Ch 2. Motion in a Straight Line Definitions 1. Kinematics - Motion Kinetic Energy - Energy associated with motion 2. Motion in physics is broken down into categories a.) Translational Motion - motion such that an object moves from one position to another along a straight line. b.) Rotational Motion - motion such that an object moves from one position to another along a circular path. c.) Vibrational Motion - motion such that an object moves back and forth in some type of periodicity. One Dimensional (x-axis only) “Dinophysics : Velocity- Raptor” Straight Line Spinning Motion in 3-D space can be complicated Up and Back

Example: Diatomic Molecule Moving Through Space. i f Net Translation X - Dir Note: In this chapter all objects are going to be considered POINT PARTICLES – No Spatial Extent – No Rotations – No Vibrations

Speed 1.Speed - How fast an object is moving regardless of what direction it is moving. One way travel = 130 mi. Total Distance Traveled = 260 mi. Total time elapsed = 5.2 hrs. or (5 hrs 12 min) Example 1 Traveling from your parking space at Conestoga to New York City and back to Conestoga. Find your average speed for the round trip. ≡ Equality by Definition Speed Calculations are EASY Always distance / time Round Trip Average Speed

Displacement - Change in position (straight line distance with direction) Must specify a coordinate system. ∆ “Delta” Delta x is the displacement or change in the x position Example : Cartesian coordinate system up back Mathematical Notation for Direction x(m) y(m) x i = 2m x 1 x f = 6m x 2 x i = x initial = Initial Position x f = x final = Final Position

Average Velocity Avg. Velocity - How fast an object is moving and in what direction it is moving.  ≡ Equality by Definition

Notation for Displacement & Velocity = x “hat”, and has a value of one. The sole purpose of is to indicate the direction Example Problem: A particle initially at position x = 5 m at time t= 2 s moves to position x = -2 m and arrives at time t = 4 s. a.) Find the displacement of the particle. b.) Find the average speed and velocity of the particle.

Example Problem 1 revisited Example 1. Traveling from your parking space at Conestoga to New York City and back to Conestoga. The straight line distance from Conestoga to Y is 97 mi. One way travel = 130 mi. Total Distance Traveled = 260 mi. Travel time Con. to NY = 2.6 hrs. Travel time NY to Con. = 2.6 hrs. a.) What was the avg speed from Conestoga to NY? b.) What was the avg velocity from Conestoga to NY? c.) What was the avg speed for the round trip? d.) What was the avg velocity for the round trip? Speed in PATH DEPENDENT. Velocity is PATH INDEPENDENT. It only depends on the initial and final positions.

Scalar vs. Vector Quantities Scalar - Quantity that has magnitude only. - Mass- Speed - Length- Energy Vector - A quantity that has both magnitude and direction. - Position- Acceleration - Velocity- Forces A number (with units) that describes how big or small Example: Length vs. Position = x “hat”, and is called a unit vector in the x-direction. It has a magnitude of one (hence the name unit) and is used solely to specify direction. ScalarVector

Concepts Check – The Negatives Q. Can velocity be negative? A. YES! Negative displacement means an object moved backwards. E.g. An object with a displacement ∆x of moved backwards 10m. A. NO! – The least speed an object can have is zero – it is at rest Q. Can speed be negative? Q. Can displacement be negative? A. NO! – The least distance an object can move is zero – it is at rest Q. Can distance be negative? A. YES! Negative velocity means an object is moving backwards. E.g. An object moving is moving backwards with a speed of 10 m/s

Position vs. Time Graph xY1Y1 Y2Y2 Time (s)Position (m) Both of these movements describe an object moving in one dimension along the x-axis! NOT up and to the right!  x=rise  t=run 0 25 m 20 m 15 m 10 m 5m5m Movement 1 Movement 2 ● ● ● ● ● ●● ● ● ● 1s1s 2s2s 3s3s 4s4s 5s5s 1s1s 2s2s 3s3s 4s4s 5s5s Movement 1 Movement 2

Position vs. Time Graph for a Complete Trip xy Time (s)Position (m) Find the average velocity as the object moves from: a.) A to B b.) B to C c.) C to D d.) A to E Slope of the secant line is v avg

Velocity vs. Time (Constant Velocity) Area xx Slope rise run v = height ∆t = base

Velocity vs. Time Graph for a Complete Trip +X − X Velocity vs. Time Graph for a Complete Trip Area = 200 m Area = -50 m Area = -250 m

∆t = 1.5 sec ● ● ● Instantaneous Velocity recall: (Average velocity) Consider the function x(t): A. B. ∆t = 0.2 sec The instantaneous velocity at the time t = t i is the limiting value we get by letting the upper value of the t f approach t i. Mathematically this is expressed as: The velocity function is the time derivative of the position function. Differentiation (Calculus)

Acceleration When the instantaneous velocity of a particle is changing with time, the particle is accelerating Units: Example: If a particle is moving with a velocity in the x-direction given by a.) What is the average acceleration over the time interval (Average Acceleration)

Example: Instantaneous Acceleration a.) Find a avg. over the time interval 5  t  8 b.) What is the acceleration at time t = 6 s ? c.) What is the acceleration when the velocity of the particle is zero? Stopped Slope of tangent – pick 2 points on the tangent line. Answer will we smaller than the answer to part a, Objects with zero velocity can be accelerating!

Positive and Negative Accelerations v(m/s) t (s) A C E D B F A→B: B→C: C→D: D→E: E→F: Moving Forward Stopped 0 Moving Backward Slowing Down, Moving Backward, Pt. B=Stop Slowing Down, Moving Forward, Pt. E=Stop Speeding Up, Moving Forward Constant Speed, Moving Forward Speeding Up, Moving Backward

Special Case: Constant Acceleration a(m/s 2 ) t (s) t i = 0t f = t a 0 We make the assumption that the acceleration does not change. Near the surface of the earth, (where most of us spend most of our time) the acceleration due to gravity is approximately constant a g = 9.8 m/s 2 Area! Slope! v(m/s) t (s) t i = 0t f = t vfvf 0 x(m) t (s) t i = 0 t f = t xfxf xixi Area! Slope! vivi 1. 2.

3. Solving for the 3 rd constant acceleration equation Solve equation 1 for t and substitute t into equation 2 to get the following equation.

tftf t/2 y(t) tftf t/2 FREE-FALL ACCELERATION(9.8 m/s 2 = 32 ft/s 2 ) Consider a ball is thrown straight up. It is in “Free Fall” the moment it leaves you hand. Plot y(t) vs. t for the example above. Plot v(t) vs. t Ground Level top Moving up Stopped 0 Moving down

x(t) Area Under Curve Slope v(t) a(t) FINAL NOTES ON CH 2. Remember, when going between the following graphs Problem Solving with the constant acceleration equations 1.Write down all three equations in the margin 2.a =  9.8 m/s 2 for free fall problems 3.Analyze the problem in terms of initial and final sections.