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 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 and 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 changing distance divided by the time taken.

Then, we have average acceleration, and the average acceleration of an object is a 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 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 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 are ready, we'll begin by looking at the equations for average velocity.

The velocity of an object is not always constant.

It often changes during any journey.

So when you're running a sprint, you'll start off slow 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 its 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 of deceleration.

I've got a large round stone and it's rolled down a steep slope and there's a marker flag positioned on that slope every 10 metres, and you can see 'em there in the diagram, O, X, Y, and Z.

And 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, and velocity Y is greater than velocity X, and velocity Z is greater than Y, and Z is greater than zero.

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 first example of an equation, and this is an equation for the 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 points 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 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, 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 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, well, we can write out the equation, velocity equal distance divided by time, or V equals to S divided by 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 got 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, though 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 would look something like this, a steeper gradient.

If an object is 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's the starting point of the graph, and we use the symbol U for initial velocity, and that's at 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 divided by two because there's two values.

So average velocity is U plus V divided by two.

In this case the average velocity is naught 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.

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, 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 up 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 find 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, on 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 you to decide, which of those two equations you should 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 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 by T, because we've got distance and time that will 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 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 equation 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 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've 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 in 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 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 velocities, 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 it's 10,800 seconds, and just use vehicles S over T give me an answer of 37 metres per second.

Well done if you've got that.

We've now reached the second part of the lesson.

We're going to look at an equation of motion and 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 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 can bind together equations for acceleration and average velocity, we can come up with an equation that links velocity, distance, and acceleration together.

Let's start by looking at the equation for acceleration.

So the acceleration of an object is the rate of change of velocity or how much of velocity is changing every second.

If we write that as an expression, you might have seen this before, acceleration is changing velocity divided by time, or in symbols, A equals delta V divided by T, the delta V is the change in velocity there.

So we've got acceleration A measured in metres per second squared, change in velocity, delta V, measured in metres per second, and time measured in seconds.

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

So we write out acceleration is changing velocity by changing time, and if we use the symbols, initial velocity U and final velocity V, we get that written as A equals V minus U divided by T.

So in symbols it's like that.

And if we wanna 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 we've just worked out, T equals V minus U over A.

So the stages I 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 over A and that gives us this expression and I multiply those two together as it says.

So multiplying would give us this value in equations, and then rearranging will give us this.

So two 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 two A S equals V squared minus U squared.

That's normally written in this form where the two A S are on the right hand side.

So V squared minus U squared equals two A S, and that's our final equation of motion.

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's 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, the equation is V squared minus U squared equals two 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, put that in, that's 10 squared there.

The initial velocity, well, it was zero metres per second, so I'll put that in further U, 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 S, 25 metres, they all go into the equation just like that.

I simplify it by doing all the calculations I can, a 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 give me 50 metres times A.

So I've simplified the equation there.

And finally I need to find A in the 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 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 and 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 have 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 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 right out the equation.

V squared minus U squared equals two A S.

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

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

So I've got zero squared minus five squared is 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, given me a final value of A of 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 1,000 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 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'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.

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

8 metres.

So we've 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 (indistinct) 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 up 3.

7 metres per second squared, calculate its velocity as it reaches Mars's surface.

So we'll go through the same processes 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've got that.

The equation can also be used to find acceleration when we've got changes in velocity when we've got the 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 in a 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 and final velocities.

I can substitute those in.

And I've got the distance of 325 metres.

I simplify that by carrying out all the parts of the calculation I can.

So I can do the nine squared minus four squared, that gives me 65, and I can do the two times 325 metres.

That gives me 650 metres.

And then I can rearrange that to find A, acceleration is 65 divided by 650.

And that's a small acceleration of 0.

1 metres per second squared.

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

So I've got a sprinter near the end of the race.

The sprinter is travelling at eight metres per second.

In the last 10 metres they accelerate, they push themselves as much as they can, 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, that's eight, I've got two times 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 that 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.

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 the 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 of 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 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 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.

It's accelerating at 2.

0 metres per second squared over a distance of 40 kilometres.

Calculate its final speed.

Well, the calculation shown here.

The key things in there is we need to convert the 40 kilometres to 40,000 metres to get the right 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.

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 selected 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 quantities 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 T.

And that's the most commonly used one.

That's V equals distance divided by time, or 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 changing velocity divided by time.

There's 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 two A S, 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 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'll 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 two A S.

And the reason you should have selected that is, well, if we look at the question we've got, we've been asked to calculate the distance.

So we've been asked to calculate S, the distance travel 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 two A S.

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 2.

0 metres per second squared for 10 seconds.

Calculate the new velocity of the car.

So I'll follow the same process 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 metres per second, we've 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 equations 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 equations needed and restart please.

Welcome back.

Well, you should have selected this equation again, V squared minus U squared equals two A S.

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 a 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 amassed to calculate the acceleration acting on the ball as it moves.

So as before, I identify 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 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 two A S.

Another one for you to do.

Which equation should be used to answer the question in the box? The engines of a 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 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've got, 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, we've reached the third task of the lesson, but what I'd like you to do is to answer these four questions please, using the equations you've seen throughout the lesson and the techniques I've shown you.

So pause the video, work out the solutions showing 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 use the equation shown there, and 10 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 use the equation and we get a time of 20 seconds.

Well done if you got that.

And here's the solution to question three.

We have to use the V squared minus U squared is two A S to solve that.

And it gives us 90 metres.

And for the final one, again, we're gonna use the same equation, V squared minus U squared is two A S.

Following through each of the stages we get a final velocity of 5.

0 metres per second.

Well done if you've got that.

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 examinations.

So we've got V equals S divided by T.

A equals delta V divided by T.

V squared minus U squared equals two A S.

And delta V, a 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.