Motion

Hello students, welcome to today's science lesson. I'm so happy to see you all here, ready to learn about something that we experience every single day of our lives. Today, we are going to study Chapter 7, which is all about Motion. Now, before we begin, I want you to look around yourself. What do you see? You see your classmates sitting, you see the fan rotating above, you might see the trees moving outside the window. Everything in this world is either at rest or in motion. In fact, students, even when you think you are sitting still, your heart is beating, your blood is flowing through your veins, and you are breathing. So, let us dive into this fascinating chapter and understand what motion really means.

So students, let's start by understanding what we mean when we say something is in motion. In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through our veins and arteries, and cars move on the road. Atoms, molecules, planets, stars, and galaxies are all in motion. We often perceive an object to be in motion when its position changes with time. However, there are situations where the motion is inferred through indirect evidences. For example, we cannot see air moving, but we can infer the motion of air by observing the movement of dust particles and the movement of leaves and branches of trees. Isn't that interesting, students?

Now, let me ask you a question. What causes the phenomena of sunrise, sunset, and changing of seasons? Is it due to the motion of the earth? If it is true, why don't we directly perceive the motion of the earth? This is a thought-provoking question, and I want you to think about it as we go through this chapter. The answer lies in understanding reference points, which we will discuss very soon.

Another interesting observation is that an object may appear to be moving for one person and stationary for some other person. Let me give you a very common example from our Indian context. Imagine you are sitting in a moving bus, students. What do you see? The roadside trees appear to be moving backwards, right? But a person standing on the road-side perceives the bus along with the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. What do these observations indicate? This tells us that motion is relative. It depends on the observer's frame of reference. This is a very important concept, students, and we will revisit it later in the chapter.

Now, let us talk about the different types of motion that we see around us. Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate. There may be situations involving a combination of these. In this chapter, we shall first learn to describe the motion of objects along a straight line. We shall also learn to express such motions through simple equations and graphs. Later, we shall discuss ways of describing circular motion. So, let us take one step at a time.

Before we proceed, let me ask you to do a small activity. Just think about the walls of your classroom. Are they at rest or in motion? Discuss this with your partner. This is Activity 7.1 from your textbook. Now, I want you to think about this. The walls appear to be at rest to us, but are they really at rest? Actually, students, the earth is rotating, so technically, the walls are also moving! But we don't perceive this motion because we are moving with the earth. This is again about the frame of reference.

Here is another interesting activity. Have you ever experienced that the train in which you are sitting appears to move while it is at rest? This happens sometimes when you are sitting in a stationary train and the train next to you starts moving. From your perspective, it feels like your train is moving in the opposite direction. This is a very common experience at railway stations in India. Please discuss and share your experience with your classmates. This is Activity 7.2.

Now, students, let me ask you something. We sometimes are endangered by the motion of objects around us, especially if that motion is erratic and uncontrolled as observed in a flooded river, a hurricane, or a tsunami. On the other hand, controlled motion can be a service to human beings such as in the generation of hydro-electric power. Do you feel the necessity to study the erratic motion of some objects and learn to control them? This is something for you to think about. Understanding motion helps us predict natural disasters, design vehicles, and even play sports better.

Now, let us move on to Section 7.1, which is about Describing Motion. This is the foundation of this chapter, so pay close attention.

We describe the location of an object by specifying a reference point. Let us understand this by an example. Let us assume that a school in a village is 2 km north of the railway station. We have specified the position of the school with respect to the railway station. In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin. This is very important, students. Whenever we talk about motion, we always need to mention the reference point.

Now, let us understand motion along a straight line, which is the simplest type of motion. We shall first learn to describe this by an example. Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point. Let A, B, and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B.

Now, students, I want you to understand two very important terms here: distance and displacement. The total path length covered by the object is OA plus AC, that is 60 km plus 35 km, which equals 95 km. This is the distance covered by the object. To describe distance, we need to specify only the numerical value and not the direction of motion. There are certain quantities which are described by specifying only their numerical values. The numerical value of a physical quantity is its magnitude. So, distance is a scalar quantity - it has only magnitude, no direction.

Now, from this example, can you find out the distance of the final position C of the object from the initial position O? This difference will give you the numerical value of the displacement of the object from O to C through A. The shortest distance measured from the initial to the final position of an object is known as the displacement. So, students, displacement is the shortest distance between the initial and final positions. It is a vector quantity, which means it has both magnitude and direction.

Let me calculate this for you. The object moves from O to A, which is 60 km, and then from A back to C, which is 25 km (since C is located between O and A). So the total path length is 60 km plus 25 km, which equals 85 km. But the displacement, which is the shortest distance from O to C, is just 35 km. Thus, the magnitude of displacement, which is 35 km, is not equal to the path length, which is 85 km. This is a very important distinction, students. Further, we will notice that the magnitude of displacement for a course of motion may be zero, but the corresponding distance covered is not zero. If we consider the object to travel back to O, the final position coincides with the initial position, and therefore, the displacement is zero. However, the distance covered in this journey is OA plus AO, which is 60 km plus 60 km, that equals 120 km. Thus, two different physical quantities - the distance and the displacement - are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time.

So, students, let me recap what we have learned so far. Distance is the total path length covered by an object, while displacement is the shortest distance between the initial and final positions. Distance is a scalar quantity, while displacement is a vector quantity. The magnitude of displacement can be less than or equal to the distance traveled, but never greater than it. And in some cases, like when an object returns to its starting point, the displacement can be zero even though the distance traveled is not zero.

Now, let us do Activity 7.3. Take a metre scale and a long rope. Walk from one corner of a basketball court to its opposite corner along its sides. Measure the distance covered by you and magnitude of the displacement. What difference would you notice between the two in this case? Students, when you walk along the sides of the basketball court from one corner to the opposite corner, the distance you cover is the sum of the lengths of the two sides. But the displacement is the straight-line distance between the two corners, which is the diagonal of the rectangle. So, the distance will be greater than the displacement. This is a simple activity that you can perform in your school playground.

Now, Activity 7.4. Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km. Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India. Students, this is an interesting activity that you must do yourself. Take a Road Map of India and locate both cities. The odometer shows the total distance traveled along the road, which is 1850 km. But the displacement is the straight-line distance between Bhubaneshwar and New Delhi. Using your map and a ruler, measure this straight-line distance and compare it with the road distance. You will find that the displacement is much less than 1850 km because roads are not straight. This activity will help you understand the practical difference between distance and displacement.

Now, let us look at the questions from this section.

Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes, students, an object can have zero displacement even if it has moved through a distance. This happens when the object returns to its starting point. For example, if you run around a circular track and come back to where you started, your displacement is zero, but you have definitely covered some distance. Another example is if you walk from your home to the market and then come back home. The total distance traveled is twice the distance from your home to the market, but your displacement is zero because you are back at your starting point.

Question 2: A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Let me solve this for you, students. First, we need to find out how many rounds the farmer completes. The perimeter of the square field is 4 times 10 m, which is 40 m. The farmer takes 40 seconds to cover this 40 m. So, his speed is 1 m/s.

Now, 2 minutes 20 seconds equals 140 seconds. In 40 seconds, he completes one round. So in 140 seconds, he will complete 140 divided by 40, which is 3.5 rounds. This means he will be exactly at the starting point after completing 3 rounds, and after half a round, he will be at the opposite corner of the square.

Now, the displacement after 3 complete rounds is zero because he returns to the starting point. After half a round, he is at the opposite corner. The displacement from the starting point to the opposite corner is the diagonal of the square. The diagonal of a square with side 10 m is 10 times the square root of 2, which is approximately 14.14 m. So, the magnitude of displacement at the end of 2 minutes 20 seconds is 14.14 m.

Question 3: Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

The correct answer is that neither (a) nor (b) is true. Displacement can be zero, as we saw in the previous question. And the magnitude of displacement is always less than or equal to the distance traveled, never greater. So, students, please remember that both these statements are false.

Now, let us move on to Section 7.1.2, which is about Uniform Motion and Non-Uniform Motion.

Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second, and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion. The time interval in this motion should be small. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example, when a car is moving on a crowded street or a person is jogging in a park. These are some instances of non-uniform motion.

So, students, the key difference is: in uniform motion, the object covers equal distances in equal intervals of time. In non-uniform motion, the object covers unequal distances in equal intervals of time.

Now, let us do Activity 7.5. The data regarding the motion of two different objects A and B are given in Table 7.1. Examine them carefully and state whether the motion of the objects is uniform or non-uniform.

Looking at the table, we can see that object A travels 10 m in the first 15 minutes, then another 10 m in the next 15 minutes, and so on. The distance traveled in each 15-minute interval is always 10 m. So, object A is in uniform motion.

For object B, the distances traveled in each 15-minute interval are: 12 m, then 7 m, then 4 m, then 12 m, then 2 m, then 4 m, then 3 m. These are not equal. So, object B is in non-uniform motion.

Now, let us move on to Section 7.2, which is about Measuring the Rate of Motion.

Look at the situations given in the figures in your textbook. If the bowling speed is 143 km per hour, what does it mean? What do you understand from the signboard that shows speed limits? Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly. The rate at which objects move can be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol m s⁻¹ or m/s. The other units of speed include centimetre per second and kilometre per hour. To specify the speed of an object, we require only its magnitude. The speed of an object need not be constant. In most cases, objects will be in non-uniform motion. Therefore, we describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is, average speed equals total distance travelled divided by total time taken.

If an object travels a distance s in time t, then its speed v is v equals s divided by t. This is equation 7.1 in your textbook.

Let us understand this by an example. A car travels a distance of 100 km in 2 hours. Its average speed is 50 km per hour. The car might not have travelled at 50 km per hour all the time. Sometimes it might have travelled faster and sometimes slower than this.

Now, let me give you another example. Suppose an object travels 32 m in 6 s. What is its average speed? We calculate it as total distance divided by total time, which is 32 m divided by 6 s, which equals 5.33 m/s. So, the average speed of the object is 5.33 m/s.

Now, students, let me ask you to do Activity 7.6. Measure the time it takes you to walk from your house to your bus stop or the school. If you consider that your average walking speed is 4 km/h, estimate the distance of the bus stop or school from your house. This is a practical activity that will help you understand how to calculate distance if you know speed and time.

Now, Activity 7.7 is very interesting. At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning. Can you answer why this happens? This is because light travels much faster than sound. So, we see the lightning first, and then after some time, we hear the thunder. You can measure this time interval using a digital wrist watch or a stop watch. Then, you can calculate the distance of the nearest point of lightning using the formula: distance equals speed of sound multiplied by time. The speed of sound in air is approximately 346 m/s. So, if you hear the thunder 3 seconds after seeing the lightning, the distance would be 346 times 3, which is 1038 m, or about 1 km. This is a great way to estimate how far away the lightning is, students!

Now, let us discuss speed with direction. The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity. Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable. It can be changed by changing the object's speed, direction of motion, or both. When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as we calculate average speed.

In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time. That is, average velocity equals initial velocity plus final velocity divided by 2. Mathematically, v_average equals (u + v) divided by 2, where v_average is the average velocity, u is the initial velocity, and v is the final velocity of the object. This is equation 7.2.

Speed and velocity have the same units, that is, m/s or m s⁻¹.

Now, let us look at some examples to understand these concepts better.

Example 7.1: An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?

Solution: Total distance travelled by the object is 16 m plus 16 m, which equals 32 m. Total time taken is 4 s plus 2 s, which equals 6 s. Average speed equals total distance travelled divided by total time taken, which is 32 m divided by 6 s, which equals 5.33 m/s. Therefore, the average speed of the object is 5.33 m/s.

Example 7.2: The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km/h and m/s.

Solution: Distance covered by the car, s equals 2400 km minus 2000 km, which equals 400 km. Time elapsed, t equals 8 h. Average speed of the car is v_average equals s divided by t, which is 400 km divided by 8 h, which equals 50 km/h.

Now, to convert this to m/s, we use the conversion: 1 km equals 1000 m, and 1 h equals 3600 s. So, 50 km/h equals 50 multiplied by 1000 m divided by 3600 s, which equals 13.9 m/s. So, the average speed of the car is 50 km/h or 13.9 m/s.

Example 7.3: Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.

Solution: Total distance covered by Usha in 1 min is 180 m. Displacement of Usha in 1 min is 0 m because she starts from one end and comes back to the same end. So, she ends up at the starting point.

Average speed equals total distance covered divided by total time taken, which is 180 m divided by 1 min. But we need to convert 1 min to seconds, which is 60 s. So, average speed equals 180 m divided by 60 s, which equals 3 m/s.

Average velocity equals displacement divided by total time taken, which is 0 m divided by 60 s, which equals 0 m/s.

So, the average speed of Usha is 3 m/s and her average velocity is 0 m/s. This is a very important example, students. It shows that average speed and average velocity can be different. In this case, the average velocity is zero because Usha returns to her starting point.

Now, let me give you some questions to answer.

Question 1: Distinguish between speed and velocity.

The main difference between speed and velocity is that speed is a scalar quantity (it has only magnitude), while velocity is a vector quantity (it has both magnitude and direction). Speed tells us how fast an object is moving, but not in which direction. Velocity tells us how fast an object is moving and in which direction.

Question 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

The magnitude of average velocity equals average speed when the object is moving in a straight line in the same direction without turning back. In other words, when the displacement is equal to the distance traveled, the magnitude of average velocity equals average speed.

Question 3: What does the odometer of an automobile measure?

The odometer measures the total distance traveled by the automobile. It does not measure displacement or direction.

Question 4: What does the path of an object look like when it is in uniform motion?

When an object is in uniform motion along a straight line, its path is a straight line. If it is in uniform circular motion, its path is a circle.

Question 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m/s.

Solution: Time taken is 5 minutes, which equals 300 seconds. Speed of light is 3 × 10⁸ m/s. Distance equals speed multiplied by time, which is 3 × 10⁸ m/s multiplied by 300 s, which equals 9 × 10¹⁰ m, or 9 × 10⁷ km. So, the spaceship was 9 × 10⁷ km away from the ground station.

Now, students, let us move on to Section 7.3, which is about Rate of Change of Velocity.

During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity of the object for any time interval is zero. However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity of the object during any time interval is not zero. Can we now express the change in velocity of an object?

To answer such a question, we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is, acceleration equals change in velocity divided by time taken.

If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is a equals (v - u) divided by t. This is equation 7.3 in your textbook.

This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m/s², or m s⁻².

If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. The motion of a freely falling body is an example of uniformly accelerated motion. On the other hand, an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration.

Now, let us look at Example 7.4.

Example 7.4: Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m/s in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5 s. Calculate the acceleration of the bicycle in both the cases.

Solution: In the first case, initial velocity, u equals 0; final velocity, v equals 6 m/s; time, t equals 30 s.

From the equation a equals (v - u) divided by t, we have:

a equals (6 m/s - 0 m/s) divided by 30 s, which equals 0.2 m/s².

In the second case: initial velocity, u equals 6 m/s; final velocity, v equals 4 m/s; time, t equals 5 s.

Then, a equals (4 m/s - 6 m/s) divided by 5 s, which equals -0.4 m/s².

The acceleration of the bicycle in the first case is 0.2 m/s² and in the second case, it is -0.4 m/s². The negative sign indicates that the acceleration is in the opposite direction to the motion, which is what happens when we apply brakes.

Now, let us do Activity 7.8. In your everyday life you come across a range of motions in which (a) acceleration is in the direction of motion, (b) acceleration is against the direction of motion, (c) acceleration is uniform, (d) acceleration is non-uniform. Can you identify one example each for the above type of motion?

Let me help you with this, students:

(a) Acceleration in the direction of motion: When a car accelerates from rest at a traffic signal, the acceleration is in the direction of motion.

(b) Acceleration against the direction of motion: When a car applies brakes, the acceleration is opposite to the direction of motion. This is also called deceleration or retardation.

(c) Uniform acceleration: A freely falling body has uniform acceleration. The acceleration due to gravity is approximately 9.8 m/s² and it is constant.

(d) Non-uniform acceleration: A car moving in city traffic has non-uniform acceleration because it speeds up and slows down frequently.

Now, let us answer some questions from this section.

Question 1: When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

(i) A body is in uniform acceleration when its velocity changes by equal amounts in equal intervals of time. (ii) A body is in non-uniform acceleration when its velocity changes by unequal amounts in equal intervals of time.

Question 2: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Solution: First, we need to convert the speeds to m/s. 80 km/h equals 80 multiplied by 1000 divided by 3600, which is 22.22 m/s. Similarly, 60 km/h equals 16.67 m/s. The change in velocity is 16.67 - 22.22, which is -5.55 m/s. Time is 5 s. So, acceleration equals change in velocity divided by time, which is -5.55 divided by 5, which is -1.11 m/s². So, the acceleration is -1.11 m/s². The negative sign indicates that the bus is decelerating.

Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes. Find its acceleration.

Solution: Initial velocity, u equals 0. Final velocity, v equals 40 km/h. We need to convert this to m/s: 40 km/h equals 40 multiplied by 1000 divided by 3600, which is 11.11 m/s. Time, t equals 10 minutes, which is 600 seconds. Acceleration equals (v - u) divided by t, which is (11.11 - 0) divided by 600, which is 0.0185 m/s². So, the acceleration of the train is 0.0185 m/s².

Now, students, let us move on to Section 7.4, which is about Graphical Representation of Motion.

Graphs provide a convenient method to present basic information about a variety of events. For example, in the telecast of a one-day cricket match, vertical bar graphs show the run rate of a team in each over. As you have studied in mathematics, a straight line graph helps in solving a linear equation having two variables.

To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity, such as distance or velocity, on another quantity, such as time.

Now, let us discuss distance-time graphs. The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice. In this graph, time is taken along the x-axis and distance is taken along the y-axis. Distance-time graphs can be employed under various conditions where objects move with uniform speed, non-uniform speed, remain at rest, etc.

We know that when an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, a graph of distance travelled against time is a straight line. The portion of the graph shows that the distance is increasing at a uniform rate. Note that, you can also use the term uniform velocity in place of uniform speed if you take the magnitude of displacement equal to the distance travelled by the object along the y-axis.

We can use the distance-time graph to determine the speed of an object. To do so, consider a small part of the distance-time graph. Draw a line parallel to the x-axis from one point and another line parallel to the y-axis from another point. These two lines meet each other to form a triangle. Now, on the graph, one side denotes the time interval while the other side corresponds to the distance. We can see from the graph that as the object moves from one point to another, it covers a certain distance in a certain time. The speed, v of the object, therefore can be represented as v equals (s₂ - s₁) divided by (t₂ - t₁). This is equation 7.4.

We can also plot the distance-time graph for accelerated motion. Table 7.2 shows the distance travelled by a car in a time interval of two seconds. Looking at the table, we can see that the distance increases as: 0, 1, 4, 9, 16, 25, 36 metres at times 0, 2, 4, 6, 8, 10, 12 seconds respectively. This follows a pattern: distance is proportional to the square of time. The distance-time graph for this motion will be a curve, not a straight line. The nature of this graph shows non-linear variation of the distance travelled by the car with time. Thus, the graph represents motion with non-uniform speed.

Now, let us discuss velocity-time graphs. The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis and the velocity is represented along the y-axis. If the object moves at uniform velocity, the height of its velocity-time graph will not change with time. It will be a straight line parallel to the x-axis.

We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement.

To know the distance moved by the car between two times using the velocity-time graph, we can draw perpendiculars from the points corresponding to those times on the graph. The distance moved can be expressed as the area of the rectangle formed.

We can also study about uniformly accelerated motion by plotting its velocity-time graph. Consider a car being driven along a straight road for testing its engine. Suppose a person sitting next to the driver records its velocity after every 5 seconds by noting the reading of the speedometer of the car. The velocity of the car at different instants of time is shown in table 7.3. Looking at the table, we can see that the velocity increases by equal amounts (2.5 m/s) in equal time intervals (5 s). So, this is uniformly accelerated motion.

In this case, the velocity-time graph for the motion of the car is a straight line. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line.

There are also velocity-time graphs for non-uniformly accelerated motion. One shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time, while another shows the velocity-time graph representing the non-uniform variation of velocity of the object with time.

You can also determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area of a rectangle under the graph. Since the magnitude of the velocity of the car is changing due to acceleration, the distance travelled by the car will be given by the area under the velocity-time graph, which in the case of uniformly accelerated motion can be calculated as the area of a rectangle plus the area of a triangle.

That is, s equals area of rectangle plus area of triangle, which equals base times height plus half times base times height.

In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape.

Now, let us do Activity 7.9. The times of arrival and departure of a train at three stations A, B, and C and the distance of stations B and C from station A are given in Table 7.4.

Station A is at 0 km, and the train arrives at 08:00 and departs at 08:15. Station B is at 120 km, and the train arrives at 11:15 and departs at 11:30. Station C is at 180 km, and the train arrives at 13:00 and departs at 13:15.

Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.

Students, to solve this activity, you need to plot the distance on the y-axis and time on the x-axis. The train starts from station A at 08:00 at 0 km. It reaches station B at 11:15 at 120 km, and reaches station C at 13:00 at 180 km. Between these points, the graph will be straight lines if we assume uniform motion. The slope of each line will give the speed of the train between those stations.

Now, Activity 7.10. Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table 7.5 shows the distance travelled by them in different times.

Time: 8:00 am - both at 0 km 8:05 am - Feroz: 1.0 km, Sania: 0.8 km 8:10 am - Feroz: 1.9 km, Sania: 1.6 km 8:15 am - Feroz: 2.8 km, Sania: 2.3 km 8:20 am - Feroz: 3.6 km, Sania: 3.0 km 8:25 am - Sania: 3.6 km (Feroz's data not given, maybe he reached earlier)

Plot the distance-time graph for their motions on the same scale and interpret.

Students, when you plot this graph, you will see that Feroz's graph has a steeper slope than Sania's graph. This means Feroz is cycling faster than Sania. The graph also shows that Feroz reaches school earlier than Sania.

Now, let us answer some questions from this section.

Question 1: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

For uniform motion, the distance-time graph is a straight line. For non-uniform motion, the distance-time graph is a curved line (not a straight line).

Question 2: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

If the distance-time graph is a straight line parallel to the time axis, it means the object is at rest. There is no change in distance with time.

Question 3: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

If the speed-time graph is a straight line parallel to the time axis, it means the object is moving with constant speed (uniform motion).

Question 4: What is the quantity which is measured by the area occupied below the velocity-time graph?

The area below the velocity-time graph gives the displacement of the object.

Now, students, let us move on to Section 7.5, which is about Equations of Motion.

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion, and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below:

First equation: v equals u plus at. This is equation 7.5.

Second equation: s equals ut plus half at squared. This is equation 7.6.

Third equation: 2as equals v squared minus u squared. This is equation 7.7.

where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. The first equation describes the velocity-time relation and the second equation represents the position-time relation. The third equation, which represents the relation between the position and the velocity, can be obtained from the first and second equations by eliminating t. These three equations can be derived by graphical method.

Now, let us look at some examples to understand how to use these equations.

Example 7.5: A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.

Solution: We have been given u equals 0; v equals 72 km/h equals 20 m/s (because 72 multiplied by 1000 divided by 3600 equals 20); and t equals 5 minutes equals 300 s.

(i) From the first equation, we know that a equals (v - u) divided by t. So, a equals (20 m/s - 0 m/s) divided by 300 s, which equals 1/15 m/s², which is approximately 0.067 m/s².

(ii) From the third equation, we have 2as equals v² - u², which equals v² - 0. Thus, s equals v² divided by 2a. So, s equals (20 m/s)² divided by (2 × (1/15) m/s²), which equals 400 divided by (2/15), which equals 400 × 15/2, which equals 3000 m, which equals 3 km.

So, the acceleration of the train is 1/15 m/s² and the distance travelled is 3 km.

Example 7.6: A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Solution: We are given that u equals 18 km/h equals 5 m/s (because 18 × 1000/3600 = 5); v equals 36 km/h equals 10 m/s; and t equals 5 s.

(i) From the first equation, we have a equals (v - u) divided by t. So, a equals (10 m/s - 5 m/s) divided by 5 s, which equals 1 m/s².

(ii) From the second equation, we have s equals ut plus half at². So, s equals 5 m/s × 5 s plus half × 1 m/s² × (5 s)². This equals 25 m plus half × 1 × 25, which equals 25 m plus 12.5 m, which equals 37.5 m.

The acceleration of the car is 1 m/s² and the distance covered is 37.5 m.

Example 7.7: The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.

Solution: We have been given a equals -6 m/s²; t equals 2 s; and v equals 0 m/s.

From the first equation, we know that v equals u plus at. So, 0 equals u plus (-6 m/s²) × 2 s. Therefore, u equals 12 m/s.

From the second equation, we get s equals ut plus half at². So, s equals (12 m/s) × (2 s) plus half × (-6 m/s²) × (2 s)². This equals 24 m plus half × (-6) × 4, which equals 24 m plus (-12 m), which equals 12 m.

Thus, the car will move 12 m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road? This is a very practical example, students. If you are driving at 12 m/s (which is about 43 km/h), and you need 12 m to stop, you should maintain at least 12 m distance from the vehicle in front of you. But in real life, you need even more distance because of reaction time and other factors.

Now, let us answer some questions from this section.

Question 1: A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Solution: u equals 0; a equals 0.1 m/s²; t equals 2 minutes equals 120 s.

(a) Using v equals u plus at, we get v equals 0 plus 0.1 × 120 equals 12 m/s.

(b) Using s equals ut plus half at², we get s equals 0 × 120 plus half × 0.1 × (120)² equals 0.05 × 14400 equals 720 m.

So, the speed acquired is 12 m/s and the distance travelled is 720 m.

Question 2: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s². Find how far the train will go before it is brought to rest.

Solution: u equals 90 km/h equals 25 m/s (90 × 1000/3600 = 25); v equals 0; a equals -0.5 m/s².

Using v² equals u² plus 2as, we get 0² equals 25² plus 2 × (-0.5) × s. So, 0 equals 625 minus s. Therefore, s equals 625 m.

So, the train will travel 625 m before coming to rest.

Question 3: A trolley, while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity 3 s after the start?

Solution: u equals 0; a equals 2 cm/s² equals 0.02 m/s²; t equals 3 s.

Using v equals u plus at, we get v equals 0 plus 0.02 × 3 equals 0.06 m/s.

So, the velocity after 3 seconds will be 0.06 m/s.

Question 4: A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?

Solution: u equals 0; a equals 4 m/s²; t equals 10 s.

Using s equals ut plus half at², we get s equals 0 × 10 plus half × 4 × 10² equals 2 × 100 equals 200 m.

So, the car will cover 200 m in 10 seconds.

Question 5: A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Solution: This is a case of vertical motion. The stone is thrown upward with velocity u equals 5 m/s. The acceleration is downward, so a equals -10 m/s². At the highest point, the final velocity v equals 0.

Using v² equals u² plus 2as, we get 0² equals 5² plus 2 × (-10) × s. So, 0 equals 25 minus 20s. Therefore, s equals 25/20 equals 1.25 m.

So, the height attained is 1.25 m.

Now, to find the time, using v equals u plus at, we get 0 equals 5 plus (-10) × t. So, t equals 5/10 equals 0.5 s.

So, the stone will take 0.5 s to reach the highest point.

Now, students, let us move on to Section 7.6, which is about Uniform Circular Motion.

When the velocity of an object changes, we say that the object is accelerating. The change in the velocity could be due to change in its magnitude or the direction of the motion or both. Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion?

Let me give you an example. Consider an athlete running around a circular track. The speed of the athlete may be constant, but the direction of motion is constantly changing. Therefore, the velocity is changing, which means the athlete is accelerating. This is a very important concept, students. Even if the speed is constant, if the direction changes, the velocity changes, and that is acceleration.

In order to keep himself on track, he quickly changes his speed at the corners. How many times will the athlete have to change his direction of motion, while he completes one round? It is clear that to move in a rectangular track once, he has to change his direction of motion four times.

Now, suppose instead of a rectangular track, the athlete is running along a hexagonal shaped path. In this situation, the athlete will have to change his direction six times while he completes one round. What if the track was not a hexagon but a regular octagon, with eight equal sides? It is observed that as the number of sides of the track increases, the athlete has to take turns more and more often. What would happen to the shape of the track as we go on increasing the number of sides indefinitely? If you do this, you will notice that the shape of the track approaches the shape of a circle and the length of each of the sides will decrease to a point. If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion.

We know that the circumference of a circle of radius r is given by 2πr. If the athlete takes t seconds to go once around the circular path of radius r, the speed v is given by v equals 2πr divided by t. This is equation 7.8.

When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

Now, let us do Activity 7.11. Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end. Now, let the stone go by releasing the thread. Can you tell the direction in which the stone moves after it is released? By repeating the activity for a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not.

If you carefully note, on being released, the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant. This shows that the direction of motion changed at every point when the stone was moving along the circular path.

This is a very interesting observation, students. When you whirl a stone tied to a string and then release it, the stone flies off in the direction of the tangent to the circle at the point of release. This is because of inertia - the stone tends to keep moving in the direction it was moving at the moment of release.

When an athlete throws a hammer or a discus in a sports meet, he/she holds the hammer or the discus in his/her hand and gives it a circular motion by rotating his/her own body. Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released, just like the piece of stone in the activity described above. There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at constant speed, and so on.

Now, students, we have covered all the sections of the chapter. Let me summarize what we have learned in this chapter.

What you have learnt: - Motion is a change of position; it can be described in terms of the distance moved or the displacement. - The motion of an object could be uniform or non-uniform depending on whether its velocity is constant or changing. - The speed of an object is the distance covered per unit time, and velocity is the displacement per unit time. - The acceleration of an object is the change in velocity per unit time. - Uniform and non-uniform motions of objects can be shown through graphs. - The motion of an object moving at uniform acceleration can be described with the help of the following equations, namely v equals u plus at, s equals ut plus half at squared, and 2as equals v squared minus u squared, where u is initial velocity of the object, which moves with uniform acceleration a for time t, v is its final velocity, and s is the distance it travelled in time t. - If an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

Now, students, let us solve the exercises at the end of the chapter.

Exercise 1: An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Solution: Diameter of the track is 200 m, so the radius is 100 m. The circumference of the track is 2πr, which equals 2π × 100, which equals 200π m, which is approximately 628 m.

The athlete completes one round in 40 s. So, the time for one round is 40 s.

Now, 2 minutes 20 s equals 140 s. In 140 s, the number of rounds completed is 140 divided by 40, which is 3.5 rounds.

So, the distance covered is 3.5 times the circumference, which is 3.5 × 200π, which equals 700π m, which is approximately 2199 m or about 2.2 km.

Now, for displacement: after 3 complete rounds, the athlete returns to the starting point, so displacement is 0. After half a round, the athlete is at the diametrically opposite point. So, the displacement is the diameter, which is 200 m.

So, at the end of 2 minutes 20 s, the distance covered is approximately 2199 m and the displacement is 200 m.

Exercise 2: Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Solution: (a) From A to B: Distance is 300 m. Time is 2 minutes 30 seconds, which is 150 s. So, average speed is 300 divided by 150, which is 2 m/s. Since he is moving in a straight line from A to B, the displacement is also 300 m. So, average velocity is also 2 m/s.

(b) From A to C: Total distance is 300 m + 100 m = 400 m. Total time is 150 s + 60 s = 210 s. So, average speed is 400 divided by 210, which is approximately 1.9 m/s. The displacement from A to C is 300 m - 100 m = 200 m (since C is 100 m back from B). So, average velocity is 200 divided by 210, which is approximately 0.95 m/s.

Exercise 3: Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul's trip?

Solution: Let the distance from home to school be d km. Time taken to go to school is d/20 hours. Time taken to return is d/30 hours. Total distance is 2d km. Total time is d/20 + d/30, which equals (3d + 2d)/60, which equals 5d/60, which equals d/12 hours.

Average speed equals total distance divided by total time, which is 2d divided by (d/12), which equals 2d × 12/d, which equals 24 km/h.

So, the average speed for Abdul's trip is 24 km/h.

Exercise 4: A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 s. How far does the boat travel during this time?

Solution: u equals 0; a equals 3.0 m/s²; t equals 8.0 s.

Using s equals ut plus half at², we get s equals 0 × 8 plus half × 3 × 8² equals 1.5 × 64 equals 96 m.

So, the boat travels 96 m during this time.

Exercise 5: A driver of a car travelling at 52 km/h applies the brakes. (a) Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

This question refers to a graph in the textbook. Without the actual graph, I can tell you that: (a) The area under the velocity-time graph represents the distance travelled. (b) If there is a horizontal line on the graph (parallel to the time axis), that part represents uniform motion.

Exercise 6: Figure 7.10 shows the distance-time graph of three objects A, B, and C. Study the graph and answer the following questions: (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

Without the actual graph, I can tell you the general approach: (a) The object with the steepest slope (most steep line) is travelling the fastest. (b) They are at the same point when their lines intersect. (c) and (d) can be answered by reading the graph at the appropriate points.

Exercise 7: A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? After what time will it strike the ground?

Solution: u equals 0; a equals 10 m/s²; s equals 20 m.

Using v² equals u² plus 2as, we get v² equals 0² plus 2 × 10 × 20 equals 400. So, v equals 20 m/s.

Using v equals u plus at, we get 20 equals 0 plus 10 × t. So, t equals 2 s.

So, the ball will strike the ground with a velocity of 20 m/s after 2 seconds.

Exercise 8: The speed-time graph for a car is shown in Figure 7.11. (a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

Again, without the actual graph, the general approach is: (a) The distance travelled in the first 4 seconds is the area under the speed-time graph from 0 to 4 seconds. (b) The horizontal part of the graph (parallel to the time axis) represents uniform motion.

Exercise 9: State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity, (b) an object moving with an acceleration but with uniform speed, (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) Yes, this is possible. For example, when a ball is thrown straight up, at the highest point, its velocity is zero, but the acceleration is still 10 m/s² downward (due to gravity).

(b) Yes, this is possible. Uniform circular motion is an example. The speed is uniform (constant), but the velocity is changing because the direction is changing, so there is acceleration.

(c) Yes, this is possible. This is exactly what happens in uniform circular motion. The object moves in a certain direction (tangential), and the acceleration is towards the center of the circle, which is perpendicular to the direction of motion.

Exercise 10: An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Solution: Radius, r equals 42250 km. Time period, t equals 24 hours.

The circumference of the orbit is 2πr, which equals 2π × 42250 km, which equals 265400 km (approximately).

Speed equals distance divided by time, which is 265400 km divided by 24 h, which equals 11058 km/h, which is approximately 3.07 km/s.

So, the satellite moves at about 3.07 km/s.

Now, students, we have covered the entire chapter. Let me do a quick recap of everything we have learned:

1. Motion is the change in position of an object with time. We describe motion using reference points.

2. Distance is the total path length covered, while displacement is the shortest distance from initial to final position. Distance is a scalar, displacement is a vector.

3. Uniform motion is when an object covers equal distances in equal time intervals. Non-uniform motion is when distances are unequal in equal time intervals.

4. Speed is distance covered per unit time. Velocity is displacement per unit time. Average speed = total distance/total time. Average velocity = total displacement/total time.

5. Acceleration is the rate of change of velocity. a = (v - u)/t. Positive acceleration is in the direction of motion, negative acceleration (deceleration) is opposite to the direction of motion.

6. Distance-time graphs: straight line for uniform motion, curved line for non-uniform motion. The slope gives speed.

7. Velocity-time graphs: straight line parallel to time axis for uniform velocity, straight line with slope for uniform acceleration. The area under the graph gives displacement.

8. Equations of motion for uniform acceleration: - v = u + at - s = ut + ½at² - v² = u² + 2as

9. Uniform circular motion: when an object moves in a circle with constant speed. Even though speed is constant, velocity changes because direction changes, so there is acceleration (centripetal acceleration).

I hope this lesson has helped you understand the chapter on Motion thoroughly. Remember, physics is all around us, and understanding motion will help you understand many phenomena in everyday life. Keep practicing the numerical problems and graphs, and you will master this chapter. Thank you for your attention, students. Keep learning, keep exploring!