Physics Intro & Kinematics Quantities Units Vectors Displacement Velocity Acceleration Kinematics Graphing Motion in 1-D.

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Presentation transcript:

Physics Intro & Kinematics Quantities Units Vectors Displacement Velocity Acceleration Kinematics Graphing Motion in 1-D

Some Physics Quantities Vector - quantity with both magnitude (size) and direction Scalar - quantity with magnitude only Vectors: Displacement Velocity Acceleration Momentum Force Scalars: Distance Speed Time Mass Energy

Mass vs. Weight On the moon, your mass would be the same, but the magnitude of your weight would be less. Mass Scalar (no direction) Measures the amount of matter in an object Weight Vector (points toward center of Earth) Force of gravity on an object

Vectors The length of the arrow represents the magnitude (how far, how fast, how strong, etc, depending on the type of vector). The arrow points in the directions of the force, motion, displacement, etc. It is often specified by an angle. Vectors are represented with arrows 42° 5 m/s

Units Quantity... Unit (symbol) Displacement & Distance... meter (m) Time... second (s) Velocity & Speed... (m/s) Acceleration... (m/s 2 ) Mass... kilogram (kg) Momentum... (kg · m/s) Force...Newton (N) Energy... Joule (J) Units are not the same as quantities!

SI Prefixes Little GuysBig Guys

Kinematics definitions Kinematics – branch of physics; study of motion Position (x) – where you are located Distance (d ) – how far you have traveled, regardless of direction Displacement (  x) – where you are in relation to where you started

Distance vs. Displacement You drive the path, and your odometer goes up by 8 miles (your distance). Your displacement is the shorter directed distance from start to stop (green arrow). What if you drove in a circle? start stop

Speed, Velocity, & Acceleration Speed (v) – how fast you go Velocity (v) – how fast and which way; the rate at which position changes Average speed ( v ) – distance / time Acceleration ( a ) – how fast you speed up, slow down, or change direction; the rate at which velocity changes

Speed vs. Velocity Speed is a scalar (how fast something is moving regardless of its direction). Ex: v = 20 mph Speed is the magnitude of velocity. Velocity is a combination of speed and direction. Ex: v = 20 mph at 15  south of west The symbol for speed is v. The symbol for velocity is type written in bold: v or hand written with an arrow: v

Speed vs. Velocity During your 8 mi. trip, which took 15 min., your speedometer displays your instantaneous speed, which varies throughout the trip. Your average speed is 32 mi/hr. Your average velocity is 32 mi/hr in a SE direction. At any point in time, your velocity vector points tangent to your path. The faster you go, the longer your velocity vector.

Acceleration Acceleration – how fast you speed up, slow down, or change direction; it’s the rate at which velocity changes. Two examples: t (s) v (mph) t (s) v (m/s) a = +2 mph / s a = -3 m / sm / s s = -3 m / s 2

Velocity & Acceleration Sign Chart V E L O C I T Y ACCELERATION ACCELERATION Moving forward; Speeding up Moving backward; Slowing down - Moving forward; Slowing down Moving backward; Speeding up

Acceleration due to Gravity 9.8 m/s 2 Near the surface of the Earth, all objects accelerate at the same rate (ignoring air resistance). a = -g = -9.8 m/s 2 Interpretation: Velocity decreases by 9.8 m/s each second, meaning velocity is becoming less positive or more negative. Less positive means slowing down while going up. More negative means speeding up while going down. This acceleration vector is the same on the way up, at the top, and on the way down!

Kinematics Formula Summary (derivations to follow) v f = v 0 + a t v avg = (v 0 + v f ) / 2  x = v 0 t + ½ a t 2 v f 2 – v 0 2 = 2 a  x For 1-D motion with constant acceleration:

Kinematics Derivations a =  v /  t (by definition) a = (v f – v 0 ) / t  v f = v 0 + a t v avg = (v 0 + v f ) / 2 will be proven when we do graphing.  x = v t = ½ (v 0 + v f ) t = ½ (v 0 + v 0 + a t) t   x = v 0 t + a t 2 (cont.)

Kinematics Derivations (cont.) v f = v 0 + a t  t = (v f – v 0 ) / a  x = v 0 t + a t 2   x = v 0 [(v f – v 0 ) / a ] + a [(v f – v 0 ) / a ] 2  v f 2 – v 0 2 = 2 a  x Note that the top equation is solved for t and that expression for t is substituted twice (in red) into the  x equation. You should work out the algebra to prove the final result on the last line.

Sample Problems 1.You’re riding a unicorn at 25 m/s and come to a uniform stop at a red light 20 m away. What’s your acceleration? 2.A brick is dropped from 100 m up. Find its impact velocity and air time. 3.An arrow is shot straight up from a pit 12 m below ground at 38 m/s. a.Find its max height above ground. b.At what times is it at ground level?

#1 Given Vo = 2.5 m/s, Vf = 0 m/s, ΔX = 20m Find a = (Vf 2 – Vo 2 ) / 2 ΔX m/s/s

#2 Given Vi = 0 m/s ΔX = 100m a = m/s/s Find Vf = (2a ΔX + Vo 2 ) ½ = m/s Find t = (Vf – Vi) /a 4.5 s

#3 Given Vo = 38 m/s a = -9.8 m/s/s At Vf = 0 ΔX = (Vf 2 – Vo 2 ) / 2a = 73.67m 73.67m – 12m = 61.67m Going up Vf = (2aΔX + Vo 2 ) 1/2 At 12m = 34.76m/s t = (Vf – Vo)/a =.33s

Practice Test #2 1. Ashlie decides to set a Judson record by climbing the steps in the Judson Jr. High School building as quickly as possible. If she averages.02 m/s climbing a total distance of 32 meters how long did it take her to climb from street level to the third floor?

2. Austin accelerates on his bicycle at.002 m/s/s from rest for 5 minutes. How fast will he be traveling at the end of 5 minutes?

3. German accelerates from rest to 135 km/hr in 25 seconds in an attempt to outrun the local Converse cop. How far did he travel during this failed attempt.

4. Jessica is an excellent swimmer and decides to swim as far as she can in Canyon Lake. If she begins her swim at 4.5 m/s and decreases her speed at a constant rate to 1.5 m/s in 10 minutes how far did she swim?

Multi-step Problems 1.How fast should you throw a kumquat straight down from 40 m up so that its impact speed would be the same as a mango’s dropped from 60 m? 2.A dune buggy accelerates uniformly at 1.5 m/s 2 from rest to 22 m/s. Then the brakes are applied and it stops 2.5 s later. Find the total distance traveled m/s m Answer:

Graphing ! x t A B C A … Starts at home (origin) and goes forward slowly B … Not moving (position remains constant as time progresses) C … Turns around and goes in the other direction quickly, passing up home 1 – D Motion

Graphing w/ Acceleration x A … Start from rest south of home; increase speed gradually B … Pass home; gradually slow to a stop (still moving north) C … Turn around; gradually speed back up again heading south D … Continue heading south; gradually slow to a stop near the starting point t A B C D

Tangent Lines t SLOPEVELOCITY Positive Negative Zero SLOPESPEED SteepFast GentleSlow FlatZero x On a position vs. time graph:

Increasing & Decreasing t x Increasing Decreasing On a position vs. time graph: Increasing means moving forward (positive direction). Decreasing means moving backwards (negative direction).

Concavity t x On a position vs. time graph: Concave up means positive acceleration. Concave down means negative acceleration.

Special Points t x P Q R Inflection Pt. P, RChange of concavity Peak or ValleyQTurning point Time Axis Intercept P, S Times when you are at “home” S

Curve Summary t x A B C D

All 3 Graphs t x v t a t

Graphing Tips Line up the graphs vertically. Draw vertical dashed lines at special points except intercepts. Map the slopes of the position graph onto the velocity graph. A red peak or valley means a blue time intercept. t x v t

Graphing Tips The same rules apply in making an acceleration graph from a velocity graph. Just graph the slopes! Note: a positive constant slope in blue means a positive constant green segment. The steeper the blue slope, the farther the green segment is from the time axis. a t v t

Real life Note how the v graph is pointy and the a graph skips. In real life, the blue points would be smooth curves and the green segments would be connected. In our class, however, we’ll mainly deal with constant acceleration. a t v t

Area under a velocity graph v t “forward area” “backward area” Area above the time axis = forward (positive) displacement. Area below the time axis = backward (negative) displacement. Net area (above - below) = net displacement. Total area (above + below) = total distance traveled.

Area The areas above and below are about equal, so even though a significant distance may have been covered, the displacement is about zero, meaning the stopping point was near the starting point. The position graph shows this too. v t “forward area” “backward area” t x

Area units Imagine approximating the area under the curve with very thin rectangles. Each has area of height  width. The height is in m/s; width is in seconds. Therefore, area is in meters! v (m/s) t (s) 12 m/s 0.5 s 12 The rectangles under the time axis have negative heights, corresponding to negative displacement.

Graphs of a ball thrown straight up x v a The ball is thrown from the ground, and it lands on a ledge. The position graph is parabolic. The ball peaks at the parabola’s vertex. The v graph has a slope of -9.8 m/s 2. Map out the slopes! There is more “positive area” than negative on the v graph. t t t

Graph Practice Try making all three graphs for the following scenario: 1. Schmedrick starts out north of home. At time zero he’s driving a cement mixer south very fast at a constant speed. 2. He accidentally runs over an innocent moose crossing the road, so he slows to a stop to check on the poor moose. 3. He pauses for a while until he determines the moose is squashed flat and deader than a doornail. 4. Fleeing the scene of the crime, Schmedrick takes off again in the same direction, speeding up quickly. 5. When his conscience gets the better of him, he slows, turns around, and returns to the crash site.

Kinematics Practice A catcher catches a 90 mph fast ball. His glove compresses 4.5 cm. How long does it take to come to a complete stop? Be mindful of your units! 2.24 ms Answer

Uniform Acceleration When object starts from rest and undergoes constant acceleration: Position is proportional to the square of time. Position changes result in the sequence of odd numbers. Falling bodies exhibit this type of motion (since g is constant). t :  x = 1  x = 3  x = 5 ( arbitrary units ) x :  x = 7

Spreadsheet Problem We’re analyzing position as a function of time, initial velocity, and constant acceleration. x,  x, and the ratio depend on t, v 0, and a.  x is how much position changes each second. The ratio (1, 3, 5, 7) is the ratio of the  x’s. Make a spreadsheet like this and determine what must be true about v 0 and/or a in order to get this ratio of odd numbers. Explain your answer mathematically.

Relationships Let’s use the kinematics equations to answer these: 1. A mango is dropped from a height h. a. If dropped from a height of 2 h, would the impact speed double? b.Would the air time double when dropped from a height of 2 h ? 2.A mango is thrown down at a speed v. a.If thrown down at 2 v from the same height, would the impact speed double? b.Would the air time double in this case?

Relationships (cont.) 3.A rubber chicken is launched straight up at speed v from ground level. Find each of the following if the launch speed is tripled (in terms of any constants and v). a.max height b.hang time c.impact speed 3 v3 v 9 v 2 / 2 g 6 v / g Answers