Lesson video

Hello there, my name's Mr. Forbes and welcome to this last lesson from the measuring and calculating motion unit.

In the lesson, we're gonna look at a range of equations that will allow us to calculate the acceleration of an object, the distance it travels, and the time it takes.

By the end of this lesson, you're going to be able to use two different equations to calculate the average velocity of an object, and you're also going to be able to use an equation of motion that links initial velocity, final velocity, acceleration, distance, and time.

Here's a list of the keywords you'll need to understand for the lesson.

First of them is initial velocity, and the initial velocity of an object is the velocity it starts at during any phase or part of the motion.

Similar to that, the final velocity of an object is the velocity it finishes that phase of the motion at.

Then we have average velocity.

And the average velocity of an object travelling in a straight line is the change in distance divided by the time taken.

Then we have average acceleration, and the average acceleration of an object is the change in velocity divided by the time it takes for that change.

And finally, we have uniform acceleration, and when the acceleration of something is constant, we call that uniform acceleration.

And here's a list of those keywords that you can return to at any point in the lesson if you want to look at their definitions again.

The lesson's in three parts, and in the first part of the lesson, we're going to look at two different equations that allow us to calculate average velocity.

In the second part of the lesson, we're going to look at a general equation of motion that links together all of the factors involved in motion.

And in the final part of the lesson, we'll look at a series of examples of how we can use different equations to calculate motion.

So when you're ready, we'll begin by looking at the equations for average velocity.

The velocity of an object is not always constant and it often changes during any journey.

So when you're running a sprint, you'll start off slowly and get faster and faster.

Or when you're in a car, you'll be speeding up and slowing down all the time at different junctions and traffic lights and so on.

Those changes are due to forces acting on the objects and that causes the object to accelerate or decelerate.

So for example, a gravitational force on a falling apple will cause it to accelerate, it speeds up as it falls towards the ground.

Or if you're in a car, the braking force will act on it to slow it down and cause it to decelerate.

So in that picture there, we've got a car trying to decelerate as it goes around the curve.

I've got a quick check for you now.

I've got a large round stone and it's placed at the top of a steep slope at an origin point O and there's a marker flag placed along the slope at different positions, X, Y, and Z, each of those 10 metres apart.

And I release the stone and let it roll down the hill.

So which of the following statements are correct? So A, the velocity at X is greater than the velocity at O.

B, the velocity at Y is greater than the velocity at X.

Is it C, the velocity at Z is greater than the velocity at Y? Or D, the velocity at Z is greater than the velocity at O.

So choose which of those statements are correct and then restart, please.

Welcome back.

Well, you should have selected all of them.

All of those statements are correct.

The stone is getting faster, so at O, it's slowest, it's faster at X, faster again at Y and faster still at Z.

So well done if you selected them all.

Now we're gonna look at our first example of an equation, and this is an equation for average velocity of an object.

So when an object's travelling in a straight line between two points, we've got an equation for the average velocity and it's equal to the distance between the point divided by the time it takes to move between the points.

So we can write that out as an expression like this.

In symbols, it's much shorter.

We can write it out as v equals s divided by t.

And throughout this lesson, we're gonna meet mainly using symbols as it's much shorter and quicker to write.

So we'll define those properly.

We've got average velocity, v, measured in metres per second, distance, s, measured in metres and time, t, measured in seconds.

Okay, it's time for the first check now, and what I'd like you to do is to calculate an average velocity for me.

So I've got a fire engine travelling nine kilometres north in 15 minutes.

I want to calculate the average velocity of the fire engine.

And this question involves converting time units and distance units.

So I'd like you to pause the video, work out your answer and then restart, please.

Welcome back.

Hopefully you selected 10 metres per second north and the way we work that out is, or we can write out the equation velocity equals distance divided a time or v equal s divided t, but we've also got to make sure we convert those distances into metres.

So that's nine times 1,000 metres and the time, we need to convert that into seconds.

So it's 15 minutes, each of which has got 60 seconds.

So that gives us a final answer of 10 metres per second.

Well done if got that.

In this lesson, we're just going to be looking at objects that accelerate uniformly, and by uniformly, I mean a constant acceleration.

If we saw that type of acceleration on a velocity-time graph like this, then it would look something like this.

We've got a low uniform acceleration there.

The acceleration is a constant value throughout the whole journey.

We can analyse situations where acceleration changes, but that's much more complex, beyond the scope of this course.

If we've got something moving at uniform acceleration with a higher uniform acceleration, on a graphic, it would look something like this, a steeper gradient.

If an object's accelerating uniformly at a constant acceleration, then we can find the average velocity using the initial velocity and final velocity.

So I've got a line here showing an object with a uniform acceleration and I can identify its initial velocity, its starting velocity here.

It's the starting point of the graph, and we use the symbol u for initial velocity and that's zero metres per second.

We can also find the final velocity.

That's the endpoint of the graph here.

And we use the symbol v for final velocity and that's six metres per second according to this graph.

And to get the average, well, the average velocity is going to be adding those two values together and dividing by two because there's two values.

So average velocity is u plus v divided by two.

And in this case, the average velocity is nought metres per second, plus six metres per second divided by two.

And that gives us an average velocity of three metres per second as you probably expected.

And that average velocity can be found even if the initial velocity is not zero.

So the average velocity can be found for any change in velocity as long as the acceleration was uniform, constant acceleration.

So I've got a second graph here and I've got a change in velocity here.

I can identify the initial velocity, u is four metres per second.

I've read that from the graph there.

I can identify the final velocity and that's one metres per second, reading that off the graph.

And so I can get my average velocity again.

Average velocity is u plus v divided by two.

Put in the values that I've read off the graph, divide that by two there and that gives me 2.

5 metres per second.

Okay, I'd like you to use the same technique as I've just used to find out average velocity, please.

So I'd like to know what's the average velocity between three seconds and six seconds for the motion shown on this graph? All of the velocities are in the same direction, so you don't have to worry about direction change.

So is it 1.

5 metres per second, 3.

0 metres per second, 3.

5 metres per second, or 5.

5 metres per second? Pause the video, work out the average velocity and then restart, please.

Welcome back.

Hopefully you selected 3.

5 metres per second.

If you look at the graph, I can now identify the two points there.

The initial velocity is 5.

5 metres per second and final velocity is 1.

5 metres per second.

So I'll write those into the equation, divide them by two, and I get a final average velocity of 3.

5 metres per second.

Well done if you got that.

So as you've seen, there are two ways of calculating average velocity.

And the one we choose depends upon the information we're given in the question.

If we're given time and distance, we can use v equals s over t.

And if we're given initial velocity and final velocity, we can use average velocity is u plus v over two.

So I'd like you to decide which of those two equations you'd use to answer each of these questions.

I don't need you to answer them, just select the correct equation.

So pause the video, read each of the three questions, and then decide which equation you'd use to calculate the average velocity.

Then restart when you're done.

Welcome back.

Well, for the first one, a rocket travelling 5,000 metres in 10 seconds, you've got a distance and a time there.

So we'd use the equation v equals s divided t because we've got distance and time, that will give us the average velocity.

For the second one, we've got two different velocities, an initial velocity and a final velocity.

So we can find the average velocity there with average velocity is u plus v over two.

And for the final one, well, we've got two different velocities there.

We can ignore the time, we don't need that at all.

We've just got an initial velocity, 20 metres per second and a stop, a final velocity of zero metres per second.

So we'd use the average velocity is u plus v over two.

Well done if you selected those three.

Okay, let's try an example of using the equations to actually calculate average velocity now.

I'll do one and then you can do one.

So I've got a car travelling 520 metres in a straight line, taking 40 seconds and I wanna work out the average velocity of the car.

What I'd do is I'd write out the equation v equals s over t because I've got distance and time, fill in the values and then I can get the velocity of 13 metres per second.

Okay, now I'd like you to try and calculate an average velocity.

A truck is travelling in a straight line at eight metres per second and it accelerates to 14 metres per second.

What's the average velocity of the truck during the acceleration, please? So pause the video, work that out and then restart.

Welcome back.

You should have selected the other equation.

Average velocity is u plus v over two.

That's initial velocity plus final velocity divided by two.

If we substitute in the values from the question there, then we can get an average velocity of 11 metres per second.

Well done if you got that.

Now we've reached the first task of the lesson and I'd like you to work out some average velocities for me and I'd like you to give your answers in metres per second for each of them.

So work out the average velocity for each of those four objects as described there.

So pause the video and restart when you've done that, please.

Welcome back.

Well, let's have a look at the first two.

We've got a car accelerating from 18 metres per second to 22 metres per second along a straight road.

I can use this expression, average velocity is u plus v divided by two, substitute those two velocities in and that gives me an average velocity of 20 metres per second.

With a fairground ride launching people up to a height of 56 metres and four seconds, well, I've got a distance and a time there.

So I have to use this calculation using distance and time and that gives me an average velocity of 40 metres per second.

Well done if you got those two.

And for the next two, the ship slowing down.

Well, I've got two velocities there.

So the equation I need is average velocity is u plus v over two, substitute the initial and final velocities in there, and then I've got 3.

7 metres per second.

And for the final one there, well, I've got distance and time, but I've gotta do some conversion.

I've got a time of three hours and a distance of 400 kilometres.

So I can calculate that distance.

It's 400,000 metres and the time is 10,800 seconds and just use v equals s over t, giving me an answer of 37 metres per second.

Well done if you got that.

It's time to move on to the second part of the lesson.

In it, we're going to look at an equation of motion, an equation that describes and links distance, time, and velocities.

So let's get started with that.

In many situations, objects move at a constant velocity.

They're not speeding up or slowing down and they're the simplest scenarios and we've already seen, we've got some equations that will allow us to calculate distance and time.

Speed equals distance divided by time, for example.

But there are other times when the object will move with a constant acceleration, it's getting faster or slower at a constant rate.

It increases or decreases at the same amount each second.

That's called uniform acceleration, an acceleration that remains constant throughout that part of the motion.

If we combine together equations for acceleration and average velocity, we can come up with an equation that links velocity, distance and acceleration together.

We get an equation for uniform or constant acceleration like this.

Final velocity squared minus initial velocity squared is two times acceleration times the distance or much shorter written out in symbols, v squared minus u squared equals 2as where acceleration a is measured in metres per second squared, initial velocity, u, and final velocity, v, are measured in metres per second and distance, s, is measured in metres.

And now let's have a look at an example of using that equation to calculate motion.

So I've got a question here.

I've got a skier sliding down a steep slope.

They start with a velocity of zero metres per second and reach a velocity of 10 metres per second in a distance of 25 metres.

And I need to calculate the acceleration of the skier.

So the stages I go through these, I write out the equation.

Equation is v squared minus u squared equals 2as.

Then I put in the values from the question carefully and that's probably the most difficult part.

So the final velocity, v, was 10 metres per second squared, put that in, that's 10 squared there.

The initial velocity, well, it was zero metres per second, so put that in for the u, and the two, that's just the number two.

Then the acceleration, which is the thing I'm trying to find out.

And finally, the distance, s, 25 metres.

They all go into the equation just like that.

I simplify it by doing all the calculations I can.

The bit on the left-hand side, that's 10 squared minus zero squared, that's gonna be 100.

Then two times 25 on the right-hand side.

That gives me 50 metres times a.

So I've simplified the equation there.

And finally, I need to find a in that expression.

And a is going to be equal to the 100 divided by the 50 and that gives me two metres per second squared.

The equation can also be used for decelerations, not just accelerations.

So another example here, I've got a ball rolling along a grass pitch.

It starts with a velocity of six metres per second and it stops after it's rolled 36 metres.

Calculate the acceleration of the ball.

The same process again and write out the equation.

I'm putting the values carefully from the question.

Looking very carefully at the v and the u, the initial velocity and the final velocities.

Putting those two in there, I've got a final velocity of zero and an initial velocity of six.

So those are both squared on the left-hand side.

We've got a distance of 36 metres and I've got that value of two.

And then simplify by carrying out those squaring.

I've got minus 36 is equal to 72 metres times a.

And finally, I can do the calculation.

To find a, it's minus 36 divided by 72.

That gives me a value of minus 9.

5 metres per second and the acceleration is negative there and that shows me that the ball is slowing down or it's decelerating.

Okay, I'll do one more example and then you can have a go.

I've got a cyclist travelling at five metres per second along the road.

They see a hazard 10 metres ahead and need to stop before reaching it.

Calculate the acceleration required.

So the process is write out the equation.

V squared minus u squared equals 2as.

Look carefully at the question, identifying each of those values.

So the final velocity is zero, and the initial velocity is five.

So I've got zero squared minus five squared, it's two times a times 10 metres.

And that gives me, when I simplify it, minus 25 is 20 metres times a and I can then get a by going minus 25 divided by 20 is equal to a, giving me a final value of a of minus 1.

25 metres per second squared.

So they are decelerating, they are slowing down.

Now it's your turn.

An aeroplane starts at one end of a 1,000 metre long runway and needs to reach a speed of 40 metres per second to take off at the other end.

Calculate the minimum acceleration required.

So pause the video, follow the same process that I've just done, and then try and find that acceleration for me, please and restart.

Okay, welcome back.

Well, hopefully your calculation looks like this.

We've substituted in the values to the equation that we've written down, gone through each of those stages of simplifying and rearranging and that gives a final acceleration of 0.

8 metres per second squared.

Well done if you got that.

The equation can also be used to find the final velocity after a constant acceleration has taken place.

So we'll look at some examples of that.

I've got a hammer and it's dropped from a height of 1.

8 metres and it accelerates at 10 metres per second squared until it hits the ground.

And the question is calculate the velocity of the hammer when it reaches the ground.

Well, as before, the stage is just the same, it's just slightly different rearrangements as we go on.

So we write the equation v squared minus u squared is 2as.

We put in the values from the question there.

We've not got v, so v squared minus, well, the initial velocity was zero.

We've got the two, we've got the acceleration, so we put in 10 metres per second squared and we've got the distance of 1.

8 metres.

So we put all those values in and that gives us v squared equals 36.

Now, this is just v squared and we've been asked to find v.

So what we've gotta do then is just take the square root of that and that gives us a final value of v equals six metres per second.

I'll do another example and then you can have a go.

So I've got a feather, it's dropped on the moon this time from a height of 1.

5 metres and it accelerates up 1.

6 metres per second squared.

And that's because gravity is weaker at the moon.

Calculate its velocity as it reaches the moon's surface.

So we'll go through the same sort of process as before.

Write out the equation, substitute in the values, making sure I use the right acceleration here.

I've got the distance and I've got the initial velocity and I'm trying to find v.

I find v squared there, it's 4.

8 and that means v, it's the square root of 4.

8.

And so v is 2.

2 metres per second.

Now it's your turn.

A hammer is dropped on Mars from a height of 2.

2 metres and accelerates at 3.

7 metres per second squared.

Calculate its velocity as it reaches Mars' surface.

So go through the same process as I've done and find its velocity, please.

Welcome back.

Well, hopefully your calculation looks something like this.

Again, exactly the same processes.

We find v squared, we then take the square root and we've got our final velocity of 4.

0 metres per second.

Well done if you got that.

One more example for you to do here.

I've got a rollercoaster, it's got an 80 metres straight track and it's launched.

It launches carriages from rest.

So they're stationary when they start and it gives them an acceleration of 10 metres per second squared.

Quite a harsh acceleration.

One of the more exciting rollercoasters this one.

Calculate the speed of the carriage at the end of the launch track.

So pause the video, work out that speed, and then restart, please.

And welcome back.

The answer is 40 metres per second.

The calculation you should have carried out looks like this.

We've got the equation, we substitute the values, we get v squared, we take the square root and that gives us 40 metres per second.

Well done if you got that.

Okay, now it's time for you to use that equation to solve three questions.

I've got three questions written out here, all of which involve the use of that equation.

I'd like you to pause the video, work out your solutions to each of those, showing all your working and then restart, please.

Okay, welcome back.

Well, let's a look at the solutions to each of those.

So for the first one, the stunt driver needs to reach a speed of 20 metres per second from a standing start along a track of length 50 metres to make a jump.

Calculate the acceleration.

Well, we write out the equation, we substitute the values that we can see in the question there to find acceleration, going through each of those stages.

And that gives an acceleration of 4.

0 metres per second squared.

For the train, again, we write up the equation, we identify the data we've been given, substitute that into the equation, go through each of the steps of the calculation and it gives us a distance of 400 metres.

Well done if you've got those.

And for the third question, a rocket accelerating from rest at 2.

0 metres per second squared, calculate its speed after it's travelled 40,000 metres.

Again, write out the equation, substitute all of those in, go through each of the stages.

So we've got v squared.

Eventually, we can calculate v is 400 metres per second.

Well done if you got that one.

And now it's time to move on to the final part of the lesson.

And in it, we're going to look at calculating motion using a range of equations, and how to select the right equation for the question.

So let's go with that.

There are a wide range of equations we use in calculations of motion.

And the one we select depends upon the question that we're being asked and we need to solve those questions by following the following stages.

We need to identify which quantity is being asked for in the question.

We need to identify what information is being provided by the question.

Using that, we can then select the correct equation.

Once we've got the equation, all we need to do really is substitute in the values and solve it.

So we're gonna go through quite a wide range of examples of that.

Let's start by having a list of the equations we're gonna use about motion.

So the first of them, v equals s divided by T and that's the most commonly used one and that's speed equals distance divided by time.

Our velocity is distance divided by time in this case.

So I've got velocity, v, distance, s, and time, t.

The second equation we use, that's the acceleration equation.

Acceleration is change in in velocity divided by time as the variables there.

And the third one is the one we've just looked at in the previous part of the lesson.

V squared minus u squared is 2as, and that's the one that involves final velocity, initial velocity, acceleration, and distance.

We can also, because we've got that delta v, we can also say that delta v, the change in velocity is the final velocity minus the initial velocity.

So we may use that one as well.

Delta v equals v minus u.

And here's our first example equation.

I've got a sparrow that's travelling at an average velocity of 3.

0 metres per second.

How far will it travel if it flies for one hour? So what I need to do is identify the quantity being asked for and looking at that question, it's how far, so we're looking for the distance, s, in this question.

We're then looking for the information that's being provided that will allow us to calculate that.

So we need to identify that information provided and well, what we've got here, we've got an average velocity of 3.

0 metres per second and we've got a time.

So we've got velocity and time there.

Once we've done that, we can identify which equation we're going to use to solve it.

So we've got velocity, distance, and time involved here.

We need to use the equation velocity equals distance divided by time or if we rearrange that, distance equals velocity times time.

Let's see if you can select the correct equation to answer a question.

Which equation should be used to answer the question shown in the box? And I've got three possible equations shown there.

So I pause the video, decide on which equation it is and then restart, please.

Welcome back.

Well, hopefully you selected the answer C there.

V squared minus u squared equals 2as.

And the reason you should have selected that is well, we look at the question, we've been asked to calculate the distance, so we've been asked to calculate s, the distance travelled there and the information we've been provided, well, we've got u, the initial velocity, we've got a, the acceleration, and we've got v, the final velocity 'cause it comes to a stop.

So once we've got the question asking us for s and we've got v, a and u provided to us, we've got to use that equation.

V squared minus u squared is 2as.

Well done if you selected that.

Let's have a look at a second example of selecting an equation.

The question here is I've got a car travelling at 3.

0 metres per second.

It accelerates at 2.

0 metres per second squared for 10 seconds.

Calculate the new velocity of the car.

So I follow the same processes again.

I identify what's being asked for and it's fairly clear in this one, calculate the new velocity.

So I'm looking for the final velocity of the car, v.

Identify the information that's been provided.

So if we look carefully, we've got initial velocity, acceleration and time.

So the initial velocity of 3.

0 millimetres per second, got acceleration, 2.

0 metres per second squared and a time of 10 seconds.

So I've got all that information.

Now I need to select the equation.

So if I look carefully, I can see I've got a change in velocity.

I've got the two different velocities.

I've got time and I've been asked to find a new velocity.

So I use this equation a equals delta v divided by t, knowing that delta v is v minus u.

Okay, let's see if you can identify which equation's needed for this question.

So we've got a question in the box here.

At the start of a race, a motorcycle accelerates from a stationary start with a constant acceleration of 3.

0 metres per second squared.

Calculate the velocity of the motorcycle after its travelled 100 metres.

So pause the video, decide which equation is needed and restart, please.

Welcome back.

Well, you should have selected this equation again.

V squared minus u squared equals 2as.

And again, we look very carefully at the question.

We've been asked to calculate the velocity, so we need to find v and the information we've been provided, we've got the initial velocity, it was stationary, we've got the acceleration, 3.

0 metres per second squared, and we've got the distance, 100 metres.

So that was the equation we needed to answer the question.

Well done if you selected that.

So let's have a look at a final example of identifying which equation to use.

I've got a question here.

A bowling ball is rolled across the grass lawn.

The ball's released at a speed of 6.

0 metres per second.

And after travelling a distance of 30 metres, it speeds decreased to 2.

0 metres per second and I'm asked to calculate the acceleration acting on the ball as it moves.

So as before, I identified what's been asked for and that's the acceleration, a.

And I identify the variables that have been provided, the information that's given to me, and that's the initial velocity.

You can see it's 6.

0 metres per second and a final velocity of 2.

0 metres per second and a distance of 30 metres.

So looking at that information, I can select the correct equation and that's going to be v squared minus u squared is 2as.

Another one for you to do.

Which equation should be used to answer the question in the box? The engines of an aeroplane can provide a maximum acceleration of 5.

0 metres per second squared.

How long will it take for it to increase its speed from 140 metres per second to 180 metres per second? Pause the video, work out your answer to that and restart, please.

Welcome back.

Well, for that one, you should have selected a equals delta v divided by t.

I'm looking for how long? So a period of time, that's t.

And the information I've been provided is the acceleration and the initial and final velocities, and that gives me the delta v.

So well done if you selected that.

Okay, it's time for the final task of the lesson.

And what I'd like you to do is answer these three questions, please.

Each of them requires you to use the equations we've used throughout the lesson.

So pause the video, work out your solution, along with all your working and then restart, please.

Welcome back, well, the solution to the first of those is 9.

0 metres per second squared.

We used the equation shown there and taking the values from the question.

And the second one, well, that involved a little bit more work because we needed to calculate the velocity in metres per second and we were given it in kilometres per hour.

Well, we can see the first part of this shows you how to do that.

It was 10 metres per second.

We then used the equation and we get a time of 20 seconds.

Well done if you got that.

And here's the third part of the.

And here's the third question and its solution.

And we've got use the v squared minus u squared is 2as to solve this, and that gives us a distance of 90 metres.

Well done if you got that one.

And here's a summary of all the equations we've used throughout the lesson that you'll need to be able to use in your examination.

So we've got v equals s divided by t.

A equals delta v divided by t.

V squared minus u squared equals 2as and delta v is change in velocity is v minus u.

And we can use all those equations in different combinations to solve a wide range of questions.

Well done for reaching the end of the lesson.

I'll see you in the next one.