope of our line which represents the acceleration 𝑎, under these conditions, is given by 𝑣 minus 𝑣 naught over 𝑡. If we multiply both sides of the equation by 𝑡, we get 𝑎𝑡 equals 𝑣 minus 𝑣 naught. And then if we add 𝑣 naught to both sides, we get 𝑣 equals 𝑣 naught plus 𝑎𝑡. So 𝑣 equals 𝑣 naught plus 𝑎𝑡. That’s our first kinematic equation.

We’ll now consider our second. This time we go back to our graph and recall that the displacement is found by calculating the area between the line and the 𝑥-axis. We can split this area into a rectangle and a triangle. The area of the rectangle is the product of its dimensions. So that’s 𝑣 naught times 𝑡. And the area of a triangle is a half times the base times the height. So here, that’s a half times 𝑣 minus 𝑣 naught times 𝑡. And so, this means that the displacement Δ𝑥, which is of course the area between the line and the 𝑥-axis, is 𝑣 naught 𝑡 plus a half times 𝑣 minus 𝑣 naught 𝑡. But earlier, we defined 𝑣 minus 𝑣 naught as being equal to 𝑎𝑡. So, we replace this with 𝑎𝑡 in our equation, and we get Δ𝑥 equals 𝑣 naught 𝑡 plus a half times 𝑎𝑡 times 𝑡 or a half 𝑎𝑡 squared. And this is our second kinematic equation.

Now, it’s outside the scope of this video to derive all of our equations. But by either rearranging the two we have or considering another graphical approach, we’re able to obtain two further formulae. Those are Δ𝑥 equals a half times 𝑣 naught plus 𝑣 times 𝑡 and 𝑣 squared equals 𝑣 naught squared plus two times 𝑎 times Δ𝑥. So, how do we apply these formulae? We begin by listing all of the measurements that we know. We then identify the one that we don’t know and are not trying to find. That will eliminate all of the equations we aren’t interested in, leaving just one behind. Let’s see what that looks like.

If a particle started moving in a straight line with an initial velocity of 25.1 centimeters per second and a uniform acceleration of 2.4 centimeters per square second, determine its velocity after nine seconds.

The question states that the particle has uniform acceleration. This is a good indication to us that we’re going to need to use the equations of uniform or constant acceleration. Those are the four kinematic equations. So, we begin by listing the equations out. Our job is to eliminate all but one of these equations. And we do so by listing the measurements that we’ve been given in the question. We’re given the initial velocity 𝑣 naught. That’s 25.1 centimeters per second. We’ve got an acceleration of 2.4 centimeters per square second and a time nine seconds.

We’re looking to calculate the velocity after nine seconds. Notice that that means we’re really not interested in Δ𝑥, the displacement of our object. And so, we go through and eliminate all of the equations that contain Δ𝑥. Those are two, equation three, and four. And so, we’re left with one equation; that’s 𝑣 equals 𝑣 naught plus 𝑎𝑡. Next, we substitute everything we know about our particle into this formula. We’re looking to calculate 𝑣, so we say that 𝑣 is equal to 𝑣 naught, which is 25.1, plus 𝑎 times 𝑡, that’s 2.4 times nine. 2.4 multiplied by nine is 21.6. So, velocity is given by 25.1 plus 21.6, which is 46.7.

Notice that we’ve been working in centimeters per second, centimeters per square second, and seconds. And so, the units for our velocity after nine seconds are centimeters per second. The velocity of the particle is therefore 46.7 centimeters per second.

In our next example, we’ll consider how to calculate the distance covered by an object.

A small ball started moving horizontally at 16.3 meters per second. It moved in a straight line with a uniform deceleration of three meters per square second. Determine the distance the ball covered in the first two seconds.

The ball has a uniform deceleration. In other words, it has a uniform negative acceleration. That’s an indication to us that we’re going to use the kinematic equations, the equations of uniform acceleration. And so, the first thing that we do is list those four equations. For an initial velocity 𝑣 naught, an acceleration of 𝑎, a velocity of 𝑣 at time 𝑡 time units, and a displacement of Δ𝑥, they are as shown.

Our next step is to eliminate all but one of these equations, and we do so by listing everything we know about the motion of our object. We’re told that the small ball starts moving at 16.3 meters per second, so that’s its initial velocity. It has a uniform deceleration of three meters per square second. In other words, it’s slowing down. So, we say its acceleration is negative three. Time 𝑡 is two seconds, and we’re looking to find the distance that the ball covers.

We recall that distance is simply the magnitude of the displacement. So, distance here is Δ𝑥. That’s change in 𝑥. And that means we’re not interested in 𝑣, the final velocity of the object. So, we go through our equations and we eliminate every equation that contains 𝑣. That’s one, three, and four, leaving us, of course, with just equation two.

Our next job is to substitute what we know about our ball into this equation. Δ𝑥 is what we’re trying to calculate. Then, 𝑣 naught 𝑡 is 16.3 times two. A half 𝑎𝑡 squared becomes a half times negative three times two squared. Then, 16.3 times two is 32.6. And a half times negative three times two squared is negative six. So, Δ𝑥 is 32.6 minus six, which is 26.6, and we’re measuring that in meters. So, the ball covers a distance of 26.6 meters in the first two seconds of its motion.

Now, it might seem that the information about the ball moving in a straight line is somewhat superfluous. However, it’s actually really important. It allows us simply to consider one direction of motion. If we had to consider two directions, that would complicate things a little more.

Let’s move on to another example.

A particle moving in a straight line was accelerating at a rate of 22 centimeters per square second in the same direction as its initial velocity. If the magnitude of its displacement 10 seconds after it started moving was 29 meters, calculate the magnitude of its initial velocity 𝑣 naught and its velocity 𝑣 at the end of this period.

We’re given that the particle is accelerating at a constant rate of 22 centimeters per square second. This means to answer this question, we’re going to need to use the kinematic equations. These are, of course, equations of constant acceleration. For a starting velocity 𝑣 naught, a velocity 𝑣 after 𝑡 time units, an acceleration 𝑎, and a displacement Δ𝑥, they are as shown.

What we do is begin by listing everything we know about our motion. We’ve already said we know that the acceleration is constant, and it’s 22 centimeters per square second. It’s in the same direction as its initial velocity. Now, we don’t know its initial velocity, but by assuming that they’re in the same direction, we can take both acceleration and 𝑣 naught to be positive. We’re also told that the magnitude of the displacement 10 seconds after it started moving was 29 meters. Remember, displacement can have a direction. So, by considering just the magnitude, we’re thinking about the distance; that’s 29 meters. Time 𝑡 is 10 seconds.

Now, the question actually asks us to calculate the magnitude of the initial velocity and its velocity at the end of the period. Let’s begin by calculating its initial velocity 𝑣 naught. In this case, we’re not interested in 𝑣, so we go through our equations and eliminate those containing 𝑣. Those are one, three, and four. Our next step would normally be to substitute everything we know about the motion of our particle into that second equation. We do have a little bit of a problem though. We notice that the units for our acceleration and our displacement are different. We need them to be the same. So, we multiply displacement by 100 and we find it’s actually equal to 2900 centimeters.

Then, substituting everything we know into this formula, and we get 2900 equals 10𝑣 naught plus a half times 22 times 10 squared. A half times 22 times 10 squared is 1100. So, we subtract 1100 from both sides, and we find that 1800 is equal to 10 times 𝑣 naught. Our final step is to divide through by 10. 1800 divided by 10 is 180. Now, we’re working in centimeters. So, our velocity, our initial velocity 𝑣 naught, is 180 centimeters per second. We might choose to give our answer in meters per second by dividing through by 100. And when we do, we find that 𝑣 naught is 1.8 meters per second.

We’re not quite finished. We’re still looking to calculate its velocity 𝑣 at the end of the period. Now that we know 𝑣 naught, we can actually use any of our equations. So, let’s use the first one. We substitute everything we know about the motion of our particle into this formula, continuing to work in centimeters and centimeters per second. When we do, we get 𝑣 is 180 plus 22 times 10. 22 times 10 is 220. And 180 plus 220 is 400. We’re still, of course, working in centimeters per second. To give our answer in meters per second, we’ll divide through by 100. And when we do, we find that the velocity 𝑣 at the end of the motion is four meters per second.

In our final example, we’ll look at finding the final velocity of the body.

A particle was moving in a straight line with a constant acceleration of two centimeters per square second. Given that its initial velocity was 60 centimeters per second, find the velocity of the body when it was 15 meters from the starting point.

We have a constant acceleration, so we’re going to use our kinematic equations. For a starting velocity 𝑣 naught, an acceleration 𝑎, and a velocity 𝑣 after 𝑡 time units, the first equation is 𝑣 equals 𝑣 naught plus 𝑎𝑡. We introduce Δ𝑥 to represent the displacement of the objects. And we get Δ𝑥 equals 𝑣 naught 𝑡 plus a half 𝑎𝑡 squared. Our third equation Δ𝑥 is equal to a half 𝑣 naught plus 𝑣 times 𝑡. And then we have one further equation. That is 𝑣 squared equals 𝑣 naught squared plus two times 𝑎 times Δ𝑥.

Let’s list what we know about the motion of our particle. We have acceleration as two centimeters per square second, an initial velocity of 60 centimeters per second, and a displacement Δ𝑥 of 15 meters. Now, in fact, our acceleration and velocity are given in centimeters per second and centimeters per square second. So, we want the units to be the same for Δ𝑥. We multiply 15 by 100 and we see that that’s 1500 centimeters. We’re looking to find the velocity of the body when it’s this distance away from the starting point. Notice that that means we’re not interested in the time that this takes.

So, we go through to our four kinematic equations and we get rid of those that contain 𝑡. Well, that’s equation one, two, and three. That leaves us with just one equation, 𝑣 squared equals 𝑣 naught squared plus two times 𝑎 times Δ𝑥. We’re going to substitute everything we know about the motion of our particle into this equation. When we do, we get 𝑣 squared equals 60 squared plus two times two times 1500. 60 squared is 3600. And two times two times 1500 is 6000. So, 𝑣 squared is equal to 9600.

It follows that we can solve for 𝑣 by finding the square root of both sides of our equation. 𝑣 is therefore equal to the square root of 9600. That’s 97.97 and so on. Correct to the nearest whole number, that’s 98. And we can therefore say that the velocity of the body when it’s 15 meters from the starting point is 98 centimeters per second.

In this video, we’ve learned that we can use kinematic equations to model motion involving constant acceleration in a straight line. The four equations we use are shown. In these equations, 𝑣 naught is the initial velocity, 𝑣 is the velocity after 𝑡 time units, 𝑎 is the constant acceleration, and Δ𝑥 is the displacement of the object. We also saw that it’s really important when working with these equations to make sure all the units are the same.