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

Vectors Vectors are represented with arrows 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. 5 m/s 42°

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

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) – rate of change in velocity

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

Acceleration – The Definition The RATE of CHANGE of VELOCITY D = Change = FINAL - INITIAL Dv = Final velocity – Initial velocity

Example Example: A Cessna Aircraft goes from 0 m/s to 60 m/s in 13 seconds. Calculate the aircraft’s acceleration. m/s 4.62 m/s/s s

Example Example: The Cessna now decides to land and goes from 60 m/s to 0 m/s in 11 s. Calculate the Cessna’s ? acceleration m/s - 5.45 m/s/s s

Free-Fall Acceleration Free fall is a condition where the only force acting on an object is gravity. If an object is in FREE FALL in the VERTICAL DIRECTION, the acceleration is due to GRAVITY. On Earth, gravitational acceleration is: It NEVER ceases to exist It ALWAYS works DOWN It is NEVER zero This is ONLY true in a vacuum (no air)

Acceleration due to Gravity Example: A person throws a ball straight upward into the air. Q1: What is the Acceleration at the TOP of its path? -9.8 m/s/s Q2: What is the VELOCITY at the TOP of its path? ZERO

-9.8 m/s/s - ALWAYS DOWNWARD Acceleration due to Gravity Q3: What is the magnitude(#value) and direction of the acceleration, HALF way up? -9.8 m/s/s - ALWAYS DOWNWARD Q4: What is the magnitude(#value) and direction of the acceleration, HALF way down? -9.8 m/s/s - ALWAYS DOWNWARD THE BOTTOM LINE: EVERYTHING will accelerate at -9.8 m/s/s in a VACUUM, that is any situation involving NO AIR.

Acceleration

Acceleration – Graphical Representation What is the acceleration from t=0s to t=3s? 60/3= 20 m/s/s What is the acceleration from t=3 s to t=5s? 0/2 = 0 m/s/s What is the acceleration from t=8s to t=9s? 0-60/1 = -60 m/s/s

Formula Summary vf = v0 + a t vavg = (v0 + vf ) / 2 For 1-D motion with constant acceleration: vf = v0 + a t vavg = (v0 + vf ) / 2 x = v0 t + ½ a t 2 vf2 – v02 = 2 a x (derivations to follow)

a = v / t (by definition) a = (vf – v0) / t  vf = v0 + a t Derivations a = v / t (by definition) a = (vf – v0) / t  vf = v0 + a t vavg = (v0 + vf ) / 2 will be proven when we do graphing. x = v t = ½ (v0 + vf) t = ½ (v0 + v0 + a t) t  x = v0 t + a t 2 (cont.)

Kinematics Derivations (cont.) vf = v0 + a t  t = (vf – v0) / a x = v0 t + a t 2  x = v0 [(vf – v0) / a] + a [(vf – v0) / a] 2  vf2 – v02 = 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 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? A brick is dropped from 100 m up. Find its impact velocity and air time. An arrow is shot straight up from a pit 12 m below ground at 38 m/s. Find its max height above ground. At what times is it at ground level?

Multi-step Problems 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? A dune buggy accelerates uniformly at 1.5 m/s2 from rest to 22 m/s. Then the brakes are applied and it stops 2.5 s later. Find the total distance traveled. Answer: 19.8 m/s Answer: 188.83 m

x t A B C Graphing ! 1 – D Motion 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

Graphing w/ Acceleration x Graphing w/ Acceleration C B t A D 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