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 are 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 in 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 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's in three parts.

And in the first part of the lesson we're going to look at two different equations that allow us to calculate average velocity.

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

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

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

The velocity of an object is not always constant.

It often changes during any journey.

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

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 equals 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 that average velocity equation in an example.

I've got a fire engine that travels nine 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 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 nine kilometres or nine times 1000 metres and we need to convert that time into seconds.

And it's 15 minutes, each of it's 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, 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 graph it 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 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 nought metres per second plus six metres per second divided by two.

And that gives 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.

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 velocity, sorry, direction change.

So is it 1.

5 metres per second, 3.

0 metres per second, 3.

5 metres per second of 5.

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

Welcome back.

Hopefully you selected 3.

5 metres per second.

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

The initial velocity is 5.

5 metres per second, and final velocity is 1.

5 metres per second.

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

5 metres per second.

Well done if you'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 equals 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 two.

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, maybe selections restart.

Welcome back.

Well hopefully for the first one you selected, V equals X divided by T.

Velocity is 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 a final velocity.

So we'd use the average of velocity is U plus V divided by two.

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 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 two.

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 equation B equals X over T.

I can write in the two values that I take from the question 520 metres and 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 eight 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, hopefully you chose this equation.

Average velocity is U plus B divided by two.

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

So it's eight metres per second plus 14 metres per second divided by two, 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, a 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 B divided by two, 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 four seconds.

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

Substitute 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 five metres per second to 2.

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

So I can use my average velocity equation, U plus V divided by two, substitute those velocities in and I get 3.

7 metres per second.

And for final one, well this was the most difficult because I've got to convert three hours into seconds.

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 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 and 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, we can find a general equation for 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 equals V minus 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 second.

So we have an equation for uniform acceleration and 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 equals two 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 2AX, and acceleration A is measured in metres per second squared.

Initial velocity is U, final velocity V and then 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 a skiing down a slope.

Let's 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'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 2AX.

Then we can put in the values, very carefully reading them from the question.

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.

The final velocity was 10 metres per second, the initial velocity was zero.

So I've got 10 squared minus zero squared on the left there.

Then I've got two 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 zero squared to zero, so that's 100 on the left.

And then I've got two 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 gonna give me two 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 that starts with a velocity of six metres per second and rolls to a stop.

So that's zero 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 and put in the values very carefully from the question.

And in this case the final velocity was zero and the initial velocity was six.

So I'll 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 six 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 five 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 zero.

I need to stop, I was going at five metres per second, so that's five squared there.

And on the right hand side I've got the distance, the 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.

5 metres per second squared.

Now it's your turn.

So I'd like you to calculate a minimum acceleration for an aeroplane that starts at one end of a 1000 metre 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 2AX, I substitute in the values.

The final velocity needs to be 40.

So that's 40 squared there.

The initial velocity was zero, that's zero squared, the distance is 1000 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 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 zero, 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 fine velocity, what I'm gonna need to do is take the square root of that.

So find V by taking a square root of 36, that gives me an answer of six metres per second.

So the hammer is travelling at six 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 zero squared is two 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 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 2AX.

Substituting in all the values 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've got that.

Another quick check now, I've got a rollercoaster and it's got an 80 metre straight launch track and it launches carriages from rest or from zero metres per second with an acceleration of 10 metres per second squared.

That's quite a large acceleration for a rollercoaster, well at least the starting point.

So I'd like you to then calculate the speed of the carriage at the end of that launch track.

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

Welcome back.

Well hopefully you selected 40 metres per second and we can see the mathematics behind that.

We write the equation, substitute the values from the question, and that gives us V squared equals 1,600.

We take the square root and that gives us the velocity of 40 metres per second.

So that carriage is moving very fast by the end of that launch track.

Well done if you've got that, okay, we've reached the second task of the lesson now, and what I'd like you to do is to answer these three questions using that equation of motion.

So pause the video, read through the questions, and work out the solution and then restart please.

Welcome back.

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

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

We've got a 50 metre long track, we need to calculate the acceleration they need.

We write out the equation, substitute in 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 write out the equation, substitute the values that we can see in the question and that gives us a velocity of 400 metres per second.

Well done if you've 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 a 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 or we've been asked for the velocities, we been asked for the acceleration or the distance.

The next thing we need to do is identify what information's provided in the question.

So what are we given to work with with? Then we need to use that to select the correct equation.

And finally, substituting 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 and that's probably the first one you learnt.

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 in this lesson, the general equation for motion, V squared minus U squared equals 2AX, where we've got final velocity, initial velocity, acceleration and distance travelled.

So here's a 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's 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 one 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.

The question is, 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 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, you should have selected this one, V squared minus U squared is 2AX.

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

0 metres per second.

The final velocity, it's 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.

9 metres per second.

It accelerates at 2.

9 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 a final velocity V here, and the information provided to us.

Well we 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 a 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, it speeds decreased to 2.

0 metres per second.

Calculate the acceleration acting on the ball as it moved.

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's being provided in the question.

Information provided here is we've got the initial velocity, U, we've got the final velocity, V, and we've got the distance, X.

And putting those, we can select this equation, V squared minus U squared equals 2AX.

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've being asked for, how long will it take? And the information we've been provided to solve that is what 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.

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

And what I'd like you to do is to use the equations you've seen throughout the lesson to answer these three questions.

So pause the video, answer the question showing 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 velocity of 36 metres per second, calculate acceleration.

Or for that one we 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 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.

It's 36,000 metres divided by one hour or 3,600 seconds.

That gives us a speed of 10 metres per second.

And so the time it'll take them to travel 200 metres is 20 seconds.

Well done if you've got that.

And here's the third question and the solution to it is shown here.

We've gotta use V squared minus U squared is 2AX to solve this one.

Putting in the values, doing all the calculations gives us X the distance travelled of 90 metres, 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 2AX.

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.