1. 1 AP Physics Chapter 2 Motion Along a Straight Line

2. 2 AP Physics Turn in Chapter 1 Homework, Worksheet, & Lab Take quiz Lecture Q&A

3. 3 Motion Kinematics: Study of motion, emphasizing on describing motion Simplest motion:  1-Dimensional  Point Particle: no rotation

4. 4 Position Position: Location, where an object is Symbol: x Unit: meter How to describe a position in 1-D?  Frame of Reference (numbered line)  Reference point: origin  Direction: o Positive direction: We can define the positive direction to be any direction we want, normally direction of motion o Negative direction: Opposite to positive direction

5. 5 Position is a ______ Direction: negative sign indicates direction only (more later) Magnitude: how far from the origin vector More on Chapter 3

6. 6 Position and Frame of Reference EastWest Downtown, LAHami 10 miles -10 +10 0 (- indicates only direction, west.)  Frame 2: Define West positive, and x = 0 at Downtown LA, then position of Hami is _____ miles. Frame 1: Define East positive, and x = 0 at Downtown LA, then position of Hami is _____ miles.  Frame 3: Define East positive, and x = 0 at Hami, then position of Hami is _____ miles. 

7. 7 So, Representing Position … The representation of a position depends on the choice of the frame of reference. The same position can be expressed in different ways in different frame of reference. But the physical meaning of position does not change, regardless the choice of frame of reference.  Hami still is 10 miles west of Downtown LA, no more or less.  If you want to go from Downtown LA to Hami, you still have to go 10 miles west.

8. 8 Displacement Displacement: Change of position Symbol:  x, Δ = change  x = x 2 - x 1 = x f – x i Unit: same as position, meter Displacement is also a vector: – Magnitude: how far – Direction: Negative sign indicates direction only, it has nothing to do with magnitude. Displacement has nothing to do with the actual path. It depends only on the initial and final positions. A –3m displacement is ___________ a 2m displacement. larger than + +2m -3m

9. 9 Distance (d) Distance  Displacement Distance is a scalar. (magnitude only) Distance is equal to Magnitude of Displacement when there is no change in direction. d = |  x| Distance is not necessarily equal to the magnitude of total displacement. Displacement cares only end points; distance cares both end points and the actual path. Representations of position and displacement depend on frame of reference, but distance does not depend on frame of reference.

10. 10 Total Distance Whenever there is a change in direction, total distance will not be the same as the magnitude of total displacement. When you go from x = 0 to 3m then back to 2m,  Your total displacement is __ m.  Your total distance traveled is __ m. 2 4

11. 11 Total Distance, d tot When there is no change in direction: When there is change in direction: where d 1 and d 2 are distances of segments in which there is no change in direction.

12. 12 Average Velocity Average velocity: ratio of displacement  x that occurs during a particular time interval  t to that time interval Standard Unit: m/s Constant velocity:

13. 13 Average speed, Average speed is distance traveled during a time interval divided by the time interval.

14. 14 Example: During a hard sneeze, your eyes might shut for 0.50 s. If you are driving a car at 90 km/h, how far does it move during that time? Given:  t = 0.50s, v = 90km/h  x = ?

15. 15 Practice: Boston Red Sox pitcher Roger Clemens could routinely throw a fastball at a horizontal speed of 160 km/h. How long did the ball take to reach home plate 18.4 m away? Given:  t = ?

16. 16 Practice: 30-11 You drive on Interstate 10 from San Antonio to Houston, half the time at 55 km/h and the other half at 90 km/h. On the way back you travel half the distance at 55 km/h and the other half at 90 km/h. What is your average speed a) from San Antonio to Houston, b) from Houston back to San Antonio, and c) for the entire trip? d) What is your average velocity for the entire trip? e) Graph x versus t for (a), assuming the motion is all in the positive x direction. Indicate how the average velocity can be found on the graph.

17. 17 Practice: a) Let T be the total number of hours from San Antonio to Houston, then the distance of the first T/2 hours is (55  T/2) km, and that of the second T/2 hours is (90  T/2) km. Total distance is (55  T/2) + (90  T/2) = 72.5T km. Then average speed is

18. 18 Practice: (2) b) Let D km be the total distance. Then the time for the first D/2 km is Similarly the time for the second D/2 km is D/180 h, then total time is Then the average speed is

19. 19 Practice: (3) c) Similar to Part b), the average speed for the entire trip is d) Total displacement is 0 for round trip. Then average velocity is

20. 20 Practice: (4) t x Average velocity can be found by finding the slope of the line (in red) connecting the end points.

21. 21 Instantaneous velocity Instantaneous velocity is the average velocity when the time interval becomes very, very small, essentially zero. Instantaneous velocity is the time-derivative of position function.

22. 22 Derivatives C is constant, and x and y are functions of t.

23. 23 Try

24. 24 Instantaneous velocity Tangent linex t As the time interval becomes smaller and smaller, average velocity becomes instantaneous velocity, which is the slope of the tangent line. titi t2t2 t1t1 t3t3 t4t4 Slope = average velocity from time t i to t 1 What does the slope of this line mean?

25. 25 Derivative and slope The derivative of a curve at any point is the slope of its tangent line at that point. Instantaneous velocity v is the time derivative of position x. On a position-time graph, the instantaneous velocity at any time is the slope of the line tangent to the curve at that time. If position graph is a straight line: slope is _________, and constant _________ is constant. velocity

26. 26 How far does the runner whose velocity-time graph is shown below travel in 16s? Displacement = Area under curve of v-t graph 0 161210 t (s) 4 8 v (m/s)

27. 27 A1A1 A1A1

28. 28 Acceleration Average acceleration Instantaneous acceleration Think of acceleration as a push as of now, though not exactly correct.

29. 29 Velocity-time graph On a velocity-time graph, the slope of the line tangent to the curve at any time is the instantaneous acceleration at that time. If the graph is a straight line, acceleration is constant. Displacement is area under curve

30. 30 Position, Velocity, and acceleration Graphs Position (vs. Time) graph v = slope a = slope  x = Area under curve  v = Area under curve Velocity (vs. Time) graph Acceleration (vs. Time) graph Derivative Integrate

31. 31 Speeding Up or Slowing down v and a are in the same direction (or have the same sign)  ___________ v:v:  speeding up v:  speeding up v:  slowing down a: v and a are in the opposite direction (or have the opposite signs)  ____________ speeding up slowing down + + - - + - - +

32. 32 Directions of acceleration and velocity Sign of a vector indicates direction only. If the signs of the velocity and acceleration of a particle are the same (same direction), the speed of the particle increases (speeding up). If the signs are opposite (opposite directions), the speed decrease (slowing down). – A negative acceleration does not necessarily mean slowing down. – Deceleration = slow down

33. 33 Speeding up or slowing down Speeding up: Slowing down: in the same opposite to  Acceleration is ___________ direction of velocity.  Acceleration is __________ the direction of velocity.  Then we can determine the sign of acceleration depending on what direction has been already defined as the positive direction

34. 34 Practice 35-79 If the position of an object is given by x = 2.0t 3, where x is measured in meters and t in seconds, find (a) the average velocity and (b) the average acceleration between t = 1.0s and t = 2.0 s. Then find (c) the instantaneous velocities and (d) the instantaneous accelerations at t = 1.0s and t = 2.0 s. (e) Compare the average and instantaneous quantities and in each case explain why the larger one is larger. (f) Graph x versus t and v versus t, and indicate on the graphs your answers to (a) through (d).

35. 35 Practice (2) a)

36. 36 Practice: (3) b)

37. 37 Practice: (4) c) d)

38. 38 Practice: (5) d) Because both the velocity and acceleration are increasing, so the final values are the largest ones.

39. 39 Practice: (6) f) a) Slope of red line is the average velocity from 1s to 2s c) Slope of blue line is the instantaneous velocity at 1 s c) Slope of green line is the instantaneous velocity at 2 s

40. 40 Practice: (7) b) Slope of red line is the average acceleration from 1s to 2s d) Slope of blue line is the instantaneous acceleration at t =1s d) Slope of green line is the instantaneous acceleration at t = 2s

41. 41 Constant Acceleration Motion a = constant Let initial time t = 0, then at any time t, velocity and position are given by

42. 42 Free-Fall Motion Assume no air resistance. (Valid when speed is not too fast.) a = g, downward (g = 9.81 m/s 2 )  Acceleration can be positive or negative, depending on what we define as the positive direction.  g is always a positive number, equivalent to 9.81 m/s 2.  Does not matter if the object is on its way up, on its way down, or at the very top.  g is acceleration due to gravity (It is not gravity.) g does not depend on mass of object.

43. 43 Terms Released, drop – If hand not moving, then v i =0. – If hand moving, then v i  0 a = g, downward, regardless of direction of velocity – on the way up – on the way down, or – at the very top  v i = 0 relative to hand  V = 0

44. 44 Signs of v and a v:a: v v:a: Define: up = + v: Define: down = + a: v v:a: +- = 0 - -- - + + ++

45. 45 Free-Fall motion equations These equation are valid only when downward is defined as the positive direction. Not valid when upward is defined as the positive direction. (Must replace every g with –g.) No need to remember these equations.

46. 46 Practice: A jumbo jet must reach a speed of 360 km/h (=225 mi/h) on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 1.80 km runway?

47. 47 Practice: A stone is thrown down from a bridge 43.9 m above the water and splashes the water 2.0 s later. What is the initial speed of the stone? With what speed the stone hit the water? Let down = +. Then

48. 48 Practice: A rock is thrown straight upward with an initial speed of 10 m/s from a window that is 20 m high. A) How much higher can it go? B) How fast is it moving when it hits the ground? Let up = +, x = 0 at window.

49. 49 Practice: A rocket-driven sled running on a straight, level track is used to investigate the physiological effects of large accelerations on humans. One such sled can attain a speed of 1600 km/h in 1.8 s starting from rest. Find (a) the acceleration (assumed constant) in g units and (b) the distance traveled.