1. Physics 201: Lecture 2, Pg 1 Lecture 2 Chapter 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. Physics 201: Lecture 2, Pg 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 A. Velocity & position B. Velocity & acceleration C. Acceleration & displacement D. Position & displacement E. Position & acceleration

3. Physics 201: Lecture 2, Pg 3 Range of Lengths Distance Length (m) Radius of Visible Universe 1 x 10 26 To Andromeda Galaxy 2 x 10 22 To nearest star 4 x 10 16 Earth to Sun 1.5 x 10 11 Radius of Earth 6.4 x 10 6 Willis Tower 4.5 x 10 2 Football Field 1 x 10 2 Tall person 2 x 10 0 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 proton 1 x 10 -15

4. Physics 201: Lecture 2, Pg 4 Range of Times Interval Time (s) Age of Universe 5 x 10 17 Age of Grand Canyon 3 x 10 14 Avg age of college student 6.3 x 10 8 One year 3.2 x 10 7 One hour 3.6 x 10 3 Light travel from Earth to Moon 1.3 x 10 0 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. Physics 201: Lecture 2, Pg 5 One-Dimension Motion (Kinematics) Position, Displacement, Distance l Position: Reflects where you are.  KEY POINT 1: Magnitude, Direction, Units  KEY POINT 2: Requires a reference point (Origin) Origins are arbitrary

6. Physics 201: Lecture 2, Pg 6 One-Dimension Motion (Kinematics) Position, Displacement, Distance l Position: Reflects where you are.  KEY POINT 1: Magnitude, Direction, Units  KEY POINT 2: Requires a reference point (Origin) Origins are arbitrary Example: Where is Boston ? New York Choose origin at New York Boston is 212 miles northeast of New York OR Boston is 150 miles east and 150 miles north of New York Boston New York (Origin)

7. Physics 201: Lecture 2, Pg 7 One-Dimension Motion (Kinematics) Position, Displacement, Distance l Getting from New York to Boston requires a PATH  Path defines what places we pass though l Displacement: Change in position  Requires a time interval  Any point on the path must be associated with a specific time ( t 1, t 2, t 3, ….) l Path 1 and Path 2 have the same change in position so they the same displacement. l However the distance travelled is different. Boston New York Path 1 Path 2

8. Physics 201: Lecture 2, Pg 8 Motion in One-Dimension (Kinematics) Position l Position along a line; references x i and t i :  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 ( x i, t i ) or (10 m, 0.0 s)  Particle representation 10 meters Pat O -x +x 10 meters

9. Physics 201: Lecture 2, Pg 9 Displacement l One second later Pat is 15 meters to the right of the lamp l At t = 1.0 s the position vector is ( x f, t f ) or (15 m, 1.0 s) l Displacement is just change in position   x ≡ x f – x i There is also a change in time   t ≡ t f – t i 10 meters 15 meters Pat xixi O xfxf ΔxΔx

10. Physics 201: Lecture 2, Pg 10 Displacement l Putting it all together  x = x f - x i = 5 meters to the right !  t = t f - t i = 1 second Relating  x to  t yields average velocity O Pat xixi 10 meters 15 meters xfxf ΔxΔx

11. Physics 201: Lecture 2, Pg 11 Average Velocity Changes in position vs Changes 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. Physics 201: Lecture 2, Pg 12 Average Speed avg l Average speed, v avg, reflects a magnitude l “How fast” without the direction. l 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. Physics 201: Lecture 2, Pg 13 Pat on tour (graphical representation) l Pat is walking from and to the lamp (at the origin). (x 1, t 1 ) = (10 m, 0.0 sec) (x 2, t 2 ) = (15 m, 1.0 sec) (x 3, t 4 ) = (30 m, 2.0 sec) (x 4, t 4 ) = (10 m, 3.0 sec) (x 5, t 5 ) = ( 0 m, 4.0 sec) Compare displacement distance avg. vel. avg. speed t = 1 s  x 1,2 = x 2 – x 1 = 5 m d = 5 m v x,avg = 5 m/s v x,avg = 5 m/s t = 2 s  x 1,3 = x 3 – x 1 = 20 m d = 20 m v x,avg = 10 m/s v x,avg = 10 m/s t = 3 s  x 1,4 = x 4 – x 1 = 0 m d = 40 m v x,avg = 0 m/s v x,avg = 13 m/s Here d = |  x 1,2 | + |  x 2,3 |+ |  x 3,4 | = 5 m + 15 m + 20 m = 40 m Speed and velocity measure different things! t (seconds) x (meters) 10 30 20 1243 0

14. Physics 201: Lecture 2, Pg 14 Calculating path distance in general

15. Physics 201: Lecture 2, Pg 15 Exercise 2 Average Velocity x (meters) t (seconds) 2 6 -2 4 What is the magnitude of the average velocity over the first 4 seconds ? (A) -1 m/s(D) not enough information to decide. (C) 1 m/s(B) 4 m/s 1243 0

16. Physics 201: Lecture 2, Pg 16 Average Velocity Exercise 3 What is the average velocity in the last second (t = 3 to 4) ? A. 2 m/s B. 4 m/s C. 1 m/s D. 0 m/s x (meters) t (seconds) 2 6 -2 4 1243

17. Physics 201: Lecture 2, Pg 17 Average Speed Exercise 4 What is the average speed over the first 4 seconds ? 0 m to -2 m to 0 m to 4 m  8 meters total A. 2 m/s B. 4 m/s C. 1 m/s D. 0 m/s x (meters) t (seconds) 2 6 -2 4 1243 turning point

18. Physics 201: Lecture 2, Pg 18 Instantaneous velocity Limiting case as the change in time  0 x t 0 As  t  0 velocity is the tangent to the curve (& path) Dashed green line is v x Yellow lines are average velocities instantaneous velocity at t = 0 s

19. Physics 201: Lecture 2, Pg 19 Instantaneous speed Just the magnitude of the instantaneous velocity

20. Physics 201: Lecture 2, Pg 20 What is the instantaneous velocity at the fourth second? (A) 4 m/s(D) not enough information to decide. (C) 1 m/s (B) 0 m/s x (meters) t (seconds) 2 6 -2 4 1243 Exercise 5 Instantaneous Velocity

21. Physics 201: Lecture 2, Pg 21 Special case: Instantaneous velocity is constant Slope is constant over a time  t. x t 0 tt xx (x i, t i ) (x f, t f )

22. Physics 201: Lecture 2, Pg 22 Special case: Instantaneous velocity is constant Slope is constant over a time  t. x t 0 tt xx (x i, t i ) (x f, t f ) Given  t, x i and v x we can deduce x f and this reflects the area under the velocity curve

23. Physics 201: Lecture 2, Pg 23 Now multiple v x ; Pat’s velocity plot t (seconds) x (meters) 10 30 20 1243 0 v x (m/s) 10 20 1 24 3 0 -10 -20 t (seconds) (x 1, t 1 ) = (10 m, 0.0 s) (x 2, t 2 ) = (15 m, 1.0 s) (x 3, t 3 ) = (30 m, 2.0 s) (x 4, t 4 ) = (10 m, 3.0 s) (x 5, t 5 ) = ( 0 m, 4.0 s)

24. Physics 201: Lecture 2, Pg 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. Physics 201: Lecture 2, Pg 25 Objects with slowly varying velocities x vxvx t t l Change of the change…. changes in velocity with time give average acceleration tt vxvx

26. Physics 201: Lecture 2, Pg 26 And finally instantaneous acceleration t axax t x vxvx t t

27. Physics 201: Lecture 2, Pg 27 Example problem l 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. l What was the average velocity? l Two legs with constant velocity but …. Slope of x(t) curve vxvx t

28. Physics 201: Lecture 2, Pg 28 Example problem l 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. l What was the average velocity? l Two legs with constant velocity but …. l We must find the total displacement (x 2 –x 0 ) l And x 1 = x 0 + v 0 (t 1 -t 0 ) x 2 = x 1 + v 1 (t 2 -t 1 ) l Displacement is (x 2 - x 1 ) + (x 1 – x 0 ) = v 1 (t 2 -t 1 ) + v 0 (t 1 -t 0 ) l x 2 –x 0 = 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 vxvx t

29. Physics 201: Lecture 2, Pg 29 Position, velocity & acceleration l All are vectors! l Cannot be used interchangeably (different units!) (e.g., position vectors cannot be added directly to velocity vectors) l 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 l Given x(t)  v x (t)  a x (t) l Given a x (t)  v x (t)  x(t)

30. Physics 201: Lecture 2, Pg 30 Assignment l Reading for Tuesday’s class » All of Chapter 2