# Video: Acceleration over a Distance

In this video, we will learn how to use an object’s initial and final velocities and the object’s displacement to define acceleration by using the formula 𝑣² = 𝑢² + 2𝑎𝑠.

11:10

### Video Transcript

In this video, we will see how to relate the distance travelled in a straight line by an object undergoing constant acceleration to that object’s initial and final velocities.

So let’s start by first recalling what we mean by acceleration. Acceleration is defined as the rate of change of velocity. And using symbols, we can say that the acceleration, 𝑎, is given by the change in velocity, Δ𝑣, divided by the time interval taken in order for that change in velocity to occur, Δ𝑡. In other words, the acceleration of an object is given by how much this velocity changes by divided by the time taken for that velocity change to occur. Now, this is all well and good. But what if the things that’s important is not the amount of time taken for that velocity change to occur, but rather the distance travelled by the object during this velocity change?

For example, let’s consider a car. Let’s say this car is initially moving at 30 metres per second to the right. Now, as it goes along this road, the driver sees a sign in the distance. This sign says that the driver must slow down to 20 metres per second by the time they pass the sign. Now, of course, normally, in the real world, road signs give speed limits in miles per hour or kilometres per hour or similar units. However, for our purposes, let’s just imagine that this 20 is referencing 20 metres per second as the speed limit.

So the driver travelling in a car currently moving at 30 metres per second has to be moving at 20 metres per second by the time they pass the sign. So naturally, they press their breaks. Now, these brakes provide a certain deceleration to the car or, in other words, an acceleration, which we’ll call 𝑎, in the opposite direction to the car’s motion. Now, this acceleration, 𝑎, depends on the strength of the brakes of the car and on how hard the driver presses the brakes. So for a particular value of 𝑎, is there enough distance, a distance that we’ll call 𝑠, between the car’s position when the driver first presses the brakes and the sign for the car to slow down to 20 metres per second?

Or, another way to think about this is that for a given value of 𝑎 or, in other words, a given strength of the brakes decelerating the car, how far before the sign does the driver have to press the brakes in order to be able to get to 20 metres per second by the time they pass the sign? Well, let’s, first of all, say that the car’s initial velocity is called 𝑢. 𝑢 is equal to 30 metres per second. And let’s say that the final velocity of the car, which we know is meant to be 20 metres per second, is called 𝑣. So 𝑣 is equal to 20 metres per second, after the car has passed the sign.

Well, technically, the car’s velocity should be 20 metres per second by the time it passes the sign rather than the way that we’ve drawn it here which has already passed the sign. But, we could just assume that it carries on at 20 metres per second once it’s passed the sign, so not a big deal. The important distance, therefore, is still 𝑠 which is the distance over which the car decelerates. So how are we going to go about calculating this distance? Well, luckily, there’s an equation that links the initial velocity of an object — in this case, the car — the acceleration of the object, the final velocity of the object, and the distance travelled over which the acceleration occurs.

The equation that we’re looking for is this one here: 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. In other words, the final velocity of an object squared is equal to the initial velocity of the object squared plus two times the acceleration of the object times the distance covered by the object during its acceleration. If, in this case, we’re given a value for 𝑎, let’s say we were told the acceleration is five metres per second squared, then we’d be able to rearrange this equation to solve for 𝑠. When we do this, we find that 𝑠 is equal to 𝑣 squared minus 𝑢 squared divided by two 𝑎. And so, at this point, we’d be able to plug in the values.

We’d be able to say that the distance travelled by the car, 𝑠, is equal to the final velocity of the car squared — so 𝑣 squared, which is 20 metres per second squared — minus the initial velocity of the car squared, 30 metres per second squared, divided by two times the acceleration, which is negative five metres per second squared. And the reason it’s negative is because it’s in the opposite direction to the object’s velocity. The object is travelling in this direction whereas the acceleration is in this direction.

Now, the reason that we haven’t put a negative sign over here when we first mentioned the acceleration is because this arrow is already dealing with that. We’ve been told that the acceleration is five metres per second squared to the left, as we’ve drawn it. However, in our calculation, we need to account for that. And hence, we put in the negative sign. And it’s a good thing that we put in that negative sign because we can see that the numerator is actually going to be negative. We’ve got 20 metres per second whole squared minus 30 metres per second whole squared. This simplifies to 400 metres squared per second squared minus 900 metres squared per second squared or, in other words, negative 500 metres squared per second squared. And therefore, this negative sign cancels with this one leaving as with a positive value of the distance travelled.

And looking at the units, we see that we’ve got metres squared per second squared in the numerator and metres per second squared in the denominator. Therefore, the per second squared are going to cancel both in numerator and denominator. And one factor of metres is going to cancel in the numerator with the factor of metres in the denominator. Overall, what we’re left with is just metres. This is a good thing because we’re trying to calculate a distance which should have the unit of metres. And so, our final answer actually evaluates to 50 metres. In other words, this car would need a distance of 50 metres in order to decelerate from 30 metres per second to 20 metres per second, given an acceleration in the opposite direction of five metres per second squared. And that is how we would go about using this equation here.

However, there are a couple of caveats to using this equation. The first caveat is that this equation only works if the acceleration of the object in question is constant. In other words, the value of 𝑎 can be positive, it can be negative. That doesn’t matter. But it must be a constant value. It must be a constant value over the entire range that we’re considering. If at any point the acceleration changes — so say, for example, we had an object initially accelerating at an acceleration 𝑎 one over a distance 𝑠 one. But then, once it completed that distance, it accelerated even faster, for example, at 𝑎 two over a range 𝑠 two. Then we would have to treat these two regimes separately.

We would first have to consider the initial and final velocities over this range using a value of 𝑎 one and then the initial and final velocities over this range using a value of 𝑎 two. In other words, we can only use this equation over a range where the acceleration is constant. And that is the first caveat of this equation.

The second caveat is that the distance 𝑠 must be the distance in a straight line. This is because, technically, this equation deals with displacement. In other words, the displacement of an object from start to finish when it’s accelerating at a rate 𝑎 one. And so, we say that 𝑠 must be a straight line. So if we’ve got a situation where we’re dealing with the initial and final velocities of an object as well as its acceleration, which happens to be a constant, and the distance it travels during this acceleration, which is in a straight line, then this equation can be used. And when it can be used, it’s actually a very powerful equation.

We’ve already seen how it can be used to calculate the deceleration of a car. In fact, this is the equation that’s used to calculate the braking distances of cars. Let’s say a car, this time very badly drawn, is traveling towards the right at a certain speed. And further up ahead, the driver sees some obstruction that means they have to break and stop before they get to this obstruction. Well, if the car has to break, then by the time they get to the instruction, the car must be stopped. In other words, the final velocity of the car must be zero. Now, this equation will help us calculate what the distance travelled by the car must be, given some initial velocity of the car, whatever that may be, and the driver stomping on the brakes as hard as they can to give the maximum value of the acceleration of the car in the opposite direction to its velocity.

Or, another way to think about it is that if the driver is travelling at an initial velocity 𝑢, would they have enough space to stop before they got to the obstruction? So that’s one situation where this equation is really useful. Another one is considering an airplane about to take off. It starts at rest, 𝑢 is equal to zero. And it’s moving along a runway which is only a certain distance in length. This equation can be used to calculate: Is the runway long enough for the plane to get to the speed it needs to be in order to take off? Let’s call that speed 𝑣, whatever that may be, knowing the fact that the plane can accelerate at a rate 𝑎. And therefore, is the runway safe for that plane to take off from? So this equation is really very powerful. Let’s look at an example question to see it for ourselves.

An object has an initial velocity of three metres per second and accelerates in the direction of its velocity along a straight line at a rate of four metres per second squared. Its velocity reaches 11 metres per second when it is at the end of the line. What is the length of the line?

Okay, so in this question, we’ve got an object. Let’s say it starts here. And let’s say it’s moving in this direction to the right with its initial velocity which we’ve been told is three metres per second. Now, we’ve been told that it accelerates in the direction of its velocity along a straight line at a rate of four metres per second squared. So the object is travelling in a straight line towards the right, as we’ve drawn it. And by the time it gets to the end of the line, we’ve been told that it’s moving at 11 metres per second. And we’ve been asked to find the length of this line, from the start position of the object to the finish position.

Let’s say that the length of this line is 𝑠. And 𝑠 is what we’re trying to find. As well as this, let’s label the initial velocity of the object as 𝑢, three metres per second. Let’s say the final velocity of the object is 𝑣, 11 metres per second. And let’s say that the acceleration of the object is 𝑎, four metres per second squared. Now, the equation that links 𝑢, 𝑣, 𝑎, and 𝑠 is this one here: 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Or, the final velocity of the object squared is equal to its initial velocity squared plus two times its acceleration times the distance it travels.

Now, it’s important to know that for this equation to apply, the acceleration must be a constant value. And in this case, we’ve been told that it’s four metres per second squared which is a constant. As well as this, the distance that the object travels in order for this equation to apply must be in a straight line. And this is the case as we’ve been told in the question. Therefore, we can use this equation to solve our problem. We already know the values of 𝑣, 𝑢, and 𝑎. All we need to do is to rearrange to solve for 𝑠. We can start by subtracting 𝑢 squared from both sides of the equation. This way, it cancels on the right-hand side. And what we’re left with is 𝑣 squared minus 𝑢 squared on the left and two 𝑎𝑠 on the right. Then, we can divide both sides of the equation by two 𝑎. This way the two on the right cancels and the 𝑎 on the right cancels. And we have 𝑣 squared minus 𝑢 squared divided by two 𝑎 is equal to 𝑠.

So at this point, all we need to do is to plug in the values. 𝑠 is equal to 𝑣, which is the final velocity, squared, so 11 metres per second squared, minus 𝑢, which is the initial velocity, squared, which is three metres per second whole squared, divided by two times the acceleration, which is four metres per second squared.

Expanding the parentheses in the numerator, we have 121 metres squared per second squared minus nine metres squared per second squared. And so, the numerator becomes 112 metres squared per second squared. Then, we divide this by two times four, which is eight, metres per second squared. And so, the per second squared in the numerator and the denominator cancel. And one factor of metres in the numerator cancels with the one factor of metres in the denominator. Overall, we’re just left with a unit of metres, which is good because we’re calculating a distance which has units of metres. And hence, we’re left with the final answer to our question. The length of the line along which the object moves is 14 metres.

Okay, now let’s summarise what we’ve talked about in this video. We’ve learned how to use the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠 for an object initially travelling at a velocity 𝑢, accelerating at a constant rate 𝑎, and moving along a straight line 𝑠, which then results in the object having a final velocity 𝑣.