Motion Forces II Describing Motion Speed Velocity u

Motion & Forces II. Describing Motion Speed & Velocity u Acceleration u u

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

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

Motion

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

Reference Point

A. Motion 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.

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

Distance Displacement

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 4 km South ◦ Answers: 7 km ; 70 meters ; 16 km

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

Displacement If an object moves in a single direction, the displacement equals the distance + the direction 4 km start here total displacement = 4 km North

Displacement If an object moves in two opposing directions, the displacement is the difference between the two. total distance = 4 + 3 = 7 km 4 km start here 3 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] total displacement = 4 km North + -3 km South = 1 km North (of original starting point)

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

Displacement – cont’d If an object moves in two directions, a triangle will be formed. If the angle is 90º , use a 2 + b 2 = c 2 to solve. 3 km displacement = 4 km 5 km 42 + 3 2 = c 2 = (16) + (9) = 25 start here c 2 = √(25) c = 5 km

Displacement – cont’d Displacement can be given in: ◦ units with direction example: x number of meters North 30 meters East + 40 meters West displacement = 10 meters West

Displacement – cont’d 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

Displacement Practice – What is the displacement? ◦ Pattie walks 8 km 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 4 km South 0 meters displacement go to website!!!!

Practice 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.

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

B. Speed & Velocity Speed ◦ rate of motion ◦ distance traveled per unit time d v t

B. Speed & Velocity Instantaneous Speed Average Speed ◦ speed at a given instant

B. Speed & Velocity 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!

B. Speed & Velocity ◦ speed in a given direction ◦ can change even when the speed is constant!

Practice Find the velocity in m/s of a swimmer who swims 110 m toward the shore in 72 s. v =d t v = 110 m = 1. 5 m/s 72 s

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

C. Acceleration Positive acceleration “speeding up” Negative acceleration “slowing down”

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

D. Calculations 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: WORK: l vi = 10 m/s t=3 s vf = 32 m/s vf - vi a=? a t a = ( v f - v i) ÷ t a = (32 m/s - 10 m/s) ÷ (3 s) a = 22 m/s ÷ 3 s a = 7. 3 m/s 2 = 7 m/s

D. Calculations 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: WORK: l v = 330 m/s t=d÷v d = 1 km = 1000 m t = (1000 m) ÷ (330 m/s) t=? t = 3. 03 s = 3 s d v t

D. Calculations 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: WORK: l t=? vi = 30 m/s vf = 0 m/s a = -3 m/s 2 t = (vf - vi) ÷ a t = (0 m/s-30 m/s)÷(-3 m/s 2) vf - vi a t t = -30 m/s ÷ -3 m/s 2 t = 10 s

E. Graphing Motion Distance-Time Graph A B slope =speed steeper slope = faster speed straight line = constant speed flat line = no motion

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

E. Graphing Motion Distance-Time Graph Acceleration is indicated by a curve on a Distance-Time graph. Changing slope = changing velocity

E. Graphing Motion Speed-Time Graph slope =acceleration § +ve = speeds up § -ve = slows down straight line = constant accel. flat line = no accel. (constant velocity)

E. Graphing Motion Speed-Time Graph 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

Motion & Forces III. Defining Force Newton’s First Law u Friction u u

A. Force ◦ a push or pull that one body exerts on another ◦ What forces are being exerted on the football? Fkick Fgrav

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

A. Force Net Force ◦ unbalanced forces that are not opposite and equal ◦ velocity changes (object accelerates) Fnet Ffriction Fpull N W N

B. 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.

B. 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

Concep. Test 1 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.

Concep. Test 2 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? . . .

Concep. Test 2 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 2. Starting at the time of collision, the 4. neither of the above door exerts a leftward force on you.

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

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

Motion & Forces IV. Force & Acceleration u Newton’s Second Law u Gravity u Air Resistance u Calculations

A. 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

A. Newton’s Second Law F m F = ma F m a F: force (N) m: mass (kg) a: accel (m/s 2) 1 N = 1 kg ·m/s 2

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

B. Gravity Who experiences more gravity the astronaut or the politician? l Which exerts more gravity the Earth or the moon? less distance more mass

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

B. Gravity Would you weigh more on Earth or Jupiter? § Jupiter because. . . greater mass greater gravity greater weight

B. Gravity 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. ”

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

C. 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

C. Air Resistance Terminal Velocity ◦ maximum velocity reached by a falling object ◦ reached when… Fair Fgrav = Fair § no net force no acceleration constant velocity Fgrav

C. Air Resistance Terminal Velocity § increasing speed increasing air resistance until… Fair = Fgrav Animation from “Multimedia Physics Studios. ”

C. Air Resistance Falling with air resistance § heavier objects fall faster because they accelerate to higher speeds before reaching terminal velocity Fgrav = Fair § larger Fgrav need larger Fair need higher speed Animation from “Multimedia Physics Studios. ”

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

D. Calculations l. A 4. 0 kg shotput is thrown with 30 N of force. What is its acceleration? GIVEN: WORK: m = 4. 0 kg F = 3. 0 N a=? a=F÷m F m a a = (30 N) ÷ (4. 0 kg) a = 7. 5 m/s 2

D. Calculations l Mrs. J. weighs 557 N. What is her mass? GIVEN: WORK: F(W) = 557 N m=? a(g) = 9. 8 m/s 2 m=F÷a F m a m = (557 N) ÷ (9. 8 m/s 2) m = 56. 8 kg m = 57 kg

Concep. Test 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.

Motion & Forces VI. Action and Reaction u Newton’s Third Law u Momentum u Conservation of Momentum

A. 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.

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

A. Newton’s Third Law l Explanation: ◦ 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

A. Newton’s Third Law Action-Reaction Pairs l The hammer exerts a force on the nail to the right. l The nail exerts an equal but opposite force on the hammer to the left.

A. Newton’s Third Law Action-Reaction Pairs The rocket exerts a downward force on the exhaust gases. l The gases exert an equal but opposite upward force on the rocket. l FG FR

A. Newton’s Third Law Action-Reaction Pairs l Both objects accelerate. l The amount of acceleration depends on the mass of the object. F m l Small mass more acceleration l Large mass less acceleration

Motion & Forces 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)

A. 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.

B. 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

C. 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.