Motion

Welcome dear students! Today we are going to learn about Motion from Class 9 Science. In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins and arteries, and cars move. 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 infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees. 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? An object may appear to be moving for one person and stationary for some other. For the passengers in a moving bus, the roadside trees appear to be moving backwards. A person standing on the road-side perceives the bus alongwith the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. What do these observations indicate? 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. [CHECKPOINT]

Activity 7.1 asks you to discuss whether the walls of your classroom are at rest or in motion. Activity 7.2 asks if you have ever experienced that the train in which you are sitting appears to move while it is at rest. Discuss and share your experience. Think and Act: 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? Section 7.1 is Describing Motion. 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. [CHECKPOINT]

Section 7.1.1 covers Motion along a straight line. The simplest type of motion is the motion along a straight line. Consider an object moving along a straight path. The object starts its journey from O which is treated as its reference point, as shown in Figure 7.1. Let A, B and C represent the position of the object at different instants. The figure shows a straight line marked with distances in kilometres from zero to sixty. Points O, C, B and A are marked at zero, twenty-five, thirty-five and sixty kilometres respectively. 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. The total path length covered by the object is OA plus AC, that is 60 km plus 35 km 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. 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. [CHECKPOINT]

Can the magnitude of the displacement be equal to the distance travelled by an object? Consider the example given in Figure 7.1. For motion of the object from O to A, the distance covered is 60 km and the magnitude of displacement is also 60 km. During its motion from O to A and back to B, the distance covered equals 60 km plus 25 km equals 85 km while the magnitude of displacement equals 35 km. Thus, the magnitude of displacement is not equal to the path length. Further, we will notice that the magnitude of the 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 equals 60 km plus 60 km 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. Activity 7.3 instructs you to 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, and note the difference between the two. [CHECKPOINT]

Activity 7.4 states that automobiles are fitted with a device that shows the distance travelled, 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. Consider the questions following this section. Question one: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Question two: 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? Question three: Which of the following is true for displacement? Option a says it cannot be zero. Option b says its magnitude is greater than the distance travelled by the object. Section 7.1.2 covers 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. [CHECKPOINT]

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. Activity 7.5 provides data regarding the motion of two different objects A and B in Table 7.1. The table lists time and distance travelled by object A and object B in metres. At 9:30 am, object A travels 10 m and object B travels 12 m. At 9:45 am, A travels 20 m and B travels 19 m. At 10:00 am, A travels 30 m and B travels 23 m. At 10:15 am, A travels 40 m and B travels 35 m. At 10:30 am, A travels 50 m and B travels 37 m. At 10:45 am, A travels 60 m and B travels 41 m. At 11:00 am, A travels 70 m and B travels 44 m. Examine them carefully and state whether the motion of the objects is uniform or non-uniform. Section 7.2 is Measuring the Rate of Motion. Look at the situations given in Figure 7.2. Part a shows a bowling speed display showing speeds: 140, 143, 135, 145, 132, 130 in km/h. Part b shows a speed limit sign showing 50 km/h for cars and 40 km/h for trucks. If the bowling speed is 143 km/h, it means the ball is travelling at that rate. The signboard indicates the maximum allowed speed for different vehicles. [CHECKPOINT]

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 (cm s^-1) and kilometre per hour (km h^-1). 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 = s/t. Let us understand this by 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? [CHECKPOINT]

Solution: Total distance travelled by the object equals 16 m plus 16 m equals 32 m. Total time taken equals 4 s plus 2 s equals 6 s. Average speed equals Total distance travelled divided by Total time taken, which is 32 m divided by 6 s equals 5.33 m s^-1. Therefore, the average speed of the object is 5.33 m s^-1. Section 7.2.1 is 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_av = (u + v) / 2, where v_av is the average velocity, u is the initial velocity and v is the final velocity of the object. Speed and velocity have the same units, that is, m s^-1 or m/s. [CHECKPOINT]

Activity 7.6 asks you to 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^-1, estimate the distance of the bus stop or school from your house. Activity 7.7 states that 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? Measure this time interval using a digital wrist watch or a stop watch. Calculate the distance of the nearest point of lightning. Speed of sound in air equals 346 m s^-1. Consider the questions for this section. Question one: Distinguish between speed and velocity. Question two: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Question three: What does the odometer of an automobile measure? Question four: What does the path of an object look like when it is in uniform motion? Question five: 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 × 10^8 m s^-1. Let us solve 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^-1 and m s^-1. [CHECKPOINT]

Solution: Distance covered by the car, s equals 2400 km minus 2000 km equals 400 km. Time elapsed, t equals 8 h. Average speed of the car is, v_av = s/t = 400 km / 8 h = 50 km h^-1. Converting to m/s: 50 km/h multiplied by 1000 m divided by 1 km multiplied by 1 h divided by 3600 s equals 13.9 m s^-1. The average speed of the car is 50 km h^-1 or 13.9 m s^-1. 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 equals 0 m. Average speed equals Total distance covered divided by Total time taken equals 180 m divided by 1 min. Converting to seconds: 180 m divided by 60 s equals 3 m s^-1. Average velocity equals Displacement divided by Total time taken equals 0 m divided by 60 s equals 0 m s^-1. The average speed of Usha is 3 m s^-1 and her average velocity is 0 m s^-1. Section 7.3 is 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? [CHECKPOINT]

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 = (v - u) / t. 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 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. Activity 7.8 asks you to identify examples from your everyday life where acceleration is in the direction of motion, against the direction of motion, uniform, and non-uniform. [CHECKPOINT]

Example 7.4: Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s^-1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s^-1 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^-1; time, t equals 30 s. From the acceleration formula, a = (v - u)/t. Substituting the given values, we get a equals 6 m s^-1 minus 0 m s^-1 divided by 30 s equals 0.2 m s^-2. In the second case: initial velocity, u equals 6 m s^-1; final velocity, v equals 4 m s^-1; time, t equals 5 s. Then, a equals 4 m s^-1 minus 6 m s^-1 divided by 5 s equals -0.4 m s^-2. The acceleration of the bicycle in the first case is 0.2 m s^-2 and in the second case, it is -0.4 m s^-2. Questions for this section: Question one: When will you say a body is in uniform acceleration and non-uniform acceleration? Question two: A bus decreases its speed from 80 km h^-1 to 60 km h^-1 in 5 s. Find the acceleration of the bus. Question three: A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^-1 in 10 minutes. Find its acceleration. Section 7.4 is Graphical Representation of Motion. Graphs provide a convenient method to present basic information about a variety of events. To describe the motion of an object, we can use line graphs showing dependence of one physical quantity, such as distance or velocity, on another quantity, such as time. [CHECKPOINT]

Subsection 7.4.1 is 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. In Figure 7.3, we see a distance-time graph of an object moving with uniform speed. It is a straight line starting from the origin and going upwards to the right. 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 OB of the graph shows that the distance is increasing at a uniform rate. We can use the distance-time graph to determine the speed of an object. To do so, consider a small part AB of the distance-time graph shown in Figure 7.3. Draw a line parallel to the x-axis from point A and another line parallel to the y-axis from point B. These two lines meet each other at point C to form a triangle ABC. Now, on the graph, AC denotes the time interval t2 minus t1 while BC corresponds to the distance s2 minus s1. We can see from the graph that as the object moves from the point A to B, it covers a distance s2 minus s1 in time t2 minus t1. The speed, v of the object, therefore can be represented as v = (s2 - s1) / (t2 - t1). [CHECKPOINT]

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. At 0 seconds, distance is 0 m. At 2 s, 1 m. At 4 s, 4 m. At 6 s, 9 m. At 8 s, 16 m. At 10 s, 25 m. At 12 s, 36 m. Figure 7.4 shows the distance-time graph for a car moving with non-uniform speed. The curve is non-linear, showing that distance increases non-linearly with time. Thus, the graph shown in Figure 7.4 represents motion with non-uniform speed. Subsection 7.4.2 is 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. Figure 7.5 shows the velocity-time graph for a car moving with uniform velocity of 40 km h^-1. 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. [CHECKPOINT]

To know the distance moved by the car between time t1 and t2 using Figure 7.5, draw perpendiculars from the points corresponding to the time t1 and t2 on the graph. The velocity of 40 km h^-1 is represented by the height AC or BD and the time t2 minus t1 is represented by the length AB. So, the distance s moved by the car in time t2 minus t1 can be expressed as s = AC × CD = [(40 km h^-1) × (t2 - t1) h] = 40 (t2 - t1) km = area of the rectangle ABDC. We can also study about uniformly accelerated motion by plotting its velocity-time graph. Table 7.3 shows the velocity of a car at regular instants of time. At 0 s, velocity is 0 m/s. At 5 s, 2.5 m/s. At 10 s, 5.0 m/s. At 15 s, 7.5 m/s. At 20 s, 10.0 m/s. At 25 s, 12.5 m/s. At 30 s, 15.0 m/s. In this case, the velocity-time graph for the motion of the car is shown in Figure 7.6. It is a straight line sloping upwards from the origin. 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. 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 moved by the car in a given interval of time. [CHECKPOINT]

If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph. Since the magnitude of the velocity of the car is changing due to acceleration, the distance s travelled by the car will be given by the area ABCDE under the velocity-time graph. That is, s = area ABCDE = area of the rectangle ABCD + area of the triangle ADE = AB × BC + 1/2 (AD × DE). In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape. Figure 7.7 shows two graphs. Part a shows a curve sloping downwards, representing decreasing velocity. Part b shows a curved line, representing non-uniform variation of velocity. Try to interpret these graphs. Activity 7.9 gives 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 in Table 7.4. Station A is at 0 km, arrives at 08:00, departs at 08:15. Station B is at 120 km, arrives at 11:15, departs at 11:30. Station C is at 180 km, arrives at 13:00, departs at 13:15. Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform. [CHECKPOINT]

Activity 7.10 involves Feroz and Sania going to school on bicycles. Table 7.5 shows the distance covered by them. At 8:00 am, both are at 0 km. At 8:05 am, Feroz is at 1.0 km, Sania at 0.8 km. At 8:10 am, Feroz at 1.9 km, Sania at 1.6 km. At 8:15 am, Feroz at 2.8 km, Sania at 2.3 km. At 8:20 am, Feroz at 3.6 km, Sania at 3.0 km. At 8:25 am, Feroz data is missing, Sania at 3.6 km. Plot the distance-time graph for their motions on the same scale and interpret. Questions for this section: Question one: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? Question two: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis? Question three: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis? Question four: What is the quantity which is measured by the area occupied below the velocity-time graph? Section 7.5 is 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. [CHECKPOINT]

For convenience, a set of three such equations are given below: v = u + at, s = ut + 1/2 at^2, and 2as = v^2 - u^2. Here, u is the initial velocity, a is uniform acceleration, t is time, v is final velocity, and s is distance. The first equation describes the velocity-time relation, the second represents the position-time relation, and the third represents the relation between position and velocity. Example 7.5: A train starting from rest attains a velocity of 72 km h^-1 in 5 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance travelled. Solution: u = 0; v = 72 km h^-1 = 20 m s^-1; t = 5 minutes = 300 s. From the first equation, a = (v - u) / t = (20 m s^-1 - 0 m s^-1) / 300 s = 1/15 m s^-2. From the third equation, 2as = v^2 - u^2 = v^2 - 0. Thus, s = v^2 / 2a = (20 m s^-1)^2 / [2 × (1/15) m s^-2] = 3000 m = 3 km. The acceleration is 1/15 m s^-2 and distance is 3 km. Example 7.6: A car accelerates uniformly from 18 km h^-1 to 36 km h^-1 in 5 s. Calculate acceleration and distance. Solution: u = 18 km h^-1 = 5 m s^-1; v = 36 km h^-1 = 10 m s^-1; t = 5 s. From the first equation, a = (10 - 5) / 5 = 1 m s^-2. From the second equation, s = ut + 1/2 at^2 = 5 × 5 + 1/2 × 1 × 25 = 25 + 12.5 = 37.5 m. [CHECKPOINT]

Example 7.7: The brakes applied to a car produce an acceleration of 6 m s^-2 in the opposite direction. If the car takes 2 s to stop, calculate distance. Solution: a = -6 m s^-2; t = 2 s; v = 0 m s^-1. From the first equation, 0 = u + (-6)(2), so u = 12 m s^-1. From the second equation, s = 12(2) + 1/2(-6)(4) = 24 - 12 = 12 m. Questions: Question one: A bus starting from rest moves with uniform acceleration of 0.1 m s^-2 for 2 minutes. Find speed acquired and distance travelled. Question two: A train at 90 km h^-1 brakes to produce uniform acceleration of -0.5 m s^-2. Find distance to rest. Question three: A trolley has acceleration of 2 cm s^-2. Find velocity after 3 s. Question four: A racing car has uniform acceleration of 4 m s^-2. What distance will it cover in 10 s? Question five: A stone is thrown vertically upward with velocity 5 m s^-1. Acceleration is 10 m s^-2 downward. Find height attained and time taken. Section 7.6 is Uniform Circular Motion. When velocity changes, the object accelerates due to change in magnitude, direction, or both. Figure 7.8 shows tracks: rectangular, hexagonal, octagonal, and circular. As the number of sides increases, the shape approaches a circle. If an athlete moves with constant speed along a circular path, velocity changes only in direction, making it accelerated motion. Circumference equals 2πr. If time taken is t, speed v = 2πr / t. When an object moves in a circular path with uniform speed, its motion is called uniform circular motion. [CHECKPOINT]

Activity 7.11: Take a piece of thread and tie a small stone at one end. Move the stone in a circular path with constant speed. Release the thread. The stone moves tangentially to the circular path, showing direction changes continuously. Examples include hammer throw, moon around earth, satellites, and cyclists on circular tracks. What you have learnt: Motion is a change of position; it can be described in terms of distance moved or displacement. Motion can be uniform or non-uniform. Speed is distance per unit time, velocity is displacement per unit time. Acceleration is change in velocity per unit time. Uniform and non-uniform motions are shown through graphs. Equations of motion for uniform acceleration are v = u + at, s = ut + 1/2 at^2, and 2as = v^2 - u^2. Uniform circular motion occurs when an object moves in a circular path with uniform speed. Exercises: Question one: An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and displacement at 2 min 20 s? Question two: Joseph jogs 300 m from A to B in 2 min 30 s, then 100 m back to C in 1 min. Find average speeds and velocities from A to B and A to C. Question three: Abdul's average speed to school is 20 km h^-1, return is 30 km h^-1. Find overall average speed. Question four: A motorboat accelerates from rest at 3.0 m s^-2 for 8.0 s. Find distance. Question five: A car travelling at 52 km h^-1 applies brakes and stops in 5 s. Draw speed-time graph, shade distance area, identify uniform motion part. [CHECKPOINT]

Question six: Figure 7.10 shows distance-time graph for objects A, B, C. Which travels fastest? Are they ever at same point? How far has C travelled when B passes A? How far has B travelled when it passes C? Question seven: A ball dropped from 20 m accelerates at 10 m s^-2. Find impact velocity and time. Question eight: Figure 7.11 shows speed-time graph curving upward to (6, 7) then horizontal to (10, 7). Find distance in first 4 s, shade area, identify uniform motion part. Question nine: State if possible with examples: constant acceleration with zero velocity, acceleration with uniform speed, acceleration perpendicular to motion. Question ten: Satellite in circular orbit of radius 42250 km takes 24 hours. Calculate speed. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]