CBSE Class 9 Science Chapter 8- Motion
- 1. Chapter-8 Motion By N.Aarthi, B.E, M.Tech SASTRA UNIVERSITY
- 2. Motion
- 3. Reference point and reference frame • To describe the position of an object we need a reference point or origin. An object may seem to be moving to one observer and stationary to another. • Example: A passenger inside a bus sees the other passengers be at rest, whereas an observer outside the bus sees the passengers are in motion. • In order to make observations easy, a convention or a common reference point or frame is needed. All objects must be in the same reference frame.
- 4. Distance and Displacement • The magnitude of the length covered by a moving object is called distance. It has no direction. • Displacement is the shortest distance between two points or the distance between the starting and final positions with respect to time. It has magnitude as well direction. • Displacement can be zero, but distance cannot.
- 6. What is Distance? • Distance is the total movement of an object without any regard to direction. • We can define distance as to how much ground an object has covered despite its starting or ending point.
- 7. Distance here will be = 4m + 3m + 5m = 12 m Distance Formula Δd=d1+d2
- 8. How is Displacement defined? • Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude. • It is represented as an arrow that points from the starting position to the final position. • For example- If an object moves from A position to B, then the object’s position changes. This change in position of an object is known as Displacement.
- 10. Displacement = Δx=xf−x0 xf = Final Position x0 = Initial Position Δx = Displacement
- 11. Examples of Distance and Displacement Question 1. John travels 250 miles to North but then back-tracks to South for 105 miles to pick up a friend. What is John’s total displacement? Answer: John’s starting position Xi= 0. Her final position Xf is the distance travelled N minus the distance South. Calculating displacement, i.e.D. D = ΔX = (Xf – Xi) D = (250 mi N – 105 mi S) – 0 D = 145 mi N
- 12. Question 2. An object moves along the grid through points A, B, C, D, E, and F as shown below. The side of square tiles measures 0.5 km. a) Calculate the distance covered by the moving object. b) Find the magnitude of the displacement of the object.
- 13. Solution: a) The distance covered by the moving object is calculated as follows: AB + BC + CD + DE + EF 3 + 1 + 1.5 + 0.5 + 0.5 = 6.5 km The distance covered by the moving object is 6.5 km.
- 14. b) The initial point is A and the final point is F, hence the magnitude of the displacement is equal to the distance AF which is calculated by applying Pythagoras’s theorem to the triangle AHF as shown in the figure below
- 15. Applying the Pythagorean formula, we get AF2=AH2+HF2 Substituting the formula, we get AF2=(0.5×4)2+(0.5×3)2=6.25 AF=square root(6.25km)=2.5km The magnitude of displacement is 2.5 km.
- 17. Magnitude • Magnitude is the size or extent of a physical quantity. In physics, we have scalar and vector quantities. • Scalar quantities are only expressed as magnitude. E.g: time, distance, mass, temperature, area, volume • Vector quantities are expressed in magnitude as well as the direction of the object. E.g: Velocity, displacement, weight, momentum, force, acceleration, etc.
- 19. Time, Average Speed and Velocity Time and speed Time is the duration of an event that is expressed in seconds. Most physical phenomena occur with respect to time. It is a scalar quantity. Speed is the rate of change of distance. If a body covers a certain distance in a certain amount of time, its speed is given by Speed = Distance/Time Average speed = Total distance travelled / Total time taken
- 21. Uniform motion and non-uniform motion • When an object covers equal distances in equal intervals of time it is in uniform motion. • When an object covers unequal distances in equal intervals of time it is said to be in non- uniform motion.
- 23. Uniform Motion Definition: This type of motion is defined as the motion of an object in which the object travels in a straight line and its velocity remains constant along that line as it covers equal distances in equal intervals of time, irrespective of the duration of the time.
- 25. Example of Uniform Motion: • If the speed of a car is 10 m/s, it means that the car covers 10 meters in one second. The speed is constant in every second. • Movement of blades of a ceiling fan.
- 26. Non Uniform Motion: Definition: This type of motion is defined as the motion of an object in which the object travels with varied speed and it does not cover same distance in equal time intervals, irrespective of the time interval duration.
- 27. If a body is involved in rectilinear motion, and if the motion is not consistent, then the acceleration of the body must be non-zero.
- 29. What is Rectilinear Motion? • When we require only one co-ordinate axis along with time to describe the motion of a particle it is said to be in linear motion or rectilinear motion. • Some examples of linear motion are a parade of soldiers, a train moving along a straight line.
- 31. Comparison table Between these two types of motions:
- 32. Uniform Circular Motion: • This type of motion is seen if an object is travelling at a constant speed around a fixed axis or center point, and the object travels around a curved path and it maintains a constant radial distance from the center point at any given time and it moves in a tangent to the curved path.
- 35. Non Uniform Circular Motion This type of motion is inconsistent because if an object is travelling at a variable angular speed around a fixed axis or center point, then the motion of that object is said to be inconsistent because the object covers a curved path and it will have some variable radial acceleration due to which its velocity would change every second.
- 36. Velocity The Rate of change of displacement is velocity. It is a vector quantity. Here the direction of motion is specified. Velocity = Displacement/Time
- 38. Speed: Speed is a scalar quantity which means it has no direction. It denotes how fast an object is moving. If the speed of the particle is high it means the particle is moving fast and if it is low, it means the particle is moving slow.
- 41. Velocity: • Velocity is a vector quantity which means it has both magnitude and direction. • It denotes the rate at which the object is moving or changing position. • The direction of the velocity vector is easy to find. Its direction is same as the direction of the moving object. • Even if the object is slowing down, and the magnitude of velocity is decreasing, its direction would still be same as the direction in which the object is moving
- 45. Average Speed The average speed of a body in a certain time interval is the distance covered by the body in that time interval divided by time. So if a particle covers a certain distance s in a time t1 to t2, then the average speed of the body is:
- 46. 1.) In travelling from Pune to Nagpur , Rahul drove his bike for 2 hours at 60 kmph and 3 hours at 70 kmph. Sol 1) We know that, Distance = Speed × Time So, in 2 hours, distance covered = 2 × 60 = 120 km in the next 3 hours, distance covered = 3 × 70 = 210 km Total distance covered = 120 + 210 = 330 km Total time = 2 + 3 = 5 hrs
- 47. Average Velocity • The average velocity of a body in a certain time interval is given as the displacement of the body in that time interval divided by time. So if a particle covers a certain displacement
- 48. Calculate the average velocity at a particular time interval of a person if he moves 7 m in 4 sec and 18 m in 6 sec along x-axis? Sol) Initial distance traveled by the person, xi = 7 m, Final distance traveled, xf = 18 m, Initial time interval ti = 4 sec, Final time interval tf = 6 sec,
- 49. Acceleration
- 51. Acceleration • The rate of change of velocity is called acceleration. It is a vector quantity. In non- uniform motion, velocity varies with time, i.e., change in velocity is not 0. It is denoted by “a” • Acceleration = Change in Velocity / Time
- 55. Motion Visualised Distance-Time graph • A distance-time graph shows how far an object has travelled in a given time. It is a simple line graph that denotes distance versus time findings on the graph. • Distance is plotted on the Y-axis. • Time is plotted on the X-axis.
- 57. • We can use distance time graph to determine the speed of an object. • Consider small part AB of graph. • Draw parallel line to x-axis from point A. • Draw another line parallel to y-axis from point B. • These 2 lines meet at C to form a triangle ABC. • AC denotes time interval t2-t1. • BC denotes distance s2-s1
- 58. • We can see that object travels from A to B, it covers distance s2-s1 in time t2-t1 • Speed v of object can be represented as
- 60. Example: • For better understanding, let us consider an example of uniform motion. A bus driver drives at a constant speed which is indicated by the speedometer and the driver measures the time taken by the bus for every kilometre. The driver notices that the bus travels 1 kilometre in every 2 minutes.
- 62. • By this table, he had a clear idea about the speed which is: ½ × 60 = 30 km/hr. • The graph is a straight line and the motion of the bus is also uniform. Also, from the graph, we can find the speed of the bus at any instant of time. The initial and final position of the car can be found as the following: • Speed = (Final Position-Initial position)/Time
- 64. Conclusion: The following things can be concluded now: • If the distance-time graph is a straight line then the motion is uniform. • If the distance-time graph of anybody is given, its speed can be calculated using the slope of the graph. • The slope of the straight-line graph is the same irrespective of the interval which is chosen. This implies that the speed of an object under uniform motion remains constant.
- 65. Velocity-Time Graph • Velocity-Time graphs show the change in velocity with respect to time. • Slope gives acceleration • The area under the curve gives displacement • Line parallel to x-axis implies constant velocity-
- 66. OA = constant acceleration, AB = constant velocity, BC = constant retardation
- 68. • Figure shows velocity time graph of car moving with uniform velocity 40 km/hr • Velocity of 40km/hr is represented by height AC or BD. • Time (t2-t1) is represented by length AB • So distance s moved by the car in time t2-t1 can be expressed as
- 70. • A person sitting next to driver records its velocity after every 5 seconds by noting the reading of speedometer of car.
- 71. • Velocity changes by equal amounts in equal intervals of time. So we get straight line graph. • We can determine distance moved by car using velocity-time graph. • Distance travelled is given by area ABCD. • Since magnitude of velocity of car is changing due to acceleration, the distance s travelled by car will be given by area ABCDE under velocity-time graph.
- 72. Case 1: Velocity-time graphs with constant velocity (zero acceleration) When the velocity is constant, the velocity time graph, with Y-axis denoting velocity and X-axis denoting time, will be like:
- 73. Case 2: Velocity-time graphs with constant acceleration • When the acceleration is constant (positive), and the initial velocity of the particle is zero, the velocity of the particle will increase linearly as predicted by the equation: • v = u + at • Since u = 0 • v = at
- 74. As shown in the figure, the velocity of the particle will increase linearly with respect to time. The slope of the graph will give the magnitude of acceleration.
- 75. Case 3: Velocity-time graphs with increasing acceleration When the acceleration is increasing with time, the velocity-time graph will be a curve as predicted from the equation: v = u + at Since u = 0 v= at Since acceleration is a function of time, the velocity- time graph will be a curve. Note: Since the acceleration is continuously increasing with time, the magnitude of the slope will also continuously increase with time.
- 76. Velocity-time graphs with increasing acceleration
- 77. Equations of Motion The motion of an object moving at uniform acceleration can be described with the help of three equations, namely • v = u + at (8.5) • s = ut + (1/2)at2 (8.6) • 2as =v2 – u2 (8.7)
- 78. Derivation of velocity-time relation by graphical method
- 79. • Initial velocity of object is u(at point A). • Velocity increases to v(at point B) • Initial velocity is given by OA. • Final velocity is given by BC. • Time interval t is given by OC. • BD=BC-CD(represents change in velocity in time interval t) • Draw AD parallel to OC.
- 80. • From graph we observe that BC=BD+DC=BD+OA Substituting BC=v and OA=u We get v=BD+u Or BD=v-u (8.8) Acceleration of object is given by
- 81. Derivation of position-time relation by graphical method
- 82. • Consider object travelled a distance s in time t under uniform acceleration. • Distance travelled by object is obtained by area enclosed within OABC under velocity time graph AB • Thus distance s travelled by object is given by s=area OABC = area of rectangle OADC + area of triangle ABD
- 84. Derivation of position-velocity relation by graphical method
- 86. Uniform circular motion • If an object moves in a circular path with uniform speed, its motion is called uniform circular motion. • Velocity is changing as direction keeps changing. • Acceleration is constant
- 87. Following are the examples of uniform circular motion: • Motion of artificial satellites around the earth is an example of uniform circular motion. The gravitational force from the earth makes the satellites stay in the circular orbit around the earth. • The motion of electrons around its nucleus. • The motion of blades of the windmills. • The tip of second’s hand of a watch with circular dial shows uniform circular motion.
- 94. Extra Questions 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Solution Yes, an object moving a certain distance can have zero total displacement. Displacement refers to the shortest distance between the initial and the final positions of the object. Even if an object moves through a considerable distance, if it eventually comes back to its initial position, the corresponding displacement of the object would be zero.
- 95. 2. A farmer moves along the boundary of a square field of side 10m 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? Solution Given that the farmer covers the entire boundary of the square field in 40 seconds, the total distance traveled by the farmer in 40 seconds is 4*(10) = 40 meters.
- 99. 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. Solution Neither of the statements are true. (a) is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero. (b) Statement (b) is false because the displacement of an object can be equal to, but never greater than the distance traveled.
- 101. 5. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Solution Since average speed is the total distance traveled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance traveled is equal to the displacement.
- 102. 6. What does the odometer of an automobile measure? Solution The odometer measures the total distance traveled by the automobile. 7. What does the path of an object look like when it is in uniform motion? Solution The path of an object in uniform motion is a straight line.
- 103. 8. 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 Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance traveled by the signal. 5 minutes = 5*60 seconds = 300 seconds. Speed of the signal = 3 × 108 m/s. Therefore, total distance = (3 × 108 m/s) * 300s = 9*1010 meters.
- 104. 9. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? Solution Uniform Acceleration: In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a uniform rate (it changes at a constant rate in any constant time interval). Non-Uniform Acceleration: In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a rate that is not uniform (it changes at a different rate for a given constant time interval).
- 105. 10. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus. Solution
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