1. PHYSICS MARJORIE MANABAT

2. SCALAR Is a quantity that has magnitude and usually a unit of measure Examples: –distance –Speed –Time –Energy –Mass

3. SCALAR Rules of algebraic addition, subtraction, multiplication and division are used mathematically –Example: –2 kg + 5 kg = 7 kg –10 m +29 m = 39 m –28 cm – 5 cm = 23 cm

4. MOTION SCALAR QUANTITY –Is completely described by its size or magnitude. –Distance, mass, volume VECTOR QUANTITIES –Is completely described by its magnitude and direction –Velocity, Force, work

5. VECTOR Is a quantity with magnitude, direction and usually a unit of measure. 3 important parts: –Arrowhead – indicates the direction of the vector –Length of the arrow – represents the magnitude of the vector –Tail – represents the origin of the vector

6. VECTOR Tailarrow head length of the arrow

7. ADDITION OF VECTORS RESULTANT (R) –The sum of two or more vectors 1.Vectors acting in the same direction The resultant of 2 vectors acting in the same direction is a vector whose magnitude is equal to the sum of their magnitudes and acts in the same direction as they do –Example –25 km N + 30 km N = 55 km N

8. ADDITION OF VECTORS RESULTANT (R) –The sum of two or more vectors –Vectors acting in the opposite direction The resultant of two vectors acting in opposite direction is a vector whose magnitude is the difference of their magnitudes and which acts in the direction of the greater vector Example –3 km/hr E – 6 km/hr W = 3 km/hr W

10. ADDITION OF VECTORS RESULTANT (R) –The sum of two or more vectors 3.Vectors acting in any direction GENERAL METHODS OF FINDING RESULTANT a.POLYGON METHOD (GRAPHICAL METHOD) b.PARALLELOGRAM METHOD c.ANALYTICAL METHOD

11. ADDITION OF VECTORS a.POLYGON METHOD (GRAPHICAL)

12. POLYGON METHOD (GRAPHICAL)

13. 1.Choose an appropriate scale and frame of reference for the given vectors 2.Draw the first vector starting from the origin of the reference frame 3.Draw the second vector from the head of the first vector and draw the third vector starting from the head of the second vector

14. POLYGON METHOD (GRAPHICAL) 4.All vectors must be connected in series, head-to-tail fashion 5.To determine the resultant vector, connect the tail of the first vector to the head of the last vector drawn. Then, use a protractor in measuring its direction.

17. POLYGON METHOD (GRAPHICAL) SAMPLE PROBLEM: –F1 = 20 N, E –F2 = 30 N, 45°North of East –F3 = 10 N, North –Find F R SCALE = 10 N = 1 cm

18. PARALLELOGRAM METHOD This method is used in determining the resultant of two vectors. Just draw a parallelogram with the two given vectors as the sides. The diagonal of the parallelogram from the same point of origin represents the resultant.

20. ADDITION OF VECTORS ANALYTICAL METHOD –Vector addition by components 1.Resolve the vectors into their components in the x and y direction. A x = A cos θ, A y = A sin θ

21. ADDITION OF VECTORS 2. Add the components in the x direction to give Rx and add the components in the y direction to give ry. That is Rx = Ax + Bx + Cx = sum of x-components Ry = Ay + By + Cy = sum of y-components

22. ADDITION OF VECTORS 3. Find the magnitude of the resultant R from the components Rx and Ry. From the Pythagorean Theorem R = √(Rx)² + (Ry)²

23. ADDITION OF VECTORS 4. Determine the direction of the resultant R using Tan θ = Ry Rx θ = inv tan Ry Rx

24. SAMPLE PROBLEM GIVEN: A = 2 cm, N; B = 3 cm, 20° N of E C = 5 cm, 40° S or E; D = 4 cm, 40° N or W Ax = A cos θ = (2cm) (cos 90°) = 0 Ay = A sin θ = (2cm) (sine 90°) = 2cm Bx = B cos θ = (3cm) (cos 20°) = 2.82.cm By = B sin θ = (3cm) (sin 20°) = 1.03 cm

25. SAMPLE PROBLEM Cx = C cos θ = (5cm) (cos 40°) = 3.83 cm Cy = C sin θ = (5cm) (sin 40°) = -3.21 cm Dx = D cos θ = (4cm) (cos 40°) = -3.06 cm Dy = D sin θ = (4cm) (sin 40°) = 2.57 cm Rx = Ax + Bx + Cx + Dx = 0 + 2.82 +3.83 + (-3.06) = 3.59 cm Ry = Ay + By + Cy + Dy = 2 + 1.03 + (-3.21) +2.57 = 2.39 cm

26. SAMPLE PROBLEM R = √(Rx)² + (Ry) ² = √(3.59 cm) ² + (2.39 cm) ² R = 4.31 cm Tan θ = Ry / Rx = 2.39 cm / 3.59 cm = 0.6657 θ = inv tan 0.6657 = 33° 39’ 11” R = 4.31 cm, 33° 39’ 11”

27. HOMEWORK A car goes 5.0 km south, 2.0 km West and 1.0 km North. –Find the resultant vector using both: POLYGON METHOD ANALYTICAL METHOD

28. MOTION Defined as a continuous change of position with respect to a certain reference point. –Occurs all around us –Motion is relative  everything moves

29. MOTION PHYSICAL QUANTITIES: 1.SPEED – is a measure of how fast something is moving –Distance covered per unit of time INSTANTANEOUS SPEED: speed at any instant; speedometer of a car AVERAGE SPEED: whole distance covered divided by the total time of travel

30. MOTION AVERAGE SPEED = total distance covered time interval Ex. We travelled 240 km in 4 hours Average speed = 240 km / 4 hours = 60 km/hr

31. MOTION 2.VELOCITY –Is a speed in a given DIRECTION –V= d, dir t int v = velocity d, dir = distance in a given direction (displacement) t int = time interval

32. SAMPLE PROBLEM Marina swims north for 15 sec. she covered a distance of 20 m. Compute her velocity ANSWER: 1.33 m/s, NORTH

33. MOTION 2.VELOCITY a.CONSTANT VELOCITY Motion at constant velocity is motion in a straight line at constant speed. Constant direction ~ direction is a straight line b.CHANGING VELOCITY If either the speed or the direction (or both) is changing, then the velocity is changing Constant speed and constant velocity are not the same

34. Exercise An ocean liner going from Miami to Tokyo stops for a few days in Honolulu. You are on board and wish to estimate the time it will take you to get to Tokyo. It is 3700 km from Miami to Honolulu and 6200 km from Honolulu to Tokyo. It took you 5 days to reach Honolulu from Miami. What is the average velocity of the ocean liner in km/hr? How long will it take to reach Tokyo from Honolulu?

35. MOTION ACCELERATION –Is a measure of how fast the velocity changes with respect to time –The body accelerates whenever there is a change in speed –a= Δv or Vf - Vi t int t int

36. SAMPLE PROBLEM A car starts from rest, ran with a speed of 7 m/s for 3 sec. what is the acceleration? a = 7 m/s – o m/s 3 sec Answer: 2.3 m/s²

37. TYPES OF MOTION In 2 forms: 1.LINEAR MOTION Refers to the movement of an object along a straight path 2.CURVILINEAR MOTION Is the motion of bodies along curved paths Ex. Projectile motion and object moving along circular paths

38. TYPE OF MOTION 1.LINEAR MOTION A.Uniformly accelerated motion – along a straight line, A vehicle maintains a constant velocity A vehicle maintains a uniform change in its velocity (constant acceleration) a= Δv or Vf - Vi t int t int

39. TYPE OF MOTION 1.LINEAR MOTION A.Uniformly accelerated motion X = V i t + at² 2 X~displacementV i ~ initial velocity t ~ time a ~ acceleration

40. SAMPLE PROBLEM AN AUTOMOBILE IS TRAVELING 65 km/h. IT BRAKES DECELERATE IT at 6 m/s². 1.How long will it take to stop the car? 2.How far will the car travel after the brakes are applied? ANSWER: 3.0 seconds; 27 meters

41. TYPE OF MOTION 1.LINEAR MOTION B.FREE FALL It is the motion of an object under the influence of gravitational pull only. Acceleration is constant; also known as gravitational acceleration or acceleration due to gravity On the earth’s surface at sea level, the magnitude is 9.8 m/s²

42. TYPE OF MOTION 1.LINEAR MOTION B.FREE FALL: 3 cases 1)Object dropped – the initial velocity of the object dropped with respect to the vertical axis is zero Vi = 0 m/s²

43. TYPE OF MOTION 1.LINEAR MOTION B.FREE FALL: 3 cases 2)Object thrown vertically downward –In this case, the initial velocity is expressed as a negative value (downward, -Vi) –Constant acceleration (g = -9.8 m/s²)

44. TYPE OF MOTION 1.LINEAR MOTION B.FREE FALL: 3 cases 3)Object thrown vertically upward –The initial velocity of the object is the result of the initial force exerted by the launcher. –As it goes up ~ gravity decreases its speed until speed becomes zero (0)  have reached the maximum height and will start to fall.

45. SAMPLE PROBLEM A STONE IS DROPPED FROM THE TOP OF NEW YORK’S EMPIRE STATE BUILDING, WHICH IS 450 m HIGH. NEGLECTING AIR RESISTANCE, –How long does it take the stone to reach the ground? –What is its speed as it strikes the ground? ANSWER: 9.6 s; 94 m/s

46. TYPE OF MOTION 2.CURVILINEAR MOTION A.PROJECTILE MOTION Projectile is when an object moves through space under the influence of earth’s gravitational force. Follows a parabolic path ~ “trajectory” –Horizontal component (X): horizontal velocity remains constant if friction is negligible –Vertical component (Y): vertical velocity changes with time.

47. TYPE OF MOTION 2.CURVILINEAR MOTION B.CIRCULAR MOTION The motion of a body along circular path All objects that turn around on axis located within the object is said to be “rotating or spinning” While an object turns about external axis the rotational motion is called “revolution”

48. TYPE OF MOTION CURVILINEAR MOTION –CENTRIPETAL FORCE Means “center-seeking” or towards the center Depends on the mass (m), the tangential speed (v) and the radius of curvature ® of the circular moving object.

49. NEWTON’S LAW OF MOTION FORCE is any push or pull FRICTION is a kind of force that acts between surfaces of materials that are moving past each other. –Occurs because of the irregularities in the surface of sliding objects.

50. NEWTON’S LAW OF MOTION 1.LAW OF INERTIA –Every object continues in a state of rest, or of motion in a straight line at constant speed, unless it is compelled to change that state by force exerted upon it. MASS a measure of inertia – the more mass, the greater its inertia, and the more force it takes to change its state of motion

51. NEWTON’S LAW OF MOTION 1.LAW OF INERTIA –MASS is not volume Volume – is a measure of space –Cubic centimeter, cubic meter and liters Mass – measure in kilograms –Ex. Pillow –Mass is not weight Measure of the amount of material in an object and depends only on the number of and kind of atoms that compose it. –Weight – depends on the object’s location

52. NEWTON’S LAW OF MOTION LAW OF ACCELERATION –The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force, is in the same direction as the net force, and is inversely proportional to the mass of the object. a = F/m Unit of force is Newton (N), 1N = (kg) (m/s²)

53. NEWTON’S LAW OF MOTION 3.LAW OF ACTION AND REACTION –Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. One force is “action force” while the other force is called “reaction force” –Equal in strength and opposite in direction

54. NEWTON’S LAW OF UNIVERSAL GRAVITATION Things such as leaves, rain and others fall because of gravity. Newton discovered that gravity is universal –It pulls everything “Every object in the universe attracts every other object with a force directly proportional to each of their masses and inversely proportional to the square of the distance separating them”