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 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.

The 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 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 lessons 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, 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.

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 them 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 at X is greater than velocity at O, and the velocity at Y is greater than velocity at X, and the velocity at Z is greater than Y, and Z is greater than 0.

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.

So now we can look at the first of the equations we use to calculate average velocity.

And we base this one on an object travelling in a straight line between two different points.

And if that's the case, the average velocity is going to be equal to the distance between those points divided by the time it takes to move between them.

And if we write that in words, we've got average velocity is change in distance divided by time taken.

In symbols that's written as v = x over t.

And we're gonna be using symbols a lot in this lesson.

So we're gonna define them clearly.

We've got average velocity is written as v, it's measured in metres per second.

Distance, x, is measured in metres, and time, t, is measured in seconds.

Let's see if you can use the average velocity equation in an example.

I've got a fire engine that travels 9 kilometres north in 15 minutes, and I'd like you to calculate the average velocity of the fire engine, and I've got four options there.

Be careful because you're gonna need to convert distance and time units to answer this question.

So pause the video, work out your answer, and then restart, please.

Welcome back, hopefully, you selected 10 metres per second north.

And to see the solution when it's this, and write out the expression velocity is distance divided by time, and we need to convert that distance into metres, so it's 9 kilometres or 9 times 1,000 metres, and we need to convert that time into seconds, and it's 15 minutes, each of which has got 60 seconds, so we get that sum there, and that gives us an 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, 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'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, its starting velocity here, it's the starting point of the graph, and we use the symbol u for initial velocity, and that's 0 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 6 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 2 because there's two values.

So average velocity is u plus v divided by 2.

In this case, the average velocity is 0 metres per second, plus 6 metres per second divided by 2, and that gives us an average velocity of 3 metres per second, as you probably expected.

And that average velocity can be found even if the initial velocity is not 0.

So the average velocity can be found for any change in velocity, as long as the acceleration was uniform, a 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 4 metres per second, I read that from the graph there, I can identify the final velocity, and that's 1 metres per second, reading that off the graph, and so I can get my average velocity, again, the average velocity is u plus v divided by 2, put in the values that I've read off the graph, divide that by 2 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 3 seconds and 6 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 the velocity, if its only a 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, and I identify the two points there, the initial velocity is 5.

5 metres per second, and the final velocity is 1.

5 metres per second.

So I'll write those into the equation, divide them by 2, and I get a final average velocity of 3.

5 metres per second.

Well done if you've got that.

So we've seen there are two different equations that we can use to calculate average velocity.

And the equation we choose to use depends upon the information given to us in the question.

If the question contains time and distance, then we can use the equation velocity equals distance divided by time, or v = x over t, like that.

If the question contains initial velocity and final velocity information, u and v, we can use this version of the equation, average velocity is u plus v divided by 2.

So what I'd like you to do now is to decide which of those two equations you would use to calculate the average velocity in each of these three scenarios? So pause the video, read through scenarios A, B, and C, and decide which equation you would use to find the average velocity.

You do not need to actually calculate the velocity, just select the equation.

So pause the video, make your selections, and restart.

Welcome back, well hopefully, for the first one you selected v equals x divided by t.

Velocity equals distance divided by time 'cause the question contains a distance and a time.

The second one, we've got two different velocities, an initial and the final velocity, so we'd use the average velocity is u plus v divided by 2.

And the third one, well actually, there's a piece of information in that that you don't need to calculate velocity, but we can use the initial velocity, it was moving at 20 metres per second, and the final velocity, the object has stopped, so the train has stopped, so it's got a final velocity of zero.

We didn't need to use the time, 60 seconds at all.

So the equation we need is average velocity is u plus v divided by 2.

Well done if you selected those three.

Okay, let's see if you can calculate some average velocity with me.

I'm going to do an example and then you can do one.

So I've got a car travelling 520 metres in a straight line and it takes 40 seconds, and we're gonna calculate the average velocity.

What I do is look at the information provided,, I've got a distance and a time so I can write out my selection of the equation, v equals x over t.

I can write in the two values that I take from the question, 520 metres in 40 seconds, and that'll give me an average velocity of 13 metres per second.

Now it's your turn, I've got a truck travelling in a straight line at 8 metres per second, and it accelerates to 14 metres per second.

I'd like to know the average velocity during that acceleration.

So pause the video, see if you can work that out, and restart.

Okay, welcome back, and hopefully, you chose this equation.

Average velocity is u plus v speed divided by 2.

You read the initial and final velocities and substituted those into that equation.

So it's 8 metres per second plus 14 metres per second divided by 2, and that gives us 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, and let's have a look at the answers to those.

Well, the car accelerating from 18 metres per second to 22 metres per second, I've got two different velocities there.

So the equation I'm gonna use is average velocity is u plus v divided by 2, and that gives me, when I substitute values in, 20 metres per second.

For the fairground ride, I've got a distance and a time, I've got 56 metres and I've got 4 seconds.

So I'm gonna be using, average velocity is distance divided by time.

Substitute the values in, that's 14 metres per second.

Well done if you got those two.

And for the next two, a ship slowing down from 5 metres per second to 2.

4 metres per second in 2 minutes, well, I've got two velocities there.

So I can use my average velocity equation, u plus v divided by 2, substitute those velocities in, and I get 3.

7 metres per second.

And for the final one, well, this one's the most difficult 'cause I've got to convert 3 hours into seconds, and I've gotta convert 400 kilometres into metres.

So 400 kilometres is 400,000 metres, and three hours, well, that's three times 60, times 60, 10,800 seconds.

I substitute those two into my equation for the average velocity, Velocity is distance divided by time, and that gives me 37 metres per second.

Well done if you've got that.

Now it's time to move on to the second part of the lesson, and in it, we're going to look at a general equation of motion, an equation that links together several factors that we can use to solve a wide range of questions.

The simplest situations involve constant velocity, where the object speed or velocity isn't changing at all, but in some situations, that velocity is going to be changing, and we're only going to be looking at constant accelerations where the acceleration doesn't change through our phase of motion, and that's called uniform acceleration, as I've mentioned before.

And we can find a general equation for the uniform equation by combining together the equations for acceleration and average velocity that we've seen before.

And that gives us an equation linking together velocity, distance, and acceleration.

So we can show where we get that equation, by starting with the acceleration equation.

The acceleration of an object is the rate of change of velocity, and that's how much the velocity is changing each second or every second.

If we write that in words or symbols, we get this, acceleration is change in velocity divided by time, or a = v - u divided by t.

And if we define those symbols properly, we've got acceleration, a, measured in metres per second squared, we've got initial velocity, u, and final velocity, v, measured in metres per second, and time, t, measured in seconds.

The two equations I'm gonna combine together are shown on the left here.

The first one x equals u plus v divided by 2 times t, and that's actually, distance equals average velocity times time, and we've used that many times before.

Remember, average velocity is u plus v divided by 2.

The second one is the one we've just seen, it's a rearranged version of the equation.

We've got t equals v minus u divided by a.

So what I can do is I can write out that second equation, but instead of writing, sorry, the first equation, but instead of writing t down, I can write the value of t is, v minus u to divided by a, from the second equation.

So I get that expression there, if I multiply those two together and rearrange a little bit, I get x equals v plus u, times v minus u, divided by 2 times a.

If I rearrange that a little bit, I get 2 a x equals v plus u times v minus u.

And finally, multiplying the bracket out, I get an expression like this, 2 a x equals v squared minus u squared, and that's our final equation, but it's usually written the other way around, it's usually written as v squared minus u squared equals 2 a x, which is identical.

So we have an equation for uniform acceleration, an equation that will help us find distance and final velocity and initial velocity, and it's this one, final velocity squared minus initial velocity squared, is 2 times acceleration times distance.

That's a very long thing to write out in full, so I usually write it out in the symbols, and the standard symbols are here, v squared minus u squared equals 2 a x, and acceleration, a, is measured in metres per second squared.

Initial velocity is, u, final velocity is, v, and they're measured in metres per second, and distance, x, is measured in metres.

Let's have a look at an example calculation using that equation of motion.

So I've got a skier and they're skiing down a slope.

They start with a velocity of 0 metres per second, and reach a velocity of 10 metres per second in a distance of 25 metres.

And I'm gonna use that equation to calculate the acceleration of the skier.

So the first thing I do is I write out the equation, v squared minus u squared equals 2 a x.

Then we can put in the values, very carefully reading them from the question, and it's important that you look very carefully at which is the initial velocity, and which is the final velocity.

So I'm gonna put in the values here, our final velocity was 10 metres per second, the initial velocity was 0, so I've got 10 squared minus 0 squared on the left there.

Then I've got 2 times a, which is the thing I'm trying to find, the acceleration, and then the distance, x, and that was 25 metres.

So I can do some simplification of that, by calculating the values 10 squared, that's 100, minus 0 squared is 0, so that's 100 on the left.

Then I've got 2 times 25 metres on the right, so that's 50 metres times a, that then can give me a way of finding a, a is going to be equal to that 100 divided by the 50, and that's going to give me 2 metres per second squared.

The equation can be used for decelerations as well.

So I've got another example here, I've got a ball, it's rolled along a grass pitch.

It starts with a velocity of 6 metres per second and rolls to a stop, so that's 0 metres per second, in a distance of 36 metres.

Calculate the acceleration of the ball.

So I follow the same stages as before, I write out the equation.

I put in the values very carefully from the question, and in this case, the final velocity was 0, and the initial velocity was 6 So I put those values in, I've got a distance of 36 metres as well.

I simplify by doing all the calculations I can, for example, the 6 squared, that gives me 36 is equal to 72 metres times a.

And finally, I can rearrange that to find a, a is minus 36 divided by 72 metres, and that gives me a final acceleration of minus 0.

5 metres per second squared.

And that's minus, it's a negative acceleration, so it's a deceleration, the ball is slowing down.

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

A cyclist is travelling at 5 metres per second along the road.

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

Calculate the acceleration required.

So I follow exactly the same stages as before, I write out the equation.

I look very carefully trying to identify each of the values I can, and putting them into that equation.

So I've got a final velocity of 0, I need to stop, I was going at 5 metres per second, so that's 5 squared there.

And on the right-hand side, I've got the distance of 10 metres.

I can then calculate, sorry, I can then simplify a little bit.

And then finally, I can try and find a, and a is minus 1.

25 metres per second squared.

Now it's your turn.

So I'd like you to calculate the minimum acceleration for an aeroplane that starts at one end of a 1,000-meter-long runway and needs to reach a speed of 40 metres per second to take off at the other end.

So pause the video, work out the minimum acceleration it needs, and then restart, please.

Okay, welcome back.

Well, you should have calculated the minimum acceleration using this.

We've got v squared minus u squared is 2 a x, I substitute in the values, the final velocity needs to be 40, so that's 40 squared there.

The initial velocity was 0, zero squared, the distance is 1,000 metres and I need to find a, so I simplify, then rearrange and that gives me a final value of a, of 0.

8 metres per second squared.

Well done if you've got that.

The equation can also be used to find the final velocity after a period of acceleration.

So let's have a look at a couple of examples of that.

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

8 metres and it's gonna accelerate at 10 metres per second squared until it hits the ground.

And what I wanna do is calculate the velocity, the velocity of the hammer, when it does reach the ground.

So I follow the same stages as always.

I write out the equation.

Then I put in the values looking very carefully, I'm trying to find the final velocity, so I leave that as v or v squared there, I've got the initial velocity of 0, I can put that in.

I've got the acceleration, that's 10 metres per second squared, and I've got the distance, 1.

8 metres.

So all those values go into the equation.

And then that simplifies, I can do most of the mathematics, it gives me v squared equals 36.

So to find v, the final velocity, what I'm gonna need to do is take the square root of that.

So I find v by taking the square root of 36, that gives me an answer of 6 metres per second.

So the hammer is travelling at 6 metres per second when it reaches the ground.

I'll try another example, now I've got a feather, it's dropped on the Moon from a height of 1.

5 metres.

It accelerates at 1.

6 metres per second squared.

That's because gravity is weaker on the Moon, and I'm gonna calculate its velocity as it reaches the Moon's surface.

So the same process as before, I write out the expression, I very carefully substitute in the values from the questions.

So I've got v squared minus 0 squared is 2, times 1.

6 metres per second squared, times 1.

5 metres.

All of that data came from the question, I then get a value of v squared is 4.

8.

So to find v, I need to take the square root, v is the square root of 4.

8.

That gives me v, the final velocity of 2.

2 metres per second.

Now it's your turn.

I'd like you to find out the final velocity of a hammer that's been dropped on Mars from a height of 2.

2 metres, and it accelerates at 3.

7 metres per second squared.

So pause the video, and calculate the velocity of that hammer as it reaches Mars' surface, and then restart, please.

Welcome back, hopefully, your solution looks something like this.

v squared minus u squared is 2 a x, substituting in all the values we read from the question, gives me v squared equals 16.

28.

So to find v, I take the square root of that and v is 4.

0 metres per second.

Well done if you got that.

The equation can be used to find the acceleration when there's a velocity change as well.

So I'm gonna look at an example of that.

I've got a coach accelerating from 4 metres per second to 9 metres per second in a distance of 325 metres.

I can calculate the acceleration of that coach.

So I use the same process as always.

Write out the equation and identify the values.

So the final velocity was 9, the initial velocity was 4 metres per second, I'll put those two values in for v and u, and then I can then write out the other part of the equation, 2 times a, and I'm looking for the a, that's the acceleration, and I put in the distance of 325 metres for the question there.

I can simplify that by doing as much of the calculation as possible, and that gives me this expression.

And then the next stage is to find the acceleration, and the acceleration is 65 divided by 650 metres, and that's 0.

1 metres per second squared.

The equation can also be used to find a final velocity after there's been a period of constant acceleration.

So let's have a look at an example of that.

Near the end of a race, a sprinter is travelling at 8 metres per second, but in the last 10 metres, they accelerate uniformly at 4 metres per second squared as they push themselves to try and cross the line.

Calculate the speed as they cross the finish line.

So I write out an expression for it.

V squared equals, sorry, v squared minus u squared equals 2 a x, and I look very careful to in putting the values, I'm looking for V squared, the final velocity.

And I've got the initial velocity, they were travelling at 8 metres per second, so I put that in.

The acceleration is 4 metres per second squared and 10 metres, so I put those values in, that gives me this expression, v squared minus 64 is 80, and I simplify that, and then I finally find v, v is gonna be equal to the square root of 64 plus 80, and that gives me 12 metres per second.

Let's see if you can do a similar calculation.

I've got a cyclist travelling at 5 metres per second at the top of a hill.

They then cover 500 metres while accelerating down the hill uniformly at 0.

2 metres per second squared.

I need you to calculate the speed of the cyclist at the bottom of the hill.

I've got four options there.

So pause the video, work out the solution, and then restart.

Welcome back, the answer to that was 15 metres per second.

And you can see the calculation behind that here, I've got the equation, substituted in the values, found v squared, then taken a square root to find v, and that's 15 metres per second.

Well done if you've got that.

Okay, it's time for the second check of the lesson.

I'd like you to answer these three questions using that equation of motion.

So pause the video, work through each of those three questions, and then restart, please.

Welcome back.

Well, here's the solutions to the first two of those.

We've got a stunt driver driving and they need to reach a speed of 20 metres per second to be able to do their jump.

They've got a 50-meter-long track, we need to calculate the acceleration they need.

We write out the equation, substitute the values, and that gives us an acceleration of 4.

0 metres per second squared needed.

For the train, similarly, write out the equation, substitute the values from the question there, and that gives us a stopping distance of 400 metres.

Well done if you've got those two.

And for the third question, we can identify the data and put it into the equation.

So first the equation, we read off the information from the question, being careful to convert the 40 kilometres to 40,000 metres, and that gives us a final velocity of 500 metres per second.

Well done if you got that! And now we're onto the final part of the lesson, and in it, we're going to be using the range of motion equations that we've seen to analyse some questions and find solutions.

So let's get on with that.

As you've seen, there's several equations that we can use when we're trying to calculate motion, and we need to be able to select the correct equation to answer the question.

The stages involved in doing that are shown here.

First of all, we need to identify what's being asked for, what quantity is being asked for are we being asked for the velocities, are we being asked for the acceleration, or the distance? The next thing we need to do is identify what information is provided in the question, so what are we given to work with? Then we need to use that to select the correct equation, and finally, substitute in those values, and calculate the equation or solve the equation.

Let's have a look at the equations that we know that involve motion.

So first of all, we've got v equals x over t, or velocity is distance divided by time.

That's probably the first one you learned.

The second equation we've seen is acceleration, and the acceleration is the change in velocity or v minus u, final velocity minus initial velocity, divided by time, and those values are shown there.

And the third equation is the one we've learned during this lesson, the general equation for motion, v squared minus u squared equals 2 a x.

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

So here's our first example of a question that involves motion, and we're gonna identify which equation to use, we're not actually going to solve it, just find the equation.

So the first stage is we identify what's being asked for and in this question, it's "how far," that's a distance, so we're gonna be looking for a distance, x, then we need to identify which information has been provided in the question.

And for this one, well, we've got a velocity of 3.

0 metres per second, so there's our velocity, and we've also got a time of 1 hour.

So we've got velocity and time.

So the equation we need to solve this is one involving those three variables, v, x, and t.

So that's v equals x divided by t, giving us x equals v times t when we rearrange it, and that's what we're gonna use to solve the question.

So what I'd like you to do now is to decide which of the three equations I've given you here, should be used to answer the question in this box.

And the question is, a cargo transporter is travelling at 4.

0 metres per second, but it detects rocks ahead.

It can decelerate at 0.

1 metres per second squared.

Calculate the distance the boat needs to come to a stop in.

So pause the video, select the equation, you don't need to answer the question, just select the correct equation, and then restart.

Welcome back, and you should have selected this one, v squared minus u squared is 2 a x.

And the reason for that, well, we've been asked to calculate the distance, that's v there, and the information we've been provided with is the initial velocity 4.

0 metres per second, the final velocity, it has come to a stop, and the acceleration, 0.

1 metres per second squared.

So those four variables there are linked together by that equation shown as Part C there, well done if you correctly selected that.

And now for a second example, we'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 we're gonna try and select the correct equation to use in this situation.

So again, identify what's been asked for.

We've been asked to calculate the new velocity, so we're looking for final velocity, v here, and information provided to us, well, you can see we've got an initial velocity, 3.

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

So we can use this equation, a equals v minus u divided by t.

And that will allow us to find, v, the new velocity.

So your go again.

I'd like you to decide which equation should be used 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, and you should have selected this equation here, again, we'll look at why, we've got a velocity that we're trying to calculate, we're trying to find v, the final velocity, we've been provided with the initial velocity, it was stationary, the acceleration, a, and the distance, x there.

So we can use that one shown as Part C there.

Well done if you selected that.

And our final example, a bowling ball is 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, its speed has decreased to 2.

0 metres per second.

Calculate the acceleration acting on the ball as it moves.

So again, the same process, identify what's being asked for.

And in this case, we've been asked to calculate the acceleration, then we look at what information is being provided in the question, the information provided here is we've got the initial velocity, u, we've got the file velocity, v, and we've got the distance, x.

And put in those, we can select this equation, v squared minus u squared equals 2 a x, and that's going to allow us to calculate a.

So it's your final turn at selecting the correct equation.

I've got another question here.

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? And I'd like you to calculate, which of those equations would you use to solve that question.

Pause the video and restart when you're done.

Welcome back, you should have selected this equation.

And again, let's have a look at why, we've got time, t, is what we're being asked for, how long will it take? And the information we've been provided to solve that is, well, we've got the acceleration, 5.

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

So well done if you selected B.

So we've reached the final task of the lesson, and what I'd like you to do is to solve these four questions.

I'd like you to pause the video, work through the processes of using those equations showing all your working, and then restart, please.

Welcome back, let's solve those.

So we've got the first one, the stunt driver going off a cliff.

After 4.

0 seconds, they've got a velocity of 36 metres per second, calculate their acceleration.

Well, for that one, we'll use the equation a equals v minus u over t.

And that gives us 9.

0 metres per second squared, well done if you've got that.

For the cyclist one, there is a bit more work to do here because we've gotta convert the speed of 36 kilometres per hour into metres per second to solve this.

So I've done that in the first part here, so it's 36,000 metres divided by 1 hour or 3,600 seconds, that gives us a speed of 10 metres per second.

And so the time it'll take him to travel 200 metres is 20 seconds, well done if you've got that.

And here's the next two.

So for the skateboarder travelling at 6 metres per second, decelerating at 0.

2 metres per second squared, the distance they travel is shown here, is 90 metres, and we had to use v squared minus u squared, equals 2 a x to solve that one.

And the final one, calculating the child going on the water slide, we've got a solution shown here.

Again, we're using v squared minus u squared is 2 axe, and that gives us a final velocity of 5.

0 metres per second.

Well done if you've got that one.

And now we've reached the end of the lesson, and here's a summary of everything we've gone through.

We've looked at a range of motion equations, and each one of them can be used to solve different types of questions, and sometimes we use them in combinations.

So we've got v equals x divided by t, a equals v minus u divided by t, and v squared minus u squared equals 2 a x.

And each of those variables is defined there as well.

So those equations can be used to solve a very wide range of motion questions.

Well done for reaching the end of the lesson, I'll see you in the next one.