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# Motion Class 9th

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A thorough detailed presentation of chapter motion for class 9th.

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### Motion Class 9th

1. 1. WHAT IS MOTION?? DEFINITION: • CHANGE IN POSITION OF AN OBJECT WITH RESPECT TO TIME. • ANY MOVEMENT (LINEAR OR CIRCULAR) OR CHANGE IN PLACE • MOTION OCCURS WHEN AN OBJECT CHANGES ITS POSITION. RELATIVE MOTION: • WHEN TWO OBJECTS ARE MOVING IN A PLANE (EITHER IN SAME DIRECTION OR OPPOSITE) EACH HAVE RELATIVE MOTION WITH RESPECT TO SECOND. E.G. A PERSON SITTING IN A TRAIN AND WATCHING A TREE, IN THIS CASE TREE IS STABLE BUT IS ASSUMED TO BE MOVING BUT WITH RESPECT TO TRAIN
2. 2. Figure Showing change in the position of the bear. Figure Showing an instance of relative motion
3. 3. DISTANCE VS. DISPLACEMENT DISTANCE • TOTAL LENGTH OF PATH COVERED BY AN OBJECT. • IT HAS ONLY MAGNITUDE WITHOUT DIRECTION. (SCALAR QUANTITY) • EITHER ZERO OR POSITIVE. DISPLACEMENT • SHORTEST DISTANCE TRAVELLED BY AN OBJECT FROM INITIAL POSITION TO FINAL POSITION. • IT HAS ONLY MAGNITUDE AND DIRECTION. (VECTOR QUANTITY) • CAN BE ZERO, POSITIVE OR NEGATIVE.
4. 4. Question: An Object is moving in a circle of radius r. Calculate the distance and displacement? (i) When it completes half the circle. (ii) When it completes one full circle. Answer: When object completes half the circle, it moves from A to B. Distance= ½ x Circumference = ½ x 2Πr =Πr Displacement=2 x Radius of Circle = Diameter of Circle = 2 r A B r o
5. 5. Question: A particle moves 3m north, then 4m east and finally 6m south. Calculate the distance travelled and the displacement. Answer: The particle starts from O and moves 3m North (=OA), 4m East (=AB) and 6m South (=BC). Distance Travelled= OA+AB+BC =3m +4m+ 6m =13m Displacement = OC = 𝑂𝐷2 + 𝐷𝐶2 = 𝐴𝐵2 + (𝐵𝐶 − 𝐵𝐷)2 = 62 + (6 − 3)2 =5m 3m 4m A B C Do N S E W
6. 6. TYPES OF MOTION BASED ON DISTANCE COVERED BASED ON DIRECTION Uniform Motion Non Uniform Motion Linear MotionCircular Motion
7. 7. UNIFORM MOTION • WHEN AN OBJECT TRAVELS EQUAL DISTANCES IN EQUAL INTERVALS OF TIME • A CAR COVERS 4M IN FIRST 2SECONDS AND ANOTHER 4M IN SECOND 2 SECONDS AND SO ON NON UNIFORM MOTION • WHEN AN OBJECT TRAVELS UNEQUAL DISTANCES IN EQUAL INTERVALS OF TIME. • A CAR COVERS 8M IN FIRST 2 SECONDS THEN 1M IN ANOTHER 2 SECONDS AND SO ON
8. 8. CIRCULAR MOTION • WHEN AN OBJECT CONTINUOUSLY CHANGING ITS DIRECTION IRRESPECTIVE OF ITS UNIFORMITY. LINEAR MOTION • WHEN AN OBJECT MOVES IN A PARTICULAR DIRECTION IRRESPECTIVE OF ITS UNIFORMITY.
9. 9. MEASURING THE RATE OF MOTION: SPEED Which one is faster?? A driver A takes 4 hours to cover a distance of 200km from Ambala to Delhi A driver B takes 3 hours to cover a distance of 200km from Ambala to Delhi Speed of a body is defined as the distance travelled by the body in unit time. Speed= Distance/ Time If s is the distance travelled by a body in time t, its speed is v=s/t SI unit of speed is m/s or m/s-1 Note: Speed is a scalar quantity. It has magnitude only. The speed can be zero or positive.
10. 10. TYPES OF SPEED: Uniform Speed Non uniform Speed When a body travels equal distances in equal intervals of time, the speed of the body is said to be uniform. When a body covers unequal distances in equal intervals of time, the speed of the body is said to be non uniform. Note: in most of cases, bodies move with non uniform speed. Therefore, we describe their rate of motion in terms of average speed.
11. 11. SPEED WITH DIRECTION: VELOCITY Velocity of a body is the distance travelled by the body in unit time in a given direction. or Velocity of a body is the speed of the body in a particular direction. Velocity = distance travelled in given direction/ time taken or Velocity = displacement/time → → v = s / t Where → v is velocity → s is displacement Note: The unit of velocity is same as that of speed i.e. m/s Velocity is a vector quantity whereas speed is scalar quantity
12. 12. AVERAGE SPEED AND AVERAGE VELOCITY During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car. It shows your speed at a particular instant in time (Instantaneous speed). Average speed is a measure of the distance travelled in a given period of time. Average velocity is a measure of displacement made by the object in a given period of time. Suppose that during your trip to school, you travelled a distance of 5 miles and the trip lasted 0.2 hours (12 minutes). The average speed of your car could be determined as
13. 13. Question: Use the diagram to determine the average speed and the average velocity of the skier during these three minutes. Using v = s/t, The skier has an average speed of (420 m) / (3 min) = 140 m/min and an average velocity of (140 m, right) / (3 min) = 46.7 m/min
14. 14. RATE OF CHANGE IN VELOCITY: ACCELERATION • CHANGE IN SPEED OR DIRECTION OF AN OBJECT IN UNIT TIME OR • RATE OF CHANGE IN VELOCITY Acceleration = change in velocity/ time final velocity – Initial velocity Acceleration = --------------------------------------- Time v – u a = --------- t • Acceleration can be zero, negative or positive i.e. vector quantity Note: change in velocity can be done by (a) change in speed (b) change in direction (c) both of them
15. 15. TYPES OF ACCELERATION: Uniform Acceleration Non-Uniform acceleration When a body travels with uniform or constant velocity, the acceleration of the body is said to be uniform. When a body travels with non-uniform velocity, the acceleration of the body is said to be non uniform. Note: In most of cases, bodies move with non uniform acceleration.
16. 16. TYPES OF ACCELERATION: Positive Acceleration Negative acceleration When acceleration is in the direction of velocity, the acceleration of the body is said to be positive. Example: when car starts from rest When acceleration is in the opposite direction of velocity, the acceleration of the body is said to be negative. Example: when car stops to rest
17. 17. MOTION IN GRAPHS CONSTANT/UNIFORM SPEED CONSIDER A CAR MOVING WITH A RIGHTWARD (+), CHANGING VELOCITY - THAT IS, A CAR THAT IS MOVING RIGHTWARD BUT SPEEDING UP OR ACCELERATING. NON UNIFORM SPEED CONSIDER A CAR MOVING WITH A CONSTANT, RIGHTWARD (+) VELOCITY - SAY OF +10 M/S Position vs Time Graphs
18. 18. Let's begin by considering the position versus time graph below. The diagram above shows this method being applied to determine the slope of the line. Note that three different calculations are performed for three different sets of two points on the line. In each case, the result is the same: the slope is 10 m/s.
19. 19. Consider the graph below. Note that the slope is not positive but rather negative; that is, the line slopes in the downward direction. Note also that the line on the graph does not pass through the origin. Test your understanding of slope calculations by determining the slope of the line below. Slope = -3.0 m/s
20. 20. TEST YOUR UNDERSTANDING Determine the velocity (i.e., slope) of the object as portrayed by the graph below. The velocity (i.e., slope) is 4 m/s.
21. 21. MOTION IN GRAPHS CONSTANT VELOCITY NOW CONSIDER A CAR MOVING WITH A RIGHTWARD (+), CHANGING COVERING EQUAL DISTANCES IN EQUAL INTERVAL OF TIME. (UNIFORM ACCELERATION) CHANGING VELOCITY NOW CONSIDER A CAR MOVING WITH A RIGHTWARD (+), CHANGING VELOCITY BUT COVERING UNEQUAL DISTANCES IN EQUAL INTERVALS OF TIME. ( NON UNIFORM ACCELERATION) CONSIDER A CAR MOVING WITH A CONSTANT, RIGHTWARD (+) VELOCITY - SAY OF +10 M/S. Velocity vs Time Graphs
22. 22. VARIOUS SCENARIOS OF MOTION
23. 23. MORE IN V-T GRAPH
24. 24. QUESTIONS ON GRAPHS!!!!! Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference to the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C). Ans: The object moves in the + direction at a constant speed - zero acceleration (interval A). The object then continues in the + direction while slowing down with a negative acceleration (interval B). Finally, the object moves at a constant speed in the + direction, slower than before (interval C).
25. 25. TEST YOUR KNOWLEDGE Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference to the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C). The object moves in the + direction while slowing down; this involves a negative acceleration (interval A). It then remains at rest (interval B). The object then moves in the - direction while speeding up; this also involves a negative acceleration (interval C). The object moves in the + direction with a constant velocity and zero acceleration (interval A). The object then slows down while moving in the + direction (i.e., it has a negative acceleration) until it finally reaches a 0 velocity (stops) (interval B). Then the object moves in the - direction while speeding up; this corresponds to a - acceleration (interval C).
26. 26. DETERMINING THE SLOPE ON A V-T GRAPH Let's begin by considering the velocity versus time graph below. Slope = acceleration For 1st second, Slope = 4-0/1-0 = 4/1 = 4m/s2 For 2nd second Slope = 8-4/2-1 = 4/1 = 4m/s2
27. 27. TEST YOUR KNOWLEDGE Consider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as portrayed by the graph. The acceleration (i.e., slope) is 4 m/s2.
28. 28. DETERMINING THE AREA ON A V-T GRAPH As learned in an earlier part of this lesson, a plot of velocity-time can be used to determine the acceleration of an object (the slope). For velocity versus time graphs, the area bound by the line and the axes represents the displacement. Rectangle Triangle Trapezoid Area = b • h Area = ½ • b • h Area = ½ • b • (h1 + h2)
29. 29. PRACTICE TIME!!!!! Area = ½ * b * h where b = 1 s and h = 10 m/s Area = ½ * (1 s) * (10 m/s) Area = 5 m Area = ½ * b * (h1 + h2) Area = ½ * (2 s) * (10 m/s + 30 m/s) Area = 40 m Area = b * h Area = (6 s) * (30 m/s) Area = 180 m
30. 30. EQUATIONS OF MOTION THE KINEMATIC EQUATIONS ARE A SET OF FOUR EQUATIONS THAT CAN BE UTILIZED TO PREDICT UNKNOWN INFORMATION ABOUT AN OBJECT'S MOTION IF OTHER INFORMATION IS KNOWN. THE EQUATIONS CAN BE UTILIZED FOR ANY MOTION THAT CAN BE DESCRIBED AS BEING EITHER A CONSTANT VELOCITY MOTION (AN ACCELERATION OF 0 M/S/S) OR A CONSTANT ACCELERATION MOTION. Eqaution (1) represent velocity-time relation Eqaution (2) represents position-time relation Equation (3) represents position-velocity relation
31. 31. FIRST EQUATION OF MOTION
32. 32. SECOND EQUATION OF MOTION
33. 33. THIRD EQUATION OF MOTION
34. 34. TEST YOUR KNOWLEDGE 1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance travelled before take off. Given: a = 3.20 m/s2 t = 32.8 s u = 0 m/s using second equation of motion, s = ut + ½ at2 = 0 x 3.28 + ½ x 3.2 x 3.28 x 3.28 = 1721 m 2. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall? Given: t = 2.60 s u = 0 m/s a = 9.8 m/s2 Using first and third equation of motion, v = u + at = 0 + 9.8 x 2.6 = 25.48 m v2 – u2 = 2as (25.48)2 – (0) 2 = 2 x 9.8 x s 649.23 – 0 = 19.6 x s 649.23/19.6 = s 34.1 m = s
35. 35. FREE FALL AND THE ACCELERATION OF GRAVITY A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall.
36. 36. Characteristics of free falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s ) A position versus time graph for a free-falling object is shown below. A velocity versus time graph for a free-falling object is shown below. Equations of motion with respect to free fall
37. 37. UNIFORM CIRCULAR MOTION Did you ever stop to count the number of controls that cause acceleration in your car? At a minimum, you have a gas pedal, a brake, and a steering wheel. That's right, the steering wheel produces an acceleration on your car, because it causes you to change direction which changes your velocity, even if it doesn't change your speed. That's what Uniform Circular Motion is all about - the changing of the velocity of an object as it travels in a circle at a constant speed.
38. 38. Uniform Circular Motion – an object moving in a circle at a constant speed.
39. 39. CIRCULAR MOTION AND CENTRIPETAL FORCE An object moving in a circle is experiencing an acceleration. Even if moving around the perimeter of the circle with a constant speed, there is still a change in velocity and subsequently an acceleration. This acceleration is directed towards the centre of the circle. In accord with Newton's second law of motion, an object which experiences an acceleration must also be experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration. This is sometimes referred to as the centripetal force
40. 40. TEST YOUR KNOWLEDGE Q: A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car. Ans: To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows: a = v2 / R a = (10.0 m/s)2 / (25.0 m) a = (100 m2/s2) / (25.0 m) a = 4 m/s2 To determine the net force acting upon the car, use the equation Fnet = m x a. The solution is as follows. Fnet = m x a = (900 kg) x (4 m/s2) Fnet = 3600 N Q: Determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the Cliff-hanger in 29.3 seconds. The radius of the barrel is 2.90 meters. Ans: Given: m = 40 kg; R = 2.90 m; T = 2.93 s (since 10 cycles takes 29.3 s). First, find speed using speed=(2𝜋R) / T = 6.22 m/s. Then find the acceleration using a = v2 / R = = (6.22 m/s)2 / (2.90 m) = 13.3 m/s2 Now use Fnet = m x a Fnet = 40kg x 13.3m/s2 = 533 N.
41. 41. “ ” THANK YOU REMEMBER YOU CAN HELPYOURSELF KEEP PRACTICING