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UNIT 2A Linear Motion
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distance | speed | direction acceleration
Unit 2A: Linear Motion (Chap 2) You can describe the motion of an object by its: distance | speed | direction acceleration
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How do you know if an object is moving? Is your book moving?
2.1 Motion Is Relative How do you know if an object is moving? Is your book moving? The book is at rest, relative to the table, BUT It’s moving at about 30 km/s relative to the sun. An object is moving if its position relative to a fixed point is changing.
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An object’s motion must be described relative to something else.
2.1 Motion Is Relative An object’s motion must be described relative to something else. shuttle 8 km/s relative to Earth below race car 300 km/h relative to the track The speeds of things on Earth are usually measured relative to the Earth’s surface.
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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. WHAT?? You have set yourself as the reference point as the car rolls slightly backward. Reference point Motion
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2.2 Speed 400 yrs ago, people described motion as simply “slow” or “fast.” Galileo was the first to measure speed by the distance covered and the time it takes. 5 mi 0.20 h distance time avg. speed = speed = avg. speed = 25 mi/h
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2.2 Speed
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Instantaneous Speed Cars do not always move at a constant speed.
You can tell the speed of the car at any instant by looking at the car’s speedometer. instantaneous speed: the speed at any instant average speed: total distance time
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320 km distance speed = time distance = 80 km = x km 1 h 4 hr
If we know average speed and travel time, the distance traveled is easy to find. Example: If your average speed is 80 km/h on a 4-hour trip, then how far did you travel? distance = 80 km = x km 1 h hr distance time speed = distance = speed x time 320 km
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2.2 Speed If a cheetah can maintain a constant speed of 25 m/s, it will cover 25 meters every second. At this rate, how far will it travel in 10 seconds? distance = (25 m) = (x m) = (1 s) 10 s In 1 minute? distance = (25 m) x (x m) = (1 s) (60 s) 250 m distance = speed x time 1500 m
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distance time 35 km 0.5 h speed = = =
The speedometer in every car also has an odometer that records the distance traveled. If the odometer reads zero at the beginning of a trip and 35 km a half hour later, what is the average speed? distance time 35 km 0.5 h speed = = = 70 km/h
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Quick Quiz! Jake walks east through a passenger car on a train that moves 10 m/s in the same direction. Jake’s speed relative to the car is 2 m/s. Jake’s speed relative to an observer at rest outside the train is ___. 2 m/s 5 m/s 8 m/s 12 m/s 2.1
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Quick Quiz. A gazelle travels 2 km in a half hour. The gazelle’s average speed is ___. 1/2 km/h 1 km/h 2 km/h 4 km/h 2.2
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∆d t m s v = (m/s) In physics, Velocity: is speed in a direction.
speed: 60 km/h velocity: 60 km/h north, or right, or down… ∆: change in… (final – initial) (df – di) t ∆d t m s v = (m/s)
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If either the speed or the direction (or both)
2.3 Velocity If either the speed or the direction (or both) changes, then the velocity changes. constant speed and constant velocity are NOT the same. The car speedometer always reads 30 km/h. Is speed constant? Is velocity constant? Y N
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∆v a = t We can change an object’s motion by changing
2.4 Acceleration We can change an object’s motion by changing its speed, its direction, or both. Acceleration is the rate at which velocity changes. ∆v t (vf – vi) t a = acceleration can increase or decrease speed,
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2.4 Acceleration We can change an object’s motion by changing its speed, its direction, or both. Acceleration: is the rate at which velocity changes. ∆v t (vf – vi) t a = acceleration can increase or decrease speed, deceleration is really negative acceleration (–a)
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2.4 Acceleration Acceleration concerns change in velocity so any a change in direction is acceleration. The car speedometer always reads 30 km/h. Is velocity constant? Is there an acceleration? N Y
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a in the same direction as v : speed up
2.4 Acceleration a in the same direction as v : speed up
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a in the same direction as v : speed up
2.4 Acceleration a in the same direction as v : speed up a in the opp. direction as v : slow down
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a in the same direction as v : speed up
2.4 Acceleration a in the same direction as v : speed up a in the opp. direction as v : slow down a at an angle to v : change direction
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∆v t m/s s m s2 a = or 10 m/s – 0 m/s 1 s 10 m/s 1 s a = = =
2.4 Acceleration v units are in distance per time: (m/s) a is the change in v per change in time. a units are v per time: (m/s per s) or (m/s2) changing v from 0 m/s to 10 m/s in 1 s, a is… ∆v t m/s s m s2 a = or 10 m/s – 0 m/s 1 s 10 m/s 1 s a = = = 10 m/s2
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2.4 Acceleration In 5 seconds a car increases its speed from 8 m/s to 18 m/s, while a truck goes from rest to 10 m/s in a straight line. Which undergoes greater acceleration? 18 m/s – 8 m/s 5 s 10 m/s 5 s acar = = = 2 m/s2 10 m/s – 0 m/s 5 s 10 m/s 5 s atruck = = = 2 m/s2
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Constant speed in a constant direction is… constant velocity.
Quick Quiz! Constant speed in a constant direction is… constant velocity. constant acceleration. instantaneous speed. average velocity. 2.3
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A vehicle undergoes acceleration when it __. gains speed.
Quick Quiz. A vehicle undergoes acceleration when it __. gains speed. decreases speed. changes direction. ALL of the above 2.4
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During each 1 s of fall, v increases by 10 m/s. This gain in v in m/s
2.5 Free Fall: How Fast Imagine there is no air resistance and that gravity is the only thing affecting a falling object. An object moving under influence of the gravitational force only is said to be in free fall. During each 1 s of fall, v increases by 10 m/s. This gain in v in m/s is a in m/s2.
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g = –10 m/s2 v = vi + at g is used for the acceleration due to gravity
2.5 Free Fall: How Fast t = 0 s, v = 0 m/s g is used for the acceleration due to gravity Although g varies slightly based on altitude, its average value is nearly 10 m/s2 t = 1 s, v = 10 m/s t = 2 s, v = 20 m/s t = 3 s, v = 30 m/s t = 4 s, v = 40 m/s g = –10 m/s2 v = vi + at (a is g) t = 5 s, v = 50 m/s
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An object is thrown straight up: It moves upward for a while.
2.5 Free Fall: How Fast An object is thrown straight up: It moves upward for a while. What is v at its highest point? Going up, vi goes to 0 m/s. a = ? It then falls downward as if it had been dropped from rest, going from 0 m/s back to vi (but downward) v = 0 m/s at hmax vo a = –10 m/s2 = g a = –10 m/s2 = g
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2.5 Free Fall: How Fast What would the speedometer reading on the falling rock be 4.5 seconds after it drops from rest? (v = ?) v = vi + at v = 0 m/s + (–10 m/s2)(4.5 s) (a is g) v = –45.0 m/s How about 8 seconds after it is thrown with an initial velocity of 20 m/s downward? v = –20 m/s + (–10 m/s2)(8 s) v = –100 m/s
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With what objects might air resistance be small enough to be ignored?
2.8 Air Resistance and Falling Objects Drop a feather and a coin and the coin reaches the floor far ahead of the feather. Why? Air resistance is responsible for these different accelerations. (not just g) In a vacuum, the feather and coin fall with exactly the same acceleration, g. With what objects might air resistance be small enough to be ignored?
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g = –10 m/s2 v = vi + at d = vit + ½at2 2.6 Free Fall: How Far
t = 0 s, v = 0 m/s, d = 0 m t = 1 s, v = 10 m/s, d = 5 m t = 2 s, v = 20 m/s, d = 20 m g = –10 m/s2 t = 3 s, v = 30 m/s, d = 45 m vavg = ( ) 2 vavg = 35 m/s 1 s d = 35 m v = vi + at (a is g) t = 4 s, v = 40 m/s, d = 80 m d = vit + ½at2 vavg = ( ) 2 vavg = 45 m/s d = 45 m 1 s t = 5 s, v = 50 m/s, d = 125 m
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vf = vi + at An apple falls to the ground in 3 s.
2.6 Free Fall: How Far An apple falls to the ground in 3 s. What is its speed upon striking the ground? vf = vi + at (a is g) v = 0 m/s + (10 m/s2)(3 s) v = 30 m/s What is its vavg during the 3 s? 1 s vavg = (vf + vi) 2 vavg = 15 m/s 2 s vavg = (0 m/s + 30 m/s) 2 3 s
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d = vit + ½at2 An apple falls to the ground in 3 s.
2.6 Free Fall: How Far An apple falls to the ground in 3 s. How high above ground was the apple when it first dropped? v = 30 m/s vavg = 15 m/s d = vit + ½at2 (a is g) 1 s d = (0 m/s)(3 s) + ½(10 m/s2)(3 s)2 2 s d = 45 m 3 s
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Linear Motion - Practice Problems
1) An angry mob lynches a physics teacher after receiving their grades. They throw the physics teacher off a tall building straight down with a velocity of 20 m/s. The teacher falls for 3.0 seconds landing on a stack cardboard boxes. From what height was he thrown? d = vi t + ½ at2
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Linear Motion - Practice Problems
2) Find the uniform acceleration that causes a car’s velocity to change from 32m/s to 96m/s in an 8.0s period. 3) A car with a velocity of 22m/s is accelerated uniformly at a rate of 1.6m/s for 6.8s. What is the final velocity? vf = vi + at vf = vi + at
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Linear Motion - Practice Problems
4) An airplane starts from rest and accelerates at a constant 3.0m/s2 for 30s before leaving the ground. How far did it move? How fast was it going at liftoff? d = vi t + ½ at2 vf = vi + at
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Linear Motion - Practice Problems
5) Your sister drops your house keys down to you from the second floor window. If you catch them 4.3 m from where your sister dropped them, what is the velocity of the keys when you catch them? d = vit + 1/2at
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gravity doesn’t act in a vacuum.
Quick Quiz! In a vacuum tube, a feather is seen to fall as fast as a coin. This is because … gravity doesn’t act in a vacuum. air resistance doesn’t act in a vacuum. greater air resistance acts on the coin. gravity is greater in a vacuum. 2.8
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Quick Quiz. If a falling object gains 10 m/s each second it falls, its acceleration can be expressed as _________. 10 m/s/s 10 m/s2 v = gt both A and B 2.5
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d = vit + ½at2 Quick Quiz. A rock falls 180 m from a cliff into the ocean. How long is it in free fall? 6 s 10 s 18 s 180 s 180 = (0)t + ½(10)t2 180 = ½(10)t2 180 = 5t2 180 = t2 5 36 = t2 √36 = t t = 6 s 2.6
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Graphs can visually describe relationships.
2.7 Graphs of Motion Equations, tables, and pictures are not the only way to describe relationships between distance, velocity, and acceleration. Graphs can visually describe relationships.
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constant velocity (a = 0)
2.7 Graphs of Motion distance vs. time: constant velocity (a = 0) slope = distance = v time distance (m) time (s)
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constant acceleration (+a)
2.7 Graphs of Motion distance vs. time: constant acceleration (+a) slope = distance = v time parabolic curve b/c time is squared (quadratic) d = ½at2 distance (m) time (s)
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distance vs. time → + → + ← – ← –
2.7 Graphs of Motion distance vs. time 1 2 dir:v : a : → + fast dir:v : a : → + slow d d t t 3 4 dir:v : a : dir:v : a : ← – fast ← – slow d d t t
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distance (m) time (s) Describe the motion. v : 0 a : 0 v : + a : 0
2.7 Graphs of Motion v : 0 a : 0 v : + a : 0 distance (m) time (s) moves forward at v = 5 m/s for 5 s. stops at 25 m (v = 0 m/s) for 5 s.
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velocity (m/s) time (s)
2.7 Graphs of Motion velocity vs. time: constant velocity (a = 0) slope = velocity = a time velocity (m/s) time (s)
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velocity (m/s) time (s)
2.7 Graphs of Motion velocity vs. time: constant acceleration (+a) velocity (m/s) slope = velocity = a time time (s)
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dir: v : a : right + (constant) dir: v : a : left – (constant) dir:
2.7 Graphs of Motion dir: v : a : right + (constant) dir: v : a : left – (constant) dir: v : a : right + (faster) + dir: v : a : right + (slower) – dir: v : a : left – (slower) + dir: v : a : left – (faster) –
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2.7 Graphs of Motion Consider the graph below. Describe the motion. (include all that are true): moving forward constant velocity positive velocity negative velocity slowing down changing directions speeding up positive acceleration constant acceleration negative acceleration + v t –
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The slope of a velocity-versus-time graph represents ____. distance
Quick Quiz! The slope of a velocity-versus-time graph represents ____. distance velocity acceleration time
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A. t = 0-1 s B. t = 1-4 s C. t = 4-9 s D. t = 9-12 s
WARM UP Consider the graph below. Describe the motion from... A. t = 0-1 s B. t = 1-4 s C. t = 4-9 s D. t = 9-12 s v = a = →, faster + v = a = →, faster + v = a = →, slower + – v = a = ←, faster – At what time is v = 0 m/s v = 0 m/s at 9 s
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Equations Summary distance time speed = ∆d t vavg = (vf + vi) 2 v =
NOT given on test distance time N speed = N given on test ∆d t vavg = (vf + vi) 2 v = v = vi + at ∆v t a = d = vit + ½at2 g = –10 m/s2
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WS Motion Graphs Begin your worksheet now. We will take all of class tomorrow to finish it.
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distance vs. time dir:v : __ dir:v : __ a : a : dir:v : __ dir:v : __
2.7 Graphs of Motion distance vs. time 1 dir:v : a : __ _____ 2 dir:v : a : __ ____ d d t t dir:v : a : 3 __ _____ 4 dir:v : a : __ _____ d d t t
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velocity vs. time dir: v : a : _____ __ (______) __ dir: v : a : _____
2.7 Graphs of Motion velocity vs. time dir: v : a : _____ __ (______) __ dir: v : a : _____ __ (______) __ dir: v : a : _____ __ (______) __ dir: v : a : _____ __ (______) __ dir: v : a : _____ __ (______) __ dir: v : a : _____ __ (______) __
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Acceleration & Velocity Graphs
VIDEO – Part 1 (7:19) Acceleration & Velocity Graphs
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Acceleration & Velocity Graphs
VIDEO – Part 2 (9:53) Acceleration & Velocity Graphs
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Acceleration & Velocity Graphs
VIDEO – Part 3 (7:45) Acceleration & Velocity Graphs
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