Lecture 2

1. Lecture 2 Chapter 2.1-2.4 • Define Position, Displacement & Distance • Distinguish Time and Time Interval • Define Velocity (Average and Instantaneous), Speed • Define Acceleration • Understand algebraically, through vectors, and graphically the relationships between position, velocity and acceleration • Comment on notation

2. Informal Reading Quiz Displacement, position, velocity & acceleration are the main quantities that we will discuss today. Which of these 4 quantities have the same units • Velocity & position • Velocity & acceleration • Acceleration & displacement • Position & displacement • Position & acceleration

3. Range of Lengths Distance Length (m) Radius of Visible Universe 1 x 1026 To Andromeda Galaxy 2 x 1022 To nearest star 4 x 1016 Earth to Sun 1.5 x 1011 Radius of Earth 6.4 x 106 Willis Tower 4.5 x 102 Football Field 1 x 102 Tall person 2 x 100 Thickness of paper 1 x 10-4 Wavelength of blue light 4 x 10-7 Diameter of hydrogen atom 1 x 10-10 Diameter of proton1 x 10-15

4. Range of Times IntervalTime (s) Age of Universe 5 x 1017 Age of Grand Canyon 3 x 1014 Avg age of college student 6.3 x 108 One year3.2 x 107 One hour3.6 x 103 Light travel from Earth to Moon 1.3 x 100 One cycle of guitar A string 2 x 10-3 One cycle of FM radio wave 6 x 10-8 One cycle of visible light 1 x 10-15 Time for light to cross a proton 1 x 10-24 World’s most accurate timepiece: Cesium fountain Atomic Clock Lose or gain one second in some 138 million years

5. One-Dimension Motion (Kinematics) Position, Displacement, Distance • Position: Reflects where you are. • KEY POINT 1: Magnitude, Direction, Units • KEY POINT 2: Requires a reference point (Origin) Origins are arbitrary

6. New York (Origin) One-Dimension Motion (Kinematics) Position, Displacement, Distance • Position: Reflects where you are. • KEY POINT 1: Magnitude, Direction, Units • KEY POINT 2: Requires a reference point (Origin) Origins are arbitrary Boston Example: Where is Boston ? Choose origin at New York Boston is 212 milesnortheastof New York OR Boston is 150 mileseast and 150 milesnorth of New York

7. Boston Path 1 New York Path 2 One-Dimension Motion (Kinematics) Position, Displacement, Distance • Getting from New York to Boston requires a PATH • Path defines what places we pass though • Displacement: Change in position • Requires a time interval • Any point on the path must be associated with a specific time ( t1, t2, t3, ….) • Path 1 and Path 2 have the same change in position so they the same displacement. • However the distance travelled is different.

8. 10 meters Pat +x -x 10 meters O Motion in One-Dimension (Kinematics) Position • Position along a line; references xi and ti : • At time = 0 seconds Pat is 10 meters to the right of the lamp • Origin  lamp • Positive direction  to the right of the lamp • Position vector ( xi , ti) or (10 m, 0.0 s) • Particle representation

9. 10 meters 15 meters xi Pat xf Δx O Displacement • One second later Pat is 15 meters to the right of the lamp • At t = 1.0 s the position vector is ( xf , tf ) or (15 m, 1.0 s) • Displacement is just change in position • x ≡ xf – xi • There is also a change in time • t ≡ tf – ti

10. 10 meters 15 meters xf xi Δx Pat Displacement • Putting it all together x = xf - xi = 5 metersto the right ! t = tf - ti= 1 second Relating x to t yields average velocity O

11. Average VelocityChanges in positionvsChanges in time • Average velocity = displacement per time increment , includes BOTH magnitude and direction • Pat’s average velocity was 5 m / s to the right

12. Average Speed • Average speed, vavg, reflects a magnitude • “How fast” without the direction. • References the total distance travelled • Pat’s average speed was 5 m / s NOTE: Serway’s notation varies from other texts (There really is no standard)

13. 30 20 x (meters) 10 0 1 2 3 4 Pat on tour (graphical representation) • Pat is walking from and to the lamp (at the origin). (x1 , t1) = (10 m, 0.0 sec) (x2 , t2) = (15 m, 1.0 sec) (x3 , t4) = (30 m, 2.0 sec) (x4 , t4) = (10 m, 3.0 sec) (x5 , t5) = ( 0 m, 4.0 sec) Compare displacement distance avg. vel. avg. speed t = 1 s x1,2 = x2 – x1 = 5 m d = 5 m vx,avg= 5 m/s vx,avg= 5 m/s t = 2 s x1,3 = x3 – x1 = 20 m d = 20 m vx,avg= 10 m/s vx,avg= 10 m/s t = 3 s x1,4 = x4 – x1 = 0 m d = 40 mvx,avg= 0 m/s vx,avg= 13 m/s Here d = |x1,2 | + |x2,3 |+ |x3,4 | = 5 m + 15 m + 20 m = 40 m Speed and velocity measure different things! t (seconds)

14. Calculating path distance in general

15. Exercise 2Average Velocity x (meters) 6 4 2 0 t (seconds) 1 2 3 4 -2 What is the magnitude of the average velocity over the first 4 seconds ? (A) -1 m/s (B) 4 m/s (C) 1 m/s (D) not enough information to decide.

16. Average Velocity Exercise 3What is the average velocity in the last second (t = 3 to 4) ? • 2 m/s • 4 m/s • 1 m/s • 0 m/s x (meters) 6 4 2 -2 t (seconds) 1 2 3 4

17. x (meters) 6 4 2 -2 t (seconds) 1 2 3 4 turning point Average Speed Exercise 4What is the average speed over the first 4 seconds ?0 m to -2 m to 0 m to 4 m  8 meters total • 2 m/s • 4 m/s • 1 m/s • 0 m/s

18. Instantaneous velocity • Limiting case as the change in time  0 x • Yellow lines are average velocities instantaneous velocity at t = 0 s t 0 • As Dt  0 velocity is the tangent to the curve (& path) • Dashed green line is vx

19. Instantaneous speed • Just the magnitude of the instantaneous velocity

20. Exercise 5Instantaneous Velocity x (meters) 6 4 2 -2 t (seconds) 1 2 3 4 What is the instantaneous velocity at the fourth second? (B) 0 m/s (A) 4 m/s (C) 1 m/s (D) not enough information to decide.

21. x (xf , tf) Dx (xi , ti) Dt t 0 Special case: Instantaneous velocity is constant • Slope is constant over a time Dt.

22. Special case: Instantaneous velocity is constant • Slope is constant over a time Dt. x (xf , tf) Dx (xi , ti) t Dt 0 • Given Dt, xi and vx we can deduce xf and this reflects the area under the velocity curve

23. 30 20 x (meters) 10 0 1 2 3 4 20 10 0 vx (m/s) 2 4 1 3 -10 -20 t (seconds) Now multiple vx; Pat’s velocity plot (x1 , t1) = (10 m, 0.0 s) (x2 , t2) = (15 m, 1.0 s) (x3 , t3) = (30 m, 2.0 s) (x4 , t4) = (10 m, 3.0 s) (x5 , t5) = ( 0 m, 4.0 s) t (seconds)

24. Home exercise 6(and some things are easier than they appear) A marathon runner runs at a steady 15 km/hr. When the runner is 7.5 km from the finish, a bird begins flying from the runner to the finish at 30 km/hr. When the bird reaches the finish line, it turns around and flies back to the runner, and then turns around again, repeating the back-and-forth trips until the runner reaches the finish line. How many kilometers does the bird travel? A. 10 km B. 15 km C. 20 km D. 30 km

25. Objects with slowly varying velocities • Change of the change…. changes in velocity with time give average acceleration x t vx Dt Dvx t

26. x t vx t And finally instantaneous acceleration t ax t

27. vx t Example problem • A car moves to the right first for 2.0 sec at 1.0 m/s and then 4.0 seconds at 2.0 m/s. • What was the average velocity? • Two legs with constant velocity but …. Slope of x(t) curve

28. vx t Example problem • A particle moves to the right first for 2.0 seconds at 1.0 m/s and then 4.0 seconds at 2.0 m/s. • What was the average velocity? • Two legs with constant velocity but …. • We must find the total displacement (x2 –x0) • And x1 = x0 + v0 (t1-t0)x2 = x1 + v1 (t2-t1) • Displacement is (x2 - x1) + (x1 – x0) = v1 (t2-t1) + v0 (t1-t0) • x2 –x0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6.0 s or 1.7 m/s Slope of x(t) curve

29. Position, velocity & acceleration • All are vectors! • Cannot be used interchangeably (different units!) (e.g., position vectors cannot be added directly to velocity vectors) • But the directions can be determined • “Change in the position” vector vs. time gives the direction of the velocity vector • “Change in the velocity” vector vs. time gives the direction of the acceleration vector • Given x(t)  vx(t)  ax (t) • Given ax (t)  vx (t)  x(t)

30. Assignment • Reading for Tuesday’s class • All of Chapter 2