1. September 15, 2009Physics I Lesson 2 Dr J. Tison 1 Kinematics: One Dimensional Motion

2. September 15, 2009Physics I Lesson 2 Dr J. Tison 2 Short History of Classical Mechanics: Greeks – 17 th and 18 th Century Aristotle and other Greek philosophers –Theorized, no experimental data Muslim physicists in 11 th and 12 th centuries –Scientific methods first applied Inertia, momentum, force and mass, gravity –Precursor to Newton’s Laws of Motion Modern interpretation of principles of motion –Keppler (-1630) Planetary motion observations by Tyco Brahe –Galileo (-1642) Acceleration of rolling ball on inclined plane –Newton (-1727) Principia… Laws of Motion

3. September 15, 2009Physics I Lesson 2 Dr J. Tison 3 Classical Mechanics Kinematics –How objects move –Equations relating Distance, velocity, acceleration, time Dynamics –Why things move –Equations relating Momentum, force, potential energy, pressure and power.

4. September 15, 2009Physics I Lesson 2 Dr J. Tison 4 Motion Types Translational –1 Dimension (this lesson) Straight line motion –2 & 3 Dimension (later lesson) Trajectory motion Rotational (later lessons) Combined Translation and Rotation (later lessons)

5. September 15, 2009Physics I Lesson 2 Dr J. Tison 5 1D Motion: Model and Concepts ‘Idealized’ particle –Mathematical point –No spatial extent (no size)

6. September 15, 2009Physics I Lesson 2 Dr J. Tison 6 1D Motion: Frame of Reference Measurements (position, distance, speed) –Requires reference point –Observer vs Object

7. September 15, 2009Physics I Lesson 2 Dr J. Tison 7 1D Motion: Frame of Reference Coordinate System –Cartesian coordinate system (x,y axes) x (horizontal motion); y (vertical motion) –Position, and distance along x or y axis –Displacement vs distance Change in position vs how far object moved

8. September 15, 2009Physics I Lesson 2 Dr J. Tison 8 Motion along a Line Distance moved along x axis –∆ x = x 2 -x 1 Time moved along x axis –∆ t = t 2 -t 1 Average speed moved along x axis –∆ v = ∆ x/ ∆ t ∆ is symbol for ‘change’:final value less initial value (displacement) 201030 x 40 0 y Distance (m) S E N W -20 - 10 30 2010 -30 Distance =a+b ba Displacement = b - a b a ba (x 1, t 1 )(x 2, t 2 ) 40

9. September 15, 2009Physics I Lesson 2 Dr J. Tison 9 Displacement Vector Vectors vs Scalars –Vector symbol V (B,I,S)V (B,I,S) –Magnitude/direction Business woman walking –Displacement vectors aa = 30m east bb = 60m west cNet displacement vector ( c )  a b  a + b = 30m west –Total distance (not vector) |b|+ |a| = 90m S E N W -20 - 10 30m2010-30 b a c

10. September 15, 2009Physics I Lesson 2 Dr J. Tison 10 Speed & Velocity Speed (scalar) –How far object travels in a given time interval average speed = (distance traveled)/(time elapsed) Velocity (vector) –How far and in what direction traveled in given time interval average velocitydisplacementaverage velocity = ( displacement )/(time elapsed) Vector = scalar (magnitude) + direction

11. September 15, 2009Physics I Lesson 2 Dr J. Tison 11 Average Speed & Average Velocity: Walking Businesswoman Average speed –∆ x(distance)/ ∆ t =  [|x 2 - x 1 | + |x 3 - x 2 |]/|t 3 - t 1 | = (30+60)/(90) = 1m/s Average velocity x –∆ x (displacement)/ ∆ t =  (x 3 - x 1 )]/(t 3 - t 1 ) = (-20-10)/(90) = -30/90 = -0.33m/s (West) S N W -20 - 10 302010 -30 b a 40m tx t 1 =0 at x 1 =10m tx t 2 =30s at x 2 = 40m tx t 3 =90s at x 3 =-20m x3x3x3x3 x1x1x1x1 x2x2x2x2

12. September 15, 2009Physics I Lesson 2 Dr J. Tison 12 Instantaneous Velocity Constant or changing velocity Instantaneous velocity is.. –Velocity at an instant of time –“Average velocity during an infinitesimal time interval” v =lim (∆t →0) [ ∆ x /∆ t] v: instantaneous v v : average lim ∆ t →0 Velocity (mph) Time (h) 20 00.10.40.20.3 40 60 20 00.10.40.20.3 40 60 Velocity (mph) Average velocity ● Instantaneous velocity lim ∆ t →0

13. September 15, 2009Physics I Lesson 2 Dr J. Tison 13 Acceleration Changing velocity = acceleration –Average acceleration = (change of velocity/(time elapsed) –a –a = ( v 2 - v 1) /(t 2 - t 2 ) = ∆ v / ∆ t Instantaneous acceleration –a =lim (∆t →0) [ ∆ v /∆ t] t =0, =0 t 1 =0, v 1 =0 t =1.0S, = 15mph t =1.0S, v = 15mph t =2s, =30mph t =2s, v 1 =30mph t =5s, =60mph t =5s, v =60mph a =

14. September 15, 2009Physics I Lesson 2 Dr J. Tison 14 Acceleration vs velocity Acceleration » speeding up –How quickly velocity changes Velocity –How quickly position changes Deceleration (a vector) » slowing down –Frame of reference » + a to right, - a to left Acceleration vector is opposite to velocity vector

15. September 15, 2009Physics I Lesson 2 Dr J. Tison 15 Constant Acceleration Equations relating x, v, a, and t –Initial conditions (t 1, x 1 ) = (t 0, x 0 ) = (0,0) (t 2, x 2 ) = (t, x) Condition of motion –Magnitude of acceleration is constant a –Motion in a straight line. » a (instantaneous) = a (average)

16. September 15, 2009Physics I Lesson 2 Dr J. Tison 16 Constant Acceleration v x x x v = ∆ x / ∆ t = ( x - x 0 )/t » x xv » x = x 0 + v t t x xx0xx0 x t v vv0vv0 v t ∆x = x-x 0 ∆t = t-t 0 = t t =0 ∆v = v-v 0 ∆t = t-t 0 = t t av v v a = ∆ v / ∆ t = ( v - v 0 )/t » v va » v = v 0 + a t

17. September 15, 2009Physics I Lesson 2 Dr J. Tison 17 Constant Acceleration x v x = x 0 + t; v –where is average velocity –From time t = 0 → t = t, vvvelocity goes from v 0 → v v vv = ( v 0 + v )/2 v vv0vv0 v t ∆v = v-v 0 ∆t = t-t 0 = t t

18. September 15, 2009Physics I Lesson 2 Dr J. Tison 18 Constant Acceleration x xx0xx0 x t ∆x = x-x 0 ∆t = t-t 0 = t t =0 x v x = x 0 + t Substitute: v vv = ( v 0 + v )/2 x xvv x = x 0 + [( v 0 + v )/2]t x vv x = x 0 + ½( v 0 t + v t) x vva x = x 0 + ½( v 0 t )+½( v 0 + a t)(t) x va x = x 0 + v 0 t +½ a t 2

19. September 15, 2009Physics I Lesson 2 Dr J. Tison 19 Kinematic Equations of Motion For Constant Acceleration v vav v a v = v 0 + a t » t = ( v - v 0 )/a x va x = x 0 + v 0 t +½ a t 2 Substitute for t x vv v a av v a x = x 0 + v 0 ( v - v 0 )/a +½ a [( v - v 0 )/a] 2 Simplify v v ax - ( v 2 - v 0 2 ) = 2 a ( x - x 0 )

20. September 15, 2009Physics I Lesson 2 Dr J. Tison 20 Kinematic Equations of Motion For Constant Acceleration v va v = v 0 + a t x va x = x 0 + v 0 t +½ a t 2 v v ax - ( v 2 - v 0 2 ) = 2 a ( x - x 0 )

21. September 15, 2009Physics I Lesson 2 Dr J. Tison 21 Falling Objects: Uniform (constant)Acceleration Acceleration due to gravity: g = 9.80m/s 2 = 32.0ft/s 2 Galileo studied falling objects – –Postulated: All objects fall with same constant acceleration in absence of air or other resistance – –Predicted: Free falling object travels a distance, y, proportional to time squared (t 2 ) y ~ t 2 – –Derived: Free fall equation y vg y = y 0 + v 0 t +½ g t 2

22. September 15, 2009Physics I Lesson 2 Dr J. Tison 22 Falling Objects: Uniform (constant)Acceleration No air resistance –In a vacuum –Feathers and boulders fall with same acceleration ~ g – x ~ t 2 = t = ∆ y ∆ y

23. September 15, 2009Physics I Lesson 2 Dr J. Tison 23 Graphical Analysis of Linear Motion Linear Equation: (y = mx + b) m is slope b is y intercept Eq of Motion for velocity x = x 0 + vt Slope: v = ∆ x/ ∆ t Y intercept: x 0 x t ∆x = x-x 0 ∆t = t x0x0

24. September 15, 2009Physics I Lesson 2 Dr J. Tison 24 Graphical Analysis of Linear Motion Eq of Motion for position v = v 0 + at Slope: a = ∆ v/ ∆ t y intercept: v 0 t ∆v = v - v 0 ∆t = t v0v0v0v0 v