2. II. Describing Motion  Motion  Speed & Velocity  Acceleration

3.  Newton’s First Law of Motion ◦ An object at rest will remain at rest and an object in motion will continue moving at a constant velocity unless acted upon by a net force. motion constant velocity net force

4.  Problem: ◦ Is your desk moving?  We need a reference point... ◦ nonmoving point from which motion is measured

6.  Motion ◦ Change in position in relation to a reference point. Reference point Motion

8. Problem:  You are a passenger in a car stopped at a stop sign. Out of the corner of your eye, you notice a tree on the side of the road begin to move forward.  You have mistakenly set yourself as the reference point.

9.  You can describe the direction of an object in motion with a reference direction.  north, south, east, west, up, or down

10.  Distance  Displacement

11.  Distance – the length traveled by an object  Practice – What is the distance? ◦ Pattie walks 4 km North, then 3 km East ◦ Joe runs 30 m East, then runs 40 meters West ◦ Aaron jogs 4 km East, 4 km North, 4 km West, and 4km South ◦ Answers: 7 km ; 70 meters ; 16 km

12.  Distance – length  Displacement – distance AND direction of a movement from the starting point

13.  If an object moves in a single direction, the displacement equals the distance + the direction start here  4 kmtotal displacement = 4km North

14.  If an object moves in two opposing directions, the displacement is the difference between the two. start here  3 km 4 km total displacement = 4 km North + -3 km South = 1 km North (of original starting point) total distance = 4 + 3 = 7 km Displacement can be positive or negative. A negative direction can be either the opposite of the original movement, or can follow the sign of a typical graph [North/East vs South/West]

15.  If an object moves in two directions, a triangle will be formed. If the angle is 90 º, use a 2 + b 2 = c 2 to solve. start here  4 km 3 km displacem ent

16.  If an object moves in two directions, a triangle will be formed. If the angle is 90 º, use a 2 + b 2 = c 2 to solve. start here  4 km 3 km 5 km displacement = 4 2 + 3 2 = c 2 c 2 = (16) + (9) = 25 c 2 = √(25) c = 5km

17.  Displacement can be given in: ◦ units with direction  example: x number of meters North  30 meters East + 40 meters West  displacement = 10 meters West

18.  Displacement can be given in: ◦ positive or negative units [like on a graph]  positive = North {up} or East {right}  negative = South {down} or West {left}  30 meters east + 40 meters west = 30 meters + (-40) meters = -10 meters

19.  Practice – What is the displacement? ◦ Pattie walks 8km North, then 3 km East  5 km displacement ◦ Joe runs 30 m East, then runs 40 meters West  10 meters [or -10] displacement ◦ Aaron jogs 4 km East, 4 km North, 4 km West, and 4km South  0 meters displacement go to website!!!!

20. Draw each scenario and show work. 2. Amy runs 2 miles south, then turns around and runs 3 miles north. a. Distance ___ b. Displacement ___ 3. Jermaine runs exactly 2 laps around a 400 meter track. 4. Joe turns around 5 times.

21. 5. Ray runs 30 feet north, 30 feet west, and then 30 feet south.

22.  Speed ◦ rate of motion ◦ distance traveled per unit time v d t

23.  Instantaneous Speed ◦ speed at a given instant  Average Speed

24.  Problem: ◦ A storm is 10 km away and is moving at a speed of 60 km/h. Should you be worried?  It depends on the storm’s direction!

25.  Velocity ◦ speed in a given direction ◦ can change even when the speed is constant!

26.  Find the velocity in m/s of a swimmer who swims 110 m toward the shore in 72 s.  v = d t  v = 110m = 1.5 m/s 72 s

27.  Acceleration ◦ the rate of change of velocity ◦ change in speed or direction a: acceleration v f : final velocity v i : initial velocity t: time a v f - v i t

28. Positive acceleration “speeding up” Negative acceleration “slowing down”

29. Your neighbor skates at a speed of 4 m/s towards home. You can skate 100 m in 20 s. Who skates faster? GIVEN: d = 100 m t = 20 s v = ? WORK : v = d ÷ t v = (100 m) ÷ (20 s) v = 5 m/s You skate faster! v d t

30. A roller coaster starts down a hill at 10 m/s. Three seconds later, its speed is 32 m/s. What is the roller coaster’s acceleration? GIVEN: v i = 10 m/s t = 3 s v f = 32 m/s a = ? WORK : a = ( v f - v i ) ÷ t a = (32m/s - 10m/s) ÷ (3s) a = 22 m/s ÷ 3 s a = 7.3 m/s 2 = 7 m/s a v f - v i t

31. Sound travels 330 m/s. If a lightning bolt strikes the ground 1 km away from you, how long will it take for you to hear it? GIVEN: v = 330 m/s d = 1km = 1000m t = ? WORK : t = d ÷ v t = (1000 m) ÷ (330 m/s) t = 3.03 s = 3 s v d t

32. How long will it take a car traveling 30 m/s to come to a stop if its acceleration is -3 m/s 2 ? GIVEN: t = ? v i = 30 m/s v f = 0 m/s a = -3 m/s 2 WORK : t = ( v f - v i ) ÷ a t = (0m/s-30m/s)÷(-3m/s 2 ) t = -30 m/s ÷ -3m/s 2 t = 10 s a v f - v i t

33.  slope =  steeper slope =  straight line =  flat line = Distance-Time Graph A B faster speed constant speed no motion speed

34.  Who started out faster? ◦ A (steeper slope)  Who had a constant speed? ◦A◦A  Describe B from 10-20 min. ◦ B stopped moving  Find their average speeds. ◦ A = (2400m) ÷ (30min) A = 80 m/min ◦ B = (1200m) ÷ (30min) B = 40 m/min Distance-Time Graph A B

35.  Acceleration is indicated by a curve on a Distance-Time graph.  Changing slope = changing velocity

36.  slope =  straight line =  flat line = Speed-Time Graph acceleration  +ve = speeds up  -ve = slows down constant accel. no accel. (constant velocity)

37. Specify the time period when the object was...  slowing down ◦ 5 to 10 seconds  speeding up ◦ 0 to 3 seconds  moving at a constant speed ◦ 3 to 5 seconds  not moving ◦ 0 & 10 seconds Speed-Time Graph

38. III. Defining Force  Force  Newton’s First Law  Friction

39.  Force ◦ a push or pull that one body exerts on another ◦ What forces are being exerted on the football? F kick F grav

40.  Balanced Forces ◦ forces acting on an object that are opposite in direction and equal in size ◦ no change in velocity

41.  Net Force ◦ unbalanced forces that are not opposite and equal ◦ velocity changes (object accelerates) F friction W F pull F net NN

42.  Newton’s First Law of Motion ◦ An object at rest will remain at rest and an object in motion will continue moving at a constant velocity unless acted upon by a net force.

43.  Newton’s First Law of Motion ◦ “Law of Inertia”  Inertia ◦ tendency of an object to resist any change in its motion ◦ increases as mass increases

44. TRUE or FALSE? The object shown in the diagram must be at rest since there is no net force acting on it. FALSE! A net force does not cause motion. A net force causes a change in motion, or acceleration. Taken from “The Physics Classroom” © Tom Henderson, 1996-2001.The Physics Classroom

45. You are a passenger in a car and not wearing your seat belt. Without increasing or decreasing its speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is the correct analysis of the situation?...

46. 1. Before and after the collision, there is a rightward force pushing you into the door. 2. Starting at the time of collision, the door exerts a leftward force on you. 3. both of the above 4. neither of the above 2. Starting at the time of collision, the door exerts a leftward force on you.

47.  Friction ◦ force that opposes motion between 2 surfaces ◦ depends on the:  types of surfaces  force between the surfaces

48.  Friction is greater... ◦ between rough surfaces ◦ when there’s a greater force between the surfaces (e.g. more weight)  Pros and Cons?

49. IV. Force & Acceleration  Newton’s Second Law  Gravity  Air Resistance  Calculations

50.  Newton’s Second Law of Motion ◦ The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. F = ma

51. F:force (N) m:mass (kg) a:accel (m/s 2 ) 1 N = 1 kg ·m/s 2 m F a F m

52.  Gravity ◦ force of attraction between any two objects in the universe ◦ increases as...  mass increases  distance decreases

53.  Who experiences more gravity - the astronaut or the politician? less distance more mass Which exerts more gravity - the Earth or the moon?

54.  Weight ◦ the force of gravity on an object MASS always the same (kg) WEIGHT depends on gravity (N) W = mg W:weight (N) m:mass (kg) g:acceleration due to gravity (m/s 2 )

55.  Would you weigh more on Earth or Jupiter? greater gravity greater weight greater mass  Jupiter because...

56.  Accel. due to gravity (g)  In the absence of air resistance, all falling objects have the same acceleration!  On Earth: g = 9.8 m/s 2 elephant feather Animation from “Multimedia Physics Studios.”Multimedia Physics Studios

58. http://www.animations.physics.uns w.edu.au/jw/Newton.htm#Newton 2 Go to the astronaut’s, David Scott, demonstration

59.  Air Resistance ◦ a.k.a. “fluid friction” or “drag” ◦ force that air exerts on a moving object to oppose its motion ◦ depends on: speed surface area shape density of fluid

60.  Terminal Velocity ◦ maximum velocity reached by a falling object ◦ reached when… F grav = F air F air F grav  no net force  no acceleration  constant velocity

61.  Terminal Velocity  increasing speed  increasing air resistance until… F air = F grav Animation from “Multimedia Physics Studios.”Multimedia Physics Studios

62.  Falling with air resistance F grav = F air Animation from “Multimedia Physics Studios.”Multimedia Physics Studios  heavier objects fall faster because they accelerate to higher speeds before reaching terminal velocity  larger F grav  need larger F air  need higher speed

63. What force would be required to accelerate a 40 kg mass by 4 m/s 2 ? GIVEN: F = ? m = 40 kg a = 4 m/s 2 WORK : F = ma F = (40 kg)(4 m/s 2 ) F = 160 N F = 200 N m F a

64. A 4.0 kg shotput is thrown with 30 N of force. What is its acceleration? GIVEN: m = 4.0 kg F = 3.0 N a = ? WORK : a = F ÷ m a = (30 N) ÷ (4.0 kg) a = 7.5 m/s 2 m F a

65. Mrs. J. weighs 557 N. What is her mass? GIVEN: F(W) = 557 N m = ? a( g ) = 9.8 m/s 2 WORK : m = F ÷ a m = (557 N) ÷ (9.8 m/s 2 ) m = 56.8 kg m = 57 kg m F a

66.  Is the following statement true or false? ◦ An astronaut has less mass on the moon since the moon exerts a weaker gravitational force.  False! Mass does not depend on gravity, weight does. The astronaut has less weight on the moon.

67. VI. Action and Reaction  Newton’s Third Law  Momentum  Conservation of Momentum

68.  Newton’s Third Law of Motion ◦ When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.

69.  Problem:  How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force? NO!!!  Aren’t these “balanced forces” resulting in no acceleration?

70. ◦ forces are equal and opposite but act on different objects ◦ they are not “balanced forces” ◦ the movement of the horse depends on the forces acting on the horse Explanation:

71.  Action-Reaction Pairs The hammer exerts a force on the nail to the right. The nail exerts an equal but opposite force on the hammer to the left.

72.  Action-Reaction Pairs The rocket exerts a downward force on the exhaust gases. The gases exert an equal but opposite upward force on the rocket. FGFG FRFR

73.  Action-Reaction Pairs Both objects accelerate. The amount of acceleration depends on the mass of the object. F m Small mass  more acceleration Large mass  less acceleration

74. I. Newton’s Laws of Motion “If I have seen far, it is because I have stood on the shoulders of giants.” - Sir Isaac Newton (referring to Galileo)

75.  Newton’s First Law of Motion ◦ An object at rest will remain at rest and an object in motion will continue moving at a constant velocity unless acted upon by a net force.

76.  Newton’s Second Law of Motion ◦ The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. F = ma

77.  Newton’s Third Law of Motion ◦ When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.