[Audio]Physics Kinematics - FortisLearn Education

[Audio]Physics Kinematics

Lesson 1: Scalar & Vector Quantities

Good Day! and welcome back to another lesson!

Today we are entering into a new chapter in our physics curriculum, which is: Kinematics. So what is Kinematics? It is the study of motion, without any reference to the forces that cause the motion. It basically means studying how things are moving, not why they’re moving. It includes concepts on scalar and vector quantities, such as distance versus displacement, speed versus velocity, and acceleration, and how these values change over time. But to start, we must understand what are scalar and vector quantities, and how are they different from one another?

In today’s lesson, we’re going to look at the differences between scalar and vector quantities.

What is a Scalar quantity? Scalar quantity is defined as a physical quantity that only has a magnitude, but no direction.

And remember magnitude is just another way of saying a numerical size, as it can be measured with a numerical value. For example, if a car travels at 20, meters per second. 20 would be the magnitude of the speed. And because speed by itself does not have a direction, we consider speed as a scalar quantity. Other scalar quantities include things like distance. mass, temperature and time. Although there are many more.

Vectors on the other hand, have both a magnitude and a direction. Therefore, a Vector quantity is defined as a physical quantity that has a magnitude, and also a direction.

Vector quantities include things like  displacement, velocity, acceleration, force, momentum and many more. We will take a closer look at each of these quantities in other lessons, so don’t worry if you’re not sure what any of them are, your aim for today is to understand the fundamental differences between scalars and vectors.

Good Job! That’s everything for today’s lesson. Now, are you ready to gain some XP and even a Gem? Proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 2: Distance VS Displacement

Good Day! and welcome back to another lesson!

In this lesson, we are going to learn what is distance and what is displacement, and the differences between them.

When we move from one place to another. The measure of the total length travelled is called the distance. 

For example, in the morning, a student have to walk from your home to school, and the best route is to move in a straight path, but because of roadworks along the way, the student have to take a detour. And the distance covered would be something like this. Luckily, the roadworks are completed before school ended, and the student went back home via the best route which is the straight path, would be something like this. 

The SI unit for Distance is denoted as lowercase, M. The distance between the student’s home and school, is 500 meters, but because of the detour, the student had to cover a distance of 900 meters in the morning.

In our example, we can say that instead of 500 meters, the student traveled  900 meters in the morning.  

Is distance, a scalar, or a vector quantity. Think about it. If you say someone travels 500 meters from this point, we can’t really tell where they ended up. Maybe here. Maybe there, and maybe they just went around in circles as it does not specify a direction, therefore distance is a scalar quantity. 

That brings us to displacement. So what is displacement? Displacement is distance with a direction. What do we mean by distance with direction? 

The student went to school from home in the morning, and went back to home from school in the afternoon. Was the distance covered in both the cases the same? Have a good look. It wasn’t.  In each of the two cases, the distance was different, the time it took for the detour in the morning was greater than the time the student could go straight back home in the afternoon. 

But the location of the school did not change. In both cases, is 500 meters east of the student’s home, the displacement in both cases is the same, the student traveled 500 meters east, east shows the direction. And because direction is specified, displacement becomes a vector quantity. 

Let’s look at this example: A car travels from, A, to B, to C, and finally to D, in a square route, distance for each path is 300 meters. Can you tell me what is the distance and displacement travelled of the car?

The distance travelled is the total length of the journey, which is 900 meters. The displacement on the other hand, is the shortest distance from the end point, to the start point, with a direction indicated, which is, 300 meters east.

If you really understood the difference between distance and displacement, you should be able to answer the next question. When a passenger travels by taxi. Is he charged based on the distance, or displacement.

Taxis charge passengers based on distance. If you go to the neighboring town, and return in the same taxi. The taxi meter shows you the charge for the entire round trip. If you now try to explain to the taxi driver that you have had no result in change in your location, and hence you must pay nothing. It will not impress them much.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 3: Speed & Average Speed

Good Day! And welcome back to another lesson!!

In this lesson, we are going to learn what is speed and how to calculate average speed.

First of all, how do you define speed? The speed of an object is the distance traveled by an object over the time it takes to complete that distance. Say the distance d is 40 kilometers, and the time it takes to cover this distance is four hours.

What will be the speed of the car then.

Yes, the speed will be 20 over four, that’s ten kilometers per hour.

Look at the value carefully. What does it tell us.

Does it mean that our object traveled ten kilometers in every hour, or in other words, does the value of speed, say that the object travels at a constant speed.

Well, we know for a fact that the car does not magically start directly at a speed of ten kilometers per hour.

The speed gradually increases. So, if the car was starting from this point, then its speed cannot be constant.

What if this is not the starting point of the car?

Well, we still cannot say whether the speed was constant or not. Because maybe the car was slow in the first half of the journey, and then speed up in the second half, the given data does not imply in any way that the car will cover ten kilometers in every hour.

So if this is not a constant speed, then what is the speed call? This speed is called average speed. Average speed is defined as the total distance covered by an object over the total time taken.

So in our example, we’re given a total distance that the car travels, and the total time taken by it to cover that distance, and hence we calculated the average speed of the car for the entire journey.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 4: Velocity (I)

Good Day! And welcome back to another lesson!

We have seen previously, that speed is the amount of distance covered in a given amount of time. It’s nothing but distance covered, over the time taken to cover that distance. Simply put, it tells us how fast, an object moves. Also, if we know the speed and the time taken to cover a distance, we can easily find the distance covered. So if I tell you that I traveled at a speed of 50 meters per second, for one minute. Can you tell me how much distance I covered.

I’m sure you’ve come up with the answer. I have covered a total distance of 3000 meters. But if I covered 3000 meters from this point. Am I here, or here, or over there?

You had no clue as to which direction I headed off. This tells us something interesting. Speed offers, only the magnitude, making it what we call in physics, a scalar quantity. It doesn’t specify the direction. It just tells us how much ground is covered in unit time. Regardless of the direction traveled. Scalar quantities have only magnitude, and do not specify direction. 

Velocity on the other hand, has both magnitude, and direction, therefore velocity is a vector quantity.

What do you think is the unit of measurement for velocity? The unit of measurement for velocity, just like speed, is meters per second. Now, you might ask why meters per second? The cars we drive, indicates speed in kilometers per hour. So why is the unit for velocity is meters per second? The established S I unit of measurement for velocity is meters per second. And S I stands for International System of Units. This helps everyone in the world, adhere to one common style of measurement.

Incidentally, speed and velocity, are both measured in meters per second.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 5: Average Velocity 

Good Day! And welcome back to another lesson!

In the previous lesson, we have learnt what is average speed. But how about average velocity, are they the same? To make it a bit easier for you. Let’s look at this example: A car travels from, A, to B, to C, and finally to D, in a square route, distance for each path is 300 meters, and it took 1 minute for the car to travel to each stop. Can you tell me what is the speed of the car?

The formula for Speed is equals to the total distance over total time, total distance in this case is 900 meters, divided by the total time taken which is 3 minutes, but do remember to convert time to its S I unit, which is 180 second. And we get 5 meters per second. 

Now, what will be the average velocity here. Is it the same as its speed? 

The average velocity depends only on the shortest path between the start point, and end point, regardless of path taken. Do you remember that we had learned about displacement? So the average velocity is equals to displacement over time. What is the displacement here. It’s 300 meters East. So the average velocity is equals to 300 meters over 180 seconds, which is equals to 1.67 meters per second. That’s the concept of average velocity.

So if you think you’ve really understood this concept, here’s a question for you. Now the car travels back to point, A, from point D, and also took 1 minute. What is the average velocity for the whole journey?

Okay, if you did not get it. Let me give you a hint. This was your start point and end point as well. They were both the same. The car drove one round, only to eventually finish at the same point as where it started. Therefore the displacement is zero. So if the displacement is zero, zero divided by the time taken, which is 240 seconds, equals to, zero.

So the average velocity is zero!

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 6: Acceleration (I)

Good Day! And welcome back to another lesson!

Have you ever wondered, when the train started to move off from a station, you’re pushed or swing towards the back of the train. This is because the train is accelerating, and your body is slowly catching up to the speed of the train.

So what is acceleration? Acceleration is the rate in which an object changes its speed. It can be quick, or it can be slower. Accelerate to speed up, decelerate is to slow down. Deceleration is also known as retardation. 

There are two things that affect acceleration. Firstly, how much the speed changes. So is it from zero to 100 kilometers per hour, or is it only from 40 kilometers per hour to 50 kilometers per hour. And secondly, how much time the change in speed takes. Have you gone from zero kilometers to 100 kilometers an hour, in 10 seconds, or 20 seconds.

So how can acceleration be measured. Since acceleration is the rate at which an object changes its speed, you can say that acceleration is the change of speed over time, the units of acceleration is meters per second squared.

A common city railway train, can reach 80 kilometers an hour in 20 seconds. This gives them an acceleration of 1.11 meters per second squared.

This calculation is because we have to turn the 80 kilometers, into meters first. Then turn the per hour, into seconds. by turning it into minutes, and then seconds, and divided by 20 seconds to get the acceleration.

How does the Formula One car, and the roller coaster, compared to the car, and the space shuttle. Work it out.

Did you get them right. The roller coaster actually has the fastest acceleration.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 7: Acceleration Calculation

Good Day! And welcome back to another lesson!

In this lesson, we are going to calculate acceleration.

Acceleration is defined as the rate of change of velocity. In terms of a formula, acceleration can be written as, change in velocity over the time taken, if the velocity of an object changes from an initial value of U, to the final value of V, in time T, then the acceleration, A , can be written as v minus u over T.

Lets say this car is at rest initially at point A.  It then travels to the right, and attains a velocity of 10 meters per second, in five seconds. What will be the acceleration here at point B. We can look at the formula to get the answer. The initial velocity u, is zero meters per second, as the object is stationary. The final velocity V is 10 meters per second, and the time taken T is five seconds. Hence the acceleration, A, will equal 10 meters per second, minus zero meters per second, divided by five seconds. So this will equal 10 meters per second divided by five seconds. 10 divided by five, is 2. Can you still remember what is the S I unit for acceleration?

Yes it is meters per second squared. So the acceleration in this case will be two meters per second squared.

And acceleration is also a vector quantity here, acceleration and velocity, are in the same direction. There are many more important concepts about acceleration which we will cover in the coming lessons.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 8: Displacement-Time Graph introduction (I)

Good Day! And welcome back to another lesson!

In this lesson, we are going to learn how to draw a displacement time graph. Notice how time goes on the horizontal x axis at the bottom, and the displacement goes on the vertical y axis.

Before we into the details, first, we must understand the concept of gradient of a line. The smallest Gradient, or minimum gradient, is when the line is at its horizontal position. The value for minimum gradient is zero. 

The largest Gradient on the other hand, or maximum gradient, is when the line is at its vertical position. The value for maximum gradient is infinity large, and can be denoted by the infinity symbol. 

So, what does the gradient represent on a displacement time graph? The gradient on a displacement time graph represents its velocity! We will learn the different types of displacement time graph in the next lesson.

With the knowledge of gradient in mind, Now, let’s add some information to our displacement time graph. A Car is driving from, A, to B, at a constant velocity. This will create a diagonal line from the origin of the graph, or point zero.

If the car drives slower, the line will be less steep, having a smaller gradient, as the velocity is smaller.

Also, because in order to reach to the same end point, B, with small velocity, the car has to drive for longer time.

If the car drives faster, the line will be steeper, having a larger gradient, as the velocity is bigger.

The car then drives back to point, A, at the same speed. Do remember that displacement is a vector quantity, and thus is direction aware. As the car is driving away from point B. So the line we draw must head in the opposite direction, we draw the line downwards, back towards the horizontal axis, when the car gets back to the horizontal axis, this means the car is back to point, A, There is no displacement anymore

If the car is stationary at any point, time will increase, but distance will not change. This will mean that we draw a straight line across the plot.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 9: Types Displacement-Time Graphs

Good Day! And welcome back to another lesson!

In this lesson, we are going to learn the different types of displacement time graph.

First, let us recall the concept of gradient. In a displacement time graph, the gradient of the graph represent its velocity. 

Minimum gradient is when the line is at the horizonal position. when the line is horizonal, its gradient is zero. We can conclude that, the flatter the line is, the smaller the gradient.

Maximum gradient is when the line is at the vertical position. when the line is vertical, its gradient is infinity large. We can conclude that, the steeper the line is, the larger the gradient.

This displacement time graph shows an object at rest or stationary. This is because the gradient of the line is zero, thus its velocity is zero. Also the displacement is constant at any time, thus the object is at rest.

This displacement time graph shows an object moving forward at an uniform velocity. This is because the graph is a straight line, thus the gradient of the line is constant, which tells us that, the velocity is also constant. The displacement of the graph is increasing, therefore, the object is moving forward at an uniform velocity.

This displacement time graph shows an object moving backward at an uniform velocity. Similar to previous example, the graph is a straight line, thus the gradient of the line is constant, which tells us that, the velocity is also constant. However, the displacement of the graph is decreasing, therefore, the object is moving backward at an uniform velocity.

This displacement time graph shows an object moving forward with an Acceleration. This graph is not a straight line, therefore the gradient is not constant. Look closely, we can see that the graph is getting steeper and steeper, which means that the gradient is increasing, thus, the velocity is increasing as well, when velocity increases, it means the object is accelerating. At the same time, the displacement of the graph is also increasing, therefore, we can conclude that, the object is moving forward with an acceleration.

This displacement time graph shows an object moving forward with a Deceleration. This graph is not a straight line, therefore the gradient is not constant. Look closely, we can see that the graph is getting flatter and fatter, which means that the gradient is decreasing, thus, the velocity is decreasing as well, when velocity decreases, it means the object is decelerating. However, the displacement of the graph is increasing, therefore, we can conclude that, the object is moving forward with a deceleration.

This displacement time graph shows an object moving backward with an Acceleration. This graph is not a straight line, therefore the gradient is not constant. Look closely, we can see that the graph is getting steeper and steeper, which means that the gradient is increasing, thus, the velocity is increasing as well, when velocity increases, it means the object is accelerating. However, the displacement of the graph is also decreasing, which indicates the object is moving backwards, therefore, we can conclude that, the object is moving backward with an acceleration.

This displacement time graph shows an object moving backward with a Deceleration. This graph is not a straight line, therefore the gradient is not constant. Look closely, we can see that the graph is getting flatter and fatter, which means that the gradient is decreasing, thus, the velocity is decreasing as well, when velocity decreases, it means the object is decelerating. At the same time, the displacement of the graph is decreasing, which indicates the object is moving backward,  therefore, we can conclude that, the object is moving backward with a deceleration.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 10: Velocity-Time Graphs (I)

Good Day! And welcome back to another lesson!

In this lesson, we are going to learn how to draw a velocity time graph.

Let’s have a look at how we can put information about the velocity onto a graph. As always, the time goes on the horizontal x axis. And now the velocity goes on the vertical y axis.

Let’s have a look at what the journey would look like plotted

A Car is at rest, with a velocity of zero.

As the car begins to move, its velocity will increase. The car is accelerating. We draw a diagonal line to show that the car’s velocity is increasing with time.

The faster the car accelerates, the steeper we draw the line. After this, the car drives at a constant velocity. This means that the line will become horizontal across the graph. In this case, the car’s velocity is no longer changing over time, because the car has reached a constant velocity. As the car reaches its destination, it begins to slow down until it has come to a stop, where its velocity, be zero again. This creates another diagonal line. This time, draw downwards towards the horizontal axis. In this path, the car is decelerating. The faster the car decelerates, the steeper the line we have to draw. So now it’s your go. Can you draw this journey. 

Did you get it right. You start by drawing a diagonal line that goes from zero to five meters per second on the velocity axis, and from zero to 10 seconds. Then because the car drives at a constant velocity of five meters per second, you should have a horizontal line at five meters per second, from 10 seconds to 30 seconds. And then, we end it by drawing it downwards diagonal line, from five meters per second to zero meters per second, and from 30 to 50 seconds, to show the car is decelerating.

Did you label your horizontal and vertical axis correctly with units.

Fun fact, the fastest running velocity of a human to date, is 10.3 meters per second. But scientists have discovered, that our ancient ancestors could actually run faster. From fossilized footprints, scientists worked out that human could run 12.5 meters per second. But we have the potential to run even faster, up to 16.7 meters per second. That means we can run 100 meters in just under six seconds. 

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 11: Velocity-Time Graphs (area under the graph)

Good Day! And welcome back to another lesson!

You should already know that velocity time graphs, look like this, and how we can use them to map out a journey.

In this lesson, we’re going to look at the area under these graphs, and what they represent. 

Let’s start by looking at a simple velocity time graph to find the area underneath the line. Multiply the value on the horizontal axes, with the value on the vertical axis, we are multiplying together, the velocity of the object, and the time it has traveled for.

Look at the unit. 80 meters. The area underneath the graph, gives us the total distance, that the object has traveled. So we have velocity, time, and distance.

The area won’t always be quite so simple to calculate. Velocity time graphs, more commonly looked like this, we can calculate the area underneath the line by cleverly splitting the area into triangles and rectangles. Remember, that the area of a triangle is, the base, multiplied by its height, divided by two.

Can you work out the distance traveled for this velocity time graph? Work out the total area under the graph.

Did you get it right? 2430 meters.

For most velocity time graphs, splitting up the area will be relatively obvious. However, you might come across some more complicated blocks, splitting up an area like this, will be less obvious. Whilst it doesn’t matter exactly how you split the area up, the fewer shapes you have, the fewer calculations you will have to do. As a general tip, try to include a triangle where you see diagonal lines, and rectangles where there are horizontal sections. Give this one a go yourself, work out the distance traveled.

Did you get it right.

This means that for the journey shown by the velocity time graph, the object traveled a total distance of 20 meters. When doing these calculations, just be sure to check the units that you’re given, because this will affect what unit you will give in your answer for the total distance. For this one, it was seconds and meters per second, so the distance in meters is correct. But for this one, it’s hours, and kilometers per hour, so the distance would be measured in kilometers.

So there we have, velocity time graphs. Velocity on Y axis, time on X axis, and the area underneath the graph is the distance traveled.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 12: Types of Velocity-time graphs

Good Day! And welcome back to another lesson!

In this lesson, we are going to learn the different types of velocity time graph.

First, let us recall the concept of gradient. In a velocity time graph, the gradient of the graph represent its acceleration. 

Minimum gradient is when the line is at the horizonal position. when the line is horizonal, its gradient is zero. We can conclude that, the flatter the line is, the smaller the gradient.

Maximum gradient is when the line is at the vertical position. when the line is vertical, its gradient is infinity large. We can conclude that, the steeper the line is, the larger the gradient.

This velocity time graph shows an object at rest or stationary. This is because the velocity constantly at zero, thus the object is at rest.

This velocity time graph shows an object is moving at constant velocity, as the velocity is constant on the graph. The gradient is zero, therefore the acceleration is zero.

This velocity time graph shows an object moving at constant acceleration. The velocity of the graph is increasing, thus, the object is accelerating. Also, the graph is a straight line, thus the gradient of the line is constant, which tells us that, the acceleration is also constant. Therefore, the object is moving forward at an constant acceleration.

This velocity time graph shows an object moving at constant deceleration. The velocity of the graph is decreasing, thus, the object is decelerating. Also, the graph is a straight line, thus the gradient of the line is constant, which tells us that, the deceleration is also constant. Therefore, the object is moving forward at an constant deceleration.

This velocity time graph shows an object moving at an Increasing Acceleration. The velocity of the graph is increasing, thus, the object is accelerating. We notice that this graph is not a straight line, therefore the gradient is not constant, which indicates that the acceleration is not constant. Look closely, we can see that the graph is getting steeper and steeper, which means that the gradient is increasing, thus, the object is accelerating at an increasing rate. Therefore, the object is moving at an increasing acceleration.

This velocity time graph shows an object moving at a decreasing Acceleration. The velocity of the graph is increasing, thus, the object is accelerating. We notice that this graph is not a straight line, therefore the gradient is not constant, which indicates that the acceleration is not constant. Look closely, we can see that the graph is getting flatter and flatter, which means that the gradient is decreasing, thus, the object is accelerating at a decreasing rate. Therefore, the object is moving at an decreasing acceleration.

This velocity time graph shows an object moving at an Increasing Deceleration. The velocity of the graph is decreasing, thus, the object is decelerating. We notice that this graph is not a straight line, therefore the gradient is not constant, which indicates that the deceleration is not constant. Look closely, we can see that the graph is getting steeper and steeper, which means that the gradient is increasing, thus, the object is decelerating at an increasing rate. Therefore, the object is moving at an increasing deceleration.

This velocity time graph shows an object moving at a Decreasing Deceleration. The velocity of the graph is decreasing, thus, the object is decelerating. We notice that this graph is not a straight line, therefore the gradient is not constant, which indicates that the deceleration is not constant. Look closely, we can see that the graph is getting flatter and flatter, which means that the gradient is decreasing, thus, the object is decelerating at a decreasing rate. Therefore, the object is moving at an decreasing deceleration.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 13: Acceleration-Time Graphs Gravity 

Good Day! And welcome back to another lesson!

In this lesson, we are going to learn the different types of acceleration time graph.

Firstly, in an acceleration time graph, the gradient represents an advanced concept called JOLT, or sometimes Jerk, however, we do not need to learn this in our level, as it is usually only significant in rocket technology.

Secondly, we can see that the graph has a positive and a negative region. If the line falls in the positive region, the object is in an accelerating motion, whereas if the line falls in the negative region, the object is in a decelerating motion.

Now let us start on the different types of acceleration time graphs.

This acceleration time graph shows an object moving at a constant speed, as the acceleration is zero, indicates speed in constant. However, it does not specify the magnitude of this constant speed, it can be 100 meters per second, negative 10 meters per second, or even not moving, at zero meters per second.

This acceleration time graph shows an object moving at a constant acceleration. The value for acceleration maintain constant over time, and the graph is in the positive region, therefore, the object is moving at constant acceleration.

This acceleration time graph shows an object moving at a constant deceleration. The value for acceleration maintain constant over time, and the graph is in the negative region, therefore, the object is moving at constant deceleration.

This acceleration time graph shows an object moving at a increasing acceleration. The value for acceleration increases over time, and the graph is in the positive region, therefore, the object is moving at increasing  acceleration.

This acceleration time graph shows an object moving at a decreasing acceleration. The value for acceleration decreases over time, and the graph is in the positive region, therefore, the object is moving at decreasing  acceleration.

This acceleration time graph shows an object moving at a increasing deceleration. The value for acceleration increases over time, and the graph is in the negative region, therefore, the object is moving at increasing  deceleration.

This acceleration time graph shows an object moving at a decreasing deceleration. The value for acceleration decreases over time, and the graph is in the negative region, therefore, the object is moving at decreasing  deceleration.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 14: Acceleration due to Gravity 

Good Day! And welcome back to another lesson!

The last concept we will learn in this chapter, is this second acceleration equation. This equation will help us to simplify the calculations on acceleration due to gravity.

We’ve already seen most of these letters. But now we have distance as well, which is measured in meters.

If we compare this to our original equation. You can see that the main difference is that this new one includes distance, instead of time. So you’ll have to choose which of these to use, depending on the question.

If you’re given the unit of distance, then you’d use this one. But if you’re given time instead. Then you have to use the original one.

In both cases though, something to be aware of is that, if an object starts from a stationary, then its initial velocity, u,  will be zero, because it’s not moving.

For an example, let’s imagine that a ball is dropped from an unknown height above the ground.

The speed of the ball, just before it hits the ground is seven meters per second.

Calculate the height from which the ball is dropped. Ignoring air resistance.

Now at first, this question seems a bit impossible, because we’re only given the ball’s velocity at the end of his journey. So his final velocity, which is seven meters per second.

However, because it’s been dropped, we can assume that it started off stationary. And so its initial velocity was zero meters per second.

Also, anything that gets dropped will accelerate downwards at 9.8 meters per second squared. Because of the force of gravity.

So we now have v, u, and, A, to find the distance though, which is S, we’re going to have to rearrange, by dividing both sides by 2 A.

And all we have to do now, is to plug in our values. So, seven squared, minus zero squared, divided by two times 9.8, which gives us 49, divided by 19.6, which is 2.5.

So the ball must have been dropped from 2.5 meters above the ground.

Good Job! That’s everything for today’s lesson. Now, proceed to the A I practice section for today’s challenge! And I’ll see you next time.

Lesson 15: Revision 1

Good Day! And welcome to question revision session!

In today’s session, we are going to look at this question on acceleration due to free fall.

A student stands near the edge of a cliff. He throws a ball upwards, The ball rises vertically a short distance and then falls. Given that Air resistance is very small as the ball moves upwards.

Question A,

The table below is for the magnitude, and the direction, of the acceleration of the ball, after it has left the student’s hand. Can you complete the table? Give the unit for the magnitude of any acceleration that you write down.

Did you get it right?

This is because acceleration due to gravity does not change its direction, nor its magnitude, as long as the gravitational force does not change. In other words, as long as the planet didn’t change, in this case, earth, the acceleration will always be 10 meter per second squared, and the direction is always downwards.

Question B,

The displacement-time graph shows the first 1 second of the motion of the ball.

Air resistance is very small in the first 1 second of the motion. On empty graph below, sketch the velocity-time graph for the first 1 second of the motion.

The initial velocity is shown by the point labelled, A.

Did you get it right?

From the displacement time graph we can see that, the ball reaches the maximum height at 0.4 seconds, because the gradient at this point is zero, as the tangent at this point is a horizontal line. Since gradient in a displacement time graph represents velocity, we can conclude that, the velocity at 0.4 seconds is, zero, thus, we will mark out a point at x equals to 0.4. Next, draw a line that connects these two points together, and extent it until t equals to 1 second.

Question C, part one

State, how the displacement time graph shows that terminal velocity is not reached, in the first 1 seconds

Did you get it right?

The gradient of a displacement-time graph gives the velocity. The gradient of displacement-time graph is not constant from zero to 1 second, which indicating that, the velocity is not constant, and still changing in the first 1 second. Hence, in the first 1 second, terminal velocity has not been reached.

Question C, part two

The ball continues to fall. The effect of air resistance becomes significant and the ball eventually falls at terminal velocity. 

Describe the velocity and acceleration of the ball as it falls at terminal velocity.

Did you get it right?

As the ball falls at terminal velocity, the velocity remains unchanged. At terminal velocity, the resultant force on the ball is zero, and hence, by Newton’s second Law, the acceleration of the ball is also zero.

Good Job! That’s everything for today’s Question revision session. Now, proceed to the A I practice section to practise today’s question to perfection! And I’ll see you next time.

Lesson 16: Revision 2

Good Day! And welcome to question revision session!

In today’s session, we are going to look at another question on acceleration due to free fall. This type of question is very frequently asked in the exams.

A stone falls from the top of a cliff into the sea, as shown. This is the speed-time graph for the stone .

Question A,

State how acceleration is found from a speed-time graph.

Did you get it right?

This is a very straight forward question, the gradient of the speed-time graph represents the acceleration.

Question B,

Describe how the acceleration of the stone changes between point A and point D. Give it a try!

Did you get it right?

Between points A to point C, we can see that the curve is getting flatter, which indicates that, the gradient is decreasing, and similar to velocity time graph, gradient on a speed time graph represents acceleration. Therefore we can conclude that the stone experiences a decreasing acceleration between points A to point C.

Between points C to point D, first, we can see that there is a drastic decrease in speed, which indicates deceleration. At the same time, the line between points C to point D is a straight line, which indicate the rate of change in speed is constant. Therefore, we can conclude that, the stone experiences a constant deceleration between points C to point D.

Question C,

Explain, in terms of the forces acting, why the acceleration changes between point A and point B

Did you get it right?

This question is testing on the topic of Newton’s second law of motion. As the stone falls, it experiences air resistance, which is a resistive force oppose the direction of the falling stone. As the speed of the falling stone increases, the magnitude of air resistance also increases, causing the resultant force to reduce, therefore the acceleration decreases.

Question D,

Explain how speed time graph shows that the stone does not reach terminal velocity in the air.

Did you get it right?

In the speed time graph, the section between point A and point C represents the motion of the stone falling in the air. To reach a terminal velocity, the graph as to be a horizontal line, which is not shown between point A and point C.

Good Job! That’s everything for today’s Question. Now, proceed to the A I practice section to practise today’s question! And I’ll see you next time.

Lesson 17: Revision 3

Good Day! And welcome to another question revision session!

In today’s session, we are going to look at a question on speed time graph. This type of question is very frequently asked in the exams.

A police car is travelling on a straight road. It increases its speed to overtake a truck, and then returns to its original speed. The speed-time graph shows the motion for the police car.

Question A,

Calculate the acceleration of the car as its speed increases.

Did you get it right?

From the graph, we can see that, the car accelerated from 6 seconds, to 10 seconds, as the gradient of the graph is positive, which indicates an acceleration on the speed time graph.

To calculate the magnitude of this acceleration, first, we must start with the acceleration formula. Please take note that, in most calculation questions, the concept of speed and velocity, is the same, as they share the same magnitude in numerical value. So, here we have the formula, a, equals to, v minus u, over t.

V, stands for final velocity, which in this case is 25 meters per second. Minus, u, the initial velocity, which is 20 meters per second. And divided by the time taken to reach from 20 meters per second, to 25 meters per second, which is between 6 seconds and 10 seconds.

And we get 1.25 meters per second squared.

Let’s move on to Question B,

Describe what does the shaded area on the figure represents.

Did you get it right?

Do you still remember that does the area under a velocity time graph, or speed time graph represent? Yes, that is correct, it represents the distance moved by the object. In this case, the shaded area represents the distance traveled by the car from 10 seconds to 16 seconds. During this time, the speed decreases from 25 meters per second, to 20 meters per second.

So far so good? Next Question, C part 1,

State what is meant by the average speed of the car

Did you get it right?

This question is asking for the definition for average speed, which is the total distance travelled by the car, divided by the total time taken for the car to travel this distance.

Question C, part 2,

The average speed of the car between 6 seconds and 16 seconds, is larger than between zero seconds to 20 seconds. Suggest why this is so.

Did you get it right?

This is because from 6 seconds to 16 seconds, a longer distance is travelled  per unit time than the rest of the period, thus, it has a larger average speed than that of the whole period. Try to calculate it yourself to support this statement.

Good Job! That’s everything for today’s Question. Now, proceed to the A I practice section to practise today’s question! And I’ll see you next time.

Lesson 18: Revision 4

Good Day! And welcome to another question revision session!

In today’s session, we are going to look at this question on velocity time graph.

A student throws a ball vertically upwards. The following sequence describes the motion of the ball.

Firstly, the ball leaves the student’s hand at time equals to zero second.

Secondly, it rises into the air until it reaches its highest point.

Thirdly, the ball then falls, and hits the ground with a velocity of positive 4 meters per second, at time equals to 0.6 seconds.

Fourthly, the ball is in contact with the ground for a very short time and bounces upwards.

Lastly, the student catches the ball at time equals to 0.8 seconds, when its velocity is negative 1 meters per second.

The effect of air resistance is negligible.

Question A,

State what is represented by the gradient of a velocity-time graph.

Did you get it right?

This is a very straight forward question, the gradient of the velocity-time graph represents the acceleration.

Question B,

On the velocity time graph below, draw the velocity-time graph for the ball from time equals to zero second, until time equals to 0.8 seconds. The initial velocity is marked as point P on the graph.

Did you get it right?

Let us go through this question from the first point.

The first point states that, the ball leaves the student’s hand at time equals to zero second, with a velocity of negative 2 meters per second. With this information, we can conclude that the starting point, with velocity equals to negative 2 meters per second, and time equals to zero, is at point P.

The second point states that, the ball rises into the air, until it reaches its highest point. This information does not give us any numerical detail, except we know that, the velocity at its highest point is, Zero.

The third point states that, the ball then falls, and hits the ground with a velocity of positive 4 meters per second, at time equals to 0.6 seconds. With this information, we can draw a second point at, velocity equals to positive 4 meters per second, at time equals to 0.6 seconds. Draw a line that connects this two points.

The fourth point states that, the ball is in contact with the ground for a very short time and bounces upwards. This tells us that, the velocity goes to negative the moment the ball starts its upward motion, as the direction of motion is opposite to the direction of the gravitational acceleration. However we do not know where exactly is this point yet, except that it is somewhere on the negative region of the line t equals to 0.6 seconds.

The last point states that, the student catches the ball at time equals to 0.8 seconds, when its velocity is negative 1 meters per second. With this information, we can deduce that the ball stops its motion at, time equals to 0.8 seconds, when its velocity is negative 1 meters per second, and we can draw a point here.

This leaves us with the last mystery, at what velocity did the ball leave the ground?

Remember the answer for question part ,A? Yes, acceleration of the ball is represented by the gradient of the velocity time graph! The acceleration is always constant regardless whether the object is moving up, or downwards, which means that, the gradient of the graph will also be the same! Therefore, the velocity of the ball before it moved upwards from the ground is, negative 3 meters per second.

Question C,

On the velocity time graph, mark with a letter X, at the point where the ball reaches its maximum height.

Did you get it right?

The second point states that, the ball rises into the air, until it reaches its highest point. The velocity at its highest point is, Zero. Therefore, point x is at where the line intersects with the x axis.

Good Job! That’s everything for today’s Question. Now, proceed to the A I practice section to practise today’s question! And I’ll see you next time.

Lesson 19: Revision 5

Good Day! And welcome to another question revision session!

In today’s session, we are going to look at this  velocity time graph of a running athlete.

This velocity time graph shows the motion of an athlete running for a 100 m race.

We can break this velocity time graph in to 4 sections.

From 0 to 1.8 seconds, the athlete starts to accelerate from rest, and the graph is straight line which shows constant acceleration.

From 1.8 seconds to 6.4 seconds, the athlete is accelerating at a decreasing rate, this can be represented by the decreasing gradient of the graph.

At 6.4 seconds, we can see that the graph is at its highest point, which shows that the athlete reaches maximum velocity.

And from 6.4 seconds to 9.8 seconds, the velocity decreases, and at the same time the gradient increases, therefore the athlete is running with an increasing deceleration.

Question A,

State a time during the race, when the acceleration of the athlete is largest.

Did you get it right?

The largest acceleration is when the gradient of the graph is at its steepest, which is from 0 seconds to 1.8 seconds

Question B,

The velocity of the athlete is always positive, but his acceleration is sometimes negative.

Explain this statement.

Did you get it right?

Acceleration is represented by the gradient of the velocity time graph. From about 6.4 seconds to about 9.8 seconds, the gradient of the graph is negative, hence, the acceleration is negative, although the velocity is positive.

Question C,

State how a student can use the velocity time graph to show that the race is 100 m long.

Did you get it right?

Distance is represented by the area under the velocity time graph, therefore the student can show the distance by estimating the area under this graph.

Question D,

Determine the average speed of the athlete during the race.

Did you get it right?

Average speed is equals to total distance over total time taken. Total distance is given at 100 meters, and total time can be found from the velocity time graph which is 9.8 seconds. Thus, the average speed is equals to 10.2 meters per second.

Good Job! That’s everything for today’s Question. Now, proceed to the A I practice section to practise today’s question! And I’ll see you next time.