Video: Acceleration over Time

In this video, we will learn how to analyze the motion of objects that change their velocity in some amount of time, by using the formula for acceleration, 𝑎 = Δ𝑣/Δ𝑡.

12:12

Video Transcript

In this video, we’re going to be looking at how we can deal with changes in velocity by learning about acceleration.

So first of all, what even is acceleration? Well, acceleration is defined as the rate of change of velocity. So what do we even mean by this? Well, let’s imagine that we’ve got a car initially at rest, which means that it has a velocity of zero metres per second. Then let’s say sometime later, the car starts moving. Let’s say that 10 seconds later, the car is now moving at 10 metres per second. And let’s say that it’s moving towards the right.

So initially, we had a car which, let’s say, at the beginning of its journey or, in other words, at zero seconds was stationary but just beginning to move. And then 10 seconds later, the car was moving right at 10 metres per second. Well, in a situation like this, we can work out the acceleration of the car. And the way we would go about doing this is by saying that the acceleration of the car, which we’ll call 𝑎, is equal to the change in velocity of the car — which is represented by Δ𝑣, Δ is the letter used to represent change in and 𝑣 represents velocity — and we divide this change in velocity by the change in time from the beginning of the journey to the end of the journey.

And this is what we mean when we say that the acceleration is defined as the rate of change of velocity. The numerator in this fraction, Δ𝑣, represents the change of velocity. And when we divide it by the amount of time taken for that change in velocity to occur, then we’re working out the rate at which this change in velocity occurs because the rate of change of something talks about how quickly that something changes over time.

So coming back to our car situation, we could say that the acceleration of the car, 𝑎, is equal to the change in velocity over its journey, which can be given by the final velocity minus the initial velocity. So that’s 10 metres per second minus zero metres per second. And we divide this by the change in time over the journey. Now the time at the end of the journey was 10 seconds. And the time at the beginning of the journey was zero seconds. We just arbitrarily decided to choose zero seconds as the beginning of the journey. But the point is that the end of the journey was 10 seconds after that, which can be found as 10 seconds, that’s the end of the journey time, minus zero seconds. That’s the beginning of the journey time.

Now, a couple of things to note here. First of all, when we’re calculating the change in velocity, we’ve just seen that this is equal to the final velocity minus the initial velocity. Similarly, the change in time is equal to the final time minus the initial time. But an important thing to note is that velocity is a vector quantity. And we can recall that vector quantities have magnitude or size and direction. So when we’re calculating the change in velocity, we’ve basically got 10 metres per second to the right minus zero metres per second. And zero metres per second doesn’t have a direction because there is no velocity. And so, the acceleration is also going to be towards the right. And acceleration, therefore, happens to be a vector quantity as well. In other words, directionality becomes important here.

But, anyway, so to calculate the acceleration of the car, let’s first evaluate all of the stuff in the parentheses in the numerator and denominator. In the numerator, we had 10 metres per second minus zero metres per second, which just becomes 10 metres per second. And the denominator, which was 10 seconds minus zero seconds, becomes 10 seconds. At which point, we have the acceleration of the car to be 10 metres per second divided by 10 seconds. In other words, the speed of the car changes by 10 metres per second in 10 seconds. And when we evaluate the numbers, we get 10 divided by 10, which is one. Which means that we can say that the acceleration of the car is one metre per second per second. In other words, the speed of the car is increasing by one metre per second every second. It’s getting faster and faster.

Now, let’s take a quick look at the units of this acceleration. We’ve got the units of velocity, that’s metres per second, divided by the unit of time, that’s seconds. And we can also write this as one metre per second, the unit of velocity, divided by seconds, the unit of time. But then, the way to evaluate all of this stuff in the parentheses is to imagine, first of all, that the seconds is actually seconds divided by one. So it’s a fraction. And then to say that the divide sign becomes a multiplication sign. And this fraction inverts.

In other words, what we had earlier, that was metres per second divided by seconds, is equivalent to metres per second multiplied by one over seconds. At which point, we’re left with metres in the numerator because metres times one is metres. And then at the denominator, we’ve got seconds times seconds which is seconds squared. And so, we can write this unit as metres per seconds squared. Hence, we can say that the acceleration of the car is one metre per seconds squared. And of course, because it’s a vector quantity we mustn’t forget the direction in which the car is accelerating. It’s accelerating to the right.

So at this point, we can see that this equation here is really quite useful. It helps us to calculate the acceleration of an object. The acceleration is given as the rate of change of velocity or the change in velocity divided by the change in time, where the change in velocity is defined as the final velocity, 𝑣 subscript final, minus the initial velocity, 𝑣 subscript initial. And similarly, the change in time is defined as the final time minus the initial time. In other words, what we’re finding here with Δ𝑡 is the time interval or the amount of time it took for the change in velocity to occur.

Now, looking at this equation a little bit more deeply, we can figure out a couple of things. And both of them are related to the fact that velocity is a vector. And therefore, acceleration is a vector as well. Firstly, we can see that if change in velocity is negative or, in other words, the object is slowing down, the final velocity is less than the initial velocity, then the acceleration is going to be negative as well.

To show what we mean by this, let’s think about a car once again. But this time, the car is initially moving at 10 metres per second to the right, right at the beginning of its journey. And it’s slowing down. So 10 seconds later, it’s at zero metres per second. In other words, it’s stationary. Well, in this case, what’s the acceleration of the car going to be?

We can say that the acceleration of the car is equal to the change in velocity — which is equal to the final velocity, zero metres per second, minus the initial velocity, 10 metres per second — and we divide this by the change in time. Well, the change in time is the final time, 10 seconds, minus the initial time, zero seconds. And so we say, that’s the change in time. Well, in this case, the numerator ends up being negative 10 metres per second. And the denominator ends up being 10 seconds. In which case, the acceleration of the car turns out to be negative one metres per second squared. Or, in other words, the car is slowing down or it’s decelerating.

Another way to think about this is that the acceleration of the car is negative, if the direction of the object’s velocity is opposite to the direction of the object’s change in velocity. And so that’s the same as saying the object is initially moving this way. But it starts speeding up in this direction. Which means it’s slowing down in this direction. Now, that can be quite confusing to think about. But the idea is that the acceleration is in the opposite direction to the object’s velocity. And hence, it’s slowing down. So we’ve seen that acceleration can be negative.

Now, another thing to realise is that because velocity is a vector quantity, we can think about a car, this time we’re looking from above onto the car, moving at, say, 10 metres per second to the right until it decides to change direction. The car is still moving at 10 metres per second along the entire length of its journey. It’s just the direction in which it’s moving at 10 metres per second changes. So is this going to lead to an acceleration?

Well, to answer that question, let’s remember, once again, that velocity is a vector quantity. And vector quantities have magnitude and direction. Therefore, although the velocity is always 10 metres per second, it’s not just that. It’s 10 metres per second in changing directions. Therefore, the velocity is changing. And if the velocity is changing, there is a Δ𝑣. And therefore, the car must be accelerating.

Now, this can also be a difficult concept to wrap your head around. The fact that even though a car is moving constantly at 10 metres per second, it is still accelerating. This can lead to some very interesting phenomena. But for now, let’s familiarise ourselves more with a basic definition of acceleration. Let’s do this by looking at an example.

An object accelerates at five metres per second squared for 0.25 seconds. How much does its velocity increase?

Okay, so in this question we’ve been told that we’ve got an object. Let’s say this blob is our object. And we’ve been told that it’s accelerating. Let’s say it’s accelerating to the right at five metres per second squared for a total time of 0.25 seconds. So let’s say that the object ends up here at the end of the 0.25 seconds. Now, we’ve been asked to find out how much its velocity increases by. In other words, we can say that, initially, the object had some velocity, which we’ll call 𝑣 one. And at the end of the 0.25 seconds, it had another velocity which we’ll call 𝑣 two.

We’ve been asked to find out the difference between 𝑣 two and 𝑣 one because we’ve been asked to find out how much the velocity of the object increases by. Or, in other words, we’ve been asked to find the change in velocity of the object because we use Δ to represent change in and 𝑣 to represent velocity, where of course this change in velocity is the same thing as the final velocity, 𝑣 two, minus the initial velocity, 𝑣 one. And we don’t know what this is. Now, it’s important to note that we don’t actually need to figure out the individual values of 𝑣 two and 𝑣 one. We just need to find the difference between them, Δ𝑣. And the way we can do this is to recall the definition of acceleration.

Acceleration, which we’ll call 𝑎, is defined as the rate of change of velocity which, in other words, is the change in velocity of an object divided by the change in time over which this velocity change is occurring. In other words, if we were to say that this object starts moving at, for example, three o’clock in the afternoon. Then 0.25 seconds later, so that’s three o’clock plus 0.25 seconds, is the final point at which we’ll consider the object. And so we’ve been asked to find out the change in velocity of the object over this 0.25 second time interval. In other words then, we can say that the change in time between the start and the finish is actually 0.25 seconds, which is what we’ve been told in the question. We’ve been told that the object is accelerating at five metres per second squared for 0.25 seconds.

So in this case, we know the acceleration and we know the time interval. And we’ve been asked to find out Δ𝑣, the change in velocity. So to do this, we need to rearrange the equation. We can do this by multiplying both sides of the equation by Δ𝑡, which means that Δ𝑡 cancels on the right-hand side. And so, we’ve only got Δ𝑣 left on the right-hand side and Δ𝑡 multiplied by 𝑎 on the left. Or, we can move the right-hand side to the left and the left-hand side to the right, which gives us Δ𝑣 is equal to Δ𝑡 times 𝑎. Or, we can write it as 𝑎 times Δ𝑡. And then, we can substitute in the values for 𝑎 and for Δ𝑡. When we do that, we get Δ𝑣 is equal to five metres per second squared multiplied by 0.25 seconds.

Considering the units very quickly, we see that we’ve got metres per second squared multiplied by seconds. In which case, a power of seconds in the numerator cancels with one of the powers of seconds in the denominator. And what we’re left with is metres divided by one power of seconds. Or, in other words, metres per second which is the unit of velocity. And this is good because we’re actually finding the change in velocity. So all that’s left to do now is to evaluate five times 0.25 which ends up being 1.25. And hence, we’ve found our final answer. The velocity of the object changes by or, more specifically, increases by 1.25 metres per second.

And the reason that we know the velocity is increasing is because we’ve been told that the object is accelerating at five metres per second squared. The fact that the acceleration is positive means that the acceleration is in the same direction as the velocity. Therefore, the object is only going to speed up. And hence, the velocity is increasing.

Okay, so having now looked at an example, let’s summarise what we’ve spoken about in this lesson.

We firstly saw that acceleration is defined as the rate of change of velocity. Symbolically, this can be written as 𝑎, the acceleration, is equal to the change in velocity, Δ𝑣, divided by the change in time or the time interval over which this change in velocity occurs, Δ𝑡. Secondly, we saw that acceleration is a vector quantity. This means that it has magnitude or size and direction. Thirdly, we saw that acceleration can be positive when the object in question is speeding up or it can be negative when the object is slowing down.

If the acceleration and the velocity of the object are in the same direction, then it speeds up and the acceleration is positive. And if the acceleration is in the opposite direction to the object’s velocity, then it’s negative and the object is slowing down. And finally, we saw that acceleration has units of metres per second squared or, of course, any other equivalent units. As long as there’s a unit of distance in the numerator and two powers of a time unit in the denominator, then that is a unit of acceleration, for example, kilometres per hour squared. So this is how we can use the concept of acceleration to deal with changes in velocity.

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