Lesson video

Hello there, my name's Mr. Forbes and welcome to this lesson from the Measuring and Calculating Motion Unit.

In the lesson, we're gonna look at a range of equations that'll 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 finishes that phase of motion.

Then we have average velocity.

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

Then we have average acceleration.

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 the definitions again.

The lesson is 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 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 update is not always constant.

It often changes during any journey.

So when you're running a sprint, we'll start off slow then 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.

Let's check if you understand about acceleration and deceleration.

I've got a large round stone and it's rolled down a steep slope.

And there is a marker flag positioned on that slope every 10 metres.

And you can see there in the diagram O, X, Y and Z.

And what I'd like you to do is decide which of those statements are correct.

We've got four statements there, A, B, C, and D.

So pause the video, read through those statements, decide which are correct, and then restart, please.

Welcome back.

Hopefully you selected all of them.

All of those statements are correct.

The velocity X is greater than velocity O, velocity Y is greater than velocity X, and velocity Z is greater than Y, and Z is greater than velocity O.

The stone's getting faster as it rolls downhill, it's accelerating.

So it's velocity is increasing at each stage O, X, Y, and Z.

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 the average velocity and update.

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

So we can write that out as special like this.

In symbols, it's much shorter, we can write it out v equals s divided by t.

And throughout this lesson we're gonna be 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.

What I'd like you to do is to calculate an average velocity for me.

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

I want you 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 what we can write up the equation velocity equal distance divided a time or v equals s divided by t.

What we've also got to make sure we convert those distances into metres, so that's nine times 1000 metres.

And the time, we need to convert that into seconds, so it's 15 minutes each of which 60 seconds.

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

Well done if you've 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.

Acceleration is a constant value throughout the whole journey.

We can analyse situations where acceleration changes.

Well, that's much more complex beyond the scope of this course.

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

If an object's accelerating uniformly, 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, 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 end point 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 divide it by two because there's two values.

So average velocity is u plus v divided by two.

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 metre per second, 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.

Putting 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 an average velocity case.

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 velocity, sorry 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 on the average velocity and then restart, please.

Welcome back.

Hopefully you selected 3.

5 metres per second.

If you look at the graph, you 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 'em by two, and I get a final average velocity of 3.

5 metres per second.

Well done if you've 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.

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

So I'd like for 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 would use the equation v equals s divided by t because we've got distance and time that can give us the average velocity.

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.

You 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 fine velocity of zero metres per second.

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

Well done if you selected those three.

Okay, let's try an example of using the equation strategy to calculate average velocity out.

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 do is 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 acceleration, sorry, the velocity 13 metres per second.

Okay, now I would 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, then restart.

Welcome back.

You should have selected the other equation.

Average velocity is u plus v of a 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 the straight road.

I can use this expression average velocity u plus v divided by two.

Substitute all two velocity 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 in four seconds while 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 14 metres per second.

Well done if you've 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, u plus v over two.

Substitute the initial and final losses in there and then I've got 3.

7 metres per second.

And for the final one there, while 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 it's 10,800 seconds.

And just use v equals s over t, give me an answer 37 metres per second.

Well done if you've got that.

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 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'll 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 a uniform acceleration, an acceleration that remains constant throughout that part of the motion.

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

So let's start by looking at an expression for acceleration.

Acceleration is the rate of change of velocity.

What that means is how much the velocity is changing every second.

If we write that as an equation, and you may have seen this before, acceleration is change in velocity divided by time, or a = v minus u divided by t.

And expressing each of those symbols clearly, acceleration is measured in metres per second squared, initial velocity, u, and final velocity, v, measured in metres per second.

And time, t, is measured in seconds.

And we can create an equation of motion by combining that equation for acceleration with the equation for average velocity, what that we saw earlier in the lesson.

So we write out acceleration is change in velocity divide by change in time.

And if we use the symbols initial velocity, u, and final velocity, v, we get that written as a equals v minus u divide by t.

So in symbols it's like that.

And if we wanna do our, find our equation, we need to do some rearrangement.

So I'm gonna rearrange that in terms of t, so I want t on the left hand side.

And doing that gives us this expression, t equals v minus u divided by a.

So I've written out the two equations I'm gonna combine together on the left here.

The first one is actually distance equals average velocity times time, but for the average velocity I've put u plus v over two, which we saw earlier in the lesson.

And the second one is the equation I've just worked out t equals v minus u over a.

So the stages that need are quite complex, so let's go through them.

First of all, substituting for t.

So I'm gonna replace the t in that first equation by the value in the second equation.

So replace it with v minus u of a and that gives us this expression then I multiply those two together as it says.

So multiplying, that would give us this value of the equations and then rearranging will give us this.

So 2 a s equals v plus u times v minus U.

If I multiply those two brackets altogether, I get v squared minus u squared.

So my equation is 2 a s equals v squared minus u squared.

That's normally written in this form where the 2 a s are on the right hand side.

So v squared minus u squared equals 2 a s, and that's our final equational option.

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 what shorter written out in symbols v squared minus u squared equals two a s.

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 2 a s.

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, but then that's 10 square there.

The initial velocity, well, it was zero metres per second, so I'll put that in there for you.

And the two, that's just the number two.

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

And finally the distance has 25 metres.

They all go into the equations 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 could be 50 metres times eight.

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 hundred divided by the 50 and that gives me two metres per second squared.

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

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.

I write out the equation, I put in 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 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.

I'll look at that value of two.

And then simplify by carrying out those scoring 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 -9.

5 metres per second.

And the acceleration is negative there and that shows me that the ball is slowing down, 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 right out the equation, v squared minus u squared equals 2 a s.

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

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

So I've got zero squared minus five squared.

It's two times eight times 10 metres.

And that gives me, when I simplify it -25 is 20 metres times a.

And I can then get a by going <v ->25 divided by 20 is equal to a.

</v> Giving me a final value of a minus 1.

25 metres per second squared.

So they are decelerating, they're slowing down.

Now it's your turn.

An aeroplane starts at one end of a 1000 metres 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, 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 9.

8 metres per second squared.

Well done if you've got that.

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

So we'll look at some examples of that.

We'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 stages 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 2 a s.

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 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 got to 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 the height of 1.

5 metres and it accelerates up 1.

6 metres per second squared, that because gravity is weak on the moon.

Calculate its velocity as it reaches the moon 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 got the distance and I've got the initial velocity and I'm trying to find v.

I find v square though, it's 4.

8 and that means v is 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, it accelerates at 3.

7 metres per second squared.

Calculate its velocity as it reaches Mars' surface.

So we'll go through the same processes that 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 route and we've got our final velocity of 4.

9 metres per second.

Well done if you've got that.

The equation can also be used to find acceleration when we've got changes in velocity when we've got two values for velocity and we're looking to find the acceleration.

So let's have a look at some examples.

A coach accelerates from four metres per second to nine metres per second, and distance of 325 metres.

And what I need to do is calculate the acceleration of the coach.

So to do that I write out the equation as always and I put in the values, and in this case, I've got the two velocities, initial, final velocities, I can substitute those in.

And I've got the distance from 325 metres.

I simplify that by carrying out all the parts of the calculation that counts so I can do the nine squared minus four squared, that gives me 65.

I could do the two times 325 metres, that gives me 650 metres.

And then I can rearrange that to find a.

The acceleration is 65 divided by 650.

That's a small acceleration of 0.

1 metres per second squared.

We can even use the equation to find the final velocity after constant acceleration.

So I've got a sprinter, near the end of the race the sprinter's travelling at eight metres per second.

In the last 10 metres, they accelerate, they push themselves as much as they can and so they accelerate uniformly at 4.

0 metres per second squared.

And I wanna calculate the speed as they cross the finish line.

So, again, I write out the equation and I very carefully substitute in the values.

I'm looking for the final velocity, v squared.

I've got the initial velocity at eight.

I've got two times and they were accelerating at four metres per second squared and they're covering that last 10 metres, so that's the distance.

So all those values go in, I simplify as much as I can.

I've got v squared minus 64 equal 80 in that case.

Switching that around to find the v, I'm going to take the square root of that, so it's a square root of 64 plus 80 and that gives me a final velocity of 12 metres per second.

For here's one for you to try.

So here's one for you to try.

We've got a cyclist travelling five metres per second at the top of the hill.

They cover a distance of 500 metres while accelerating uniformly down the hill at 0.

2 metres per second.

Calculate the speed of the cyclist at the bottom of the hill and you've got four options to choose from there.

So pause the video, work out that speed at the bottom of the hill, and then restart, please.

And welcome back.

And the answer was 15 metres per second.

And you can see the mathematics behind it here, starting with the equation, substituting in the values, getting an expression for v squared, and finally finding a square root of that, and that gives me 15 metres per second.

Well done if you've got that.

Okay, now it's time for the second task for the lesson.

And in it you're going to use that equation of motion to answer three different questions.

So I've got three questions here.

Each of which need to be solved using that equation.

So pause the video, work out the calculations that they ask, and then restart, please.

Okay, welcome back.

Well, let's have 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 the track of length 50 metres to make a jump, calculate the acceleration.

While we write out the equation, you substitute the values that we can see in the question there to find the acceleration going through each of those stages, and that gives an acceleration of 4.

0 metres per second squared.

For the train, again, we identified the, well, sorry, 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 that gives us a distance of 400 metres.

Well done if you've got those.

And for the third, we've got a rocket that's travelling at 300 metres per second.

That's accelerating at 2.

0 metres per second squared over a distance of 40 kilometres, calculate its final speed.

Well, calculation is shown here.

The key things in though is we need to convert the 40 kilometres to 40,000 metres to get our answer.

And again we get the v squared and then find the square root.

That gives us a final velocity of 500 metres per second.

Well done if you got that.

Now we can move on to the final part of the lesson.

And in it we're going do look at using a range of different equations involving speed, distance, time, and acceleration to solve problems. So let's go on with that.

As you've seen, there are a wide range of equations we use when we're analysing motion, and selecting the right one depends upon the question being asked.

So we need to be able to look at the question, decide on the equation before we can answer it.

To do that, we have several stages and these are, we need to identify what quantity is being asked for in the question.

We then need to identify what information's being provided in the question.

Then we need to select the correct equation.

Once we've selected that equation, we can substitute values from the question into it, and finally, we can solve the equation.

So let's have a recap of the equations we need to be able to use to calculate motion.

So the first one is velocity is distance divided by time, and we've used that quite a lot in previous lessons.

The next one is the equation for acceleration and that's the final velocity minus the initial velocity divided by time.

And the equation that we've learned this lesson v squared minus u squared equals 2 a s.

Where we've got final velocity, initial velocity, acceleration, and distance.

Now I've got a question here and I'm gonna select the correct equation to solve it.

So a sparrow is travelling at an average velocity of 3.

0 metres per second, how far will it travel if it flies for one hour? All I'm gonna do is identify the equation.

So the first stage is to identify the quantity being asked for.

And in this one it's how far, so how far is the bird travelling? So that's the distance, s.

The next stage is to identify the information provided in the question that help will solve it.

And that information is shown here.

We've got a velocity and we've got a time.

So we've got, so we're asked for the distance we've been provided with a velocity and a time.

So we need to select the equation that'll allow us to find that distance.

And that equation is this one v equals s over t because all three of those variables are involved in the question.

So I can use that and say s equals v times t, and that'll allow me to calculate s, the distance.

Now I'd like you to select an equation.

So which equation should be used to answer the question in the box though? A cargo transporter is travelling at 4.

0 metres per second, when it detects rocks ahead.

It can decelerate at 0.

1 metres per second squared.

Calculate the distance of both needs to come to a stopping.

So pause the video, select which equation will allow you to solve that, and then restart, please.

Welcome back.

You should have selected the bottom equation v squared minus u squared equals 2 a s.

And the reason for that, well, we've been asked for the distance s.

We've been provided with the initial velocity and the final velocity, that's v and u.

And we've also been provided with the acceleration, so we'll be able to answer that question using equation c though.

Well done if you've got that.

Let's have a look at another example.

So 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 velocity of the car.

So again, I identified the quantity being asked for and I've been asked for the new velocity or final velocity, v, there.

And what information's been provided well I've being provided with the initial velocity.

It was travelling at 3.

0 metres per second, the acceleration 3.

0 metres per second squared and the time accelerated for 10 seconds.

So I can select the correct equation based upon that.

Got a equals v minus u divided by t, and that will allow me to find the new velocity, v.

Okay, a second one for you to decide on.

I've got three equations again here.

Which of those will allow you to answer this question? 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 it's travelled 100 metres.

So pause the video, decide which equation to use, and then restart, please.

Welcome back, you should have selected the bottom equation again here, v squared minus u square is 2 a s.

We've been asked to calculate the velocity of the motorcycle, so the final velocity, v.

And we've been provided with the initial velocity, it was stationary.

The distance, it travelled a hundred metres, and the acceleration, which was 3.

0 metres per second squared.

So the equation to you use with that one shown are point c there.

Well done if you've got that.

And one final example of select the equation here I've got a bowling ball that's rolled across a grass lawn.

The ball is 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.

Calculate the acceleration acting on the ball as it moves.

So we identify the quantity being asked for and we've been asked to calculate the acceleration, a.

We identified the information provided and if you look carefully, we've been given the initial velocity 6.

0 metres per second, the final velocity 2.

0 metres per second, and the distance it travelled 30 metres.

So the equation we need to select is this one v squared minus u squared equals 2 a s.

And the last quick check for you to select the correct equation.

So which equation is needed to answer this question? The engines of an aeroplane can produce 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? So pause the video, set the correct equation, and then restart, please.

Welcome back.

You should have selected a equals v minus u divided by t.

We've been asked to find how long or what length of time it takes, so that's t.

And we've been provided with the acceleration, 5.

0 metres per second squared and the initial and final velocities there.

So well done if you selected that.

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

And I'd like you to do is to use the equations of motion that we've looked at during this lesson and answer the following four questions showing all your working, please.

So pause the video, solve all four of these, and then restart.

Welcome back.

Well to the first question, we are calculating an acceleration and your calculation should look something like this, giving you an acceleration of 9.

0 metres per second squared.

And for the second one, we're calculating an average speed.

And this is a little bit more complex because we need to actually calculate the velocity metres per second first.

And that's what I've done on the left here.

The 36,000 metres divided by 3,600 seconds, which is one hour.

And that gives us a 10 metres per second velocity average speed.

And that will give us a time to travel 200 metres of 20 seconds.

Well done if you've got that.

And let's go through the solutions of the next two questions.

For question three, you showed us found a distance of 90 metres using the equation, v squared minus u squared equals two a s.

Well done if you've got that one.

And for the final one, the most complex of all, what we've got the solution here, v squared minus u squared is 2 a s, and we get v squared -16, and we then take the scoring, we get a final velocity of 5.

0 metres per second.

Well done if you've got that.

And we're at the end of the lesson.

So here's a summary of everything we've learned.

There's a range of equations we used to describe motion, and they're shown here v equals s divided by t, a equals v minus u divided by t, and v squared minus u squared equals 2 a s.

And you can see all of the different variables described there.

Those equations can be used in different combinations to solve a very wide range of questions by identifying the data provided and what's been asked for.

Well done for reaching the end of the lesson.

I'll see you in the next one.