1. Motion in One Dimension Chapter 2

2. Mechanics Mechanics is the study of motion. Kinematics is the science of describing the motion of objects with words, diagrams, numbers, graphs, and equations. Terms:  Distance  Position  Displacement  Speed  Velocity  Average speed  Average Velocity  Acceleration  Freefall  Acceleration due to gravity

3. 2.1 Displacement and Velocity One-dimensional motion takes place in one direction (straight line motion). Frame of reference: –a system for specifying the precise location of objects in space and time –used to measure changes in position –remains fixed for the motion problem –an origin (starting point) from which motion and time are measured For the gecko the meter stick is the frame of reference.

4. 2.1 Displacement and Velocity Displacement –the change in position of an object and the direction of change –The length and direction of the straight line drawn from an object’s initial position (x i ) to the object’s final position (x f ). –  x = x f –x i or  x = x f –x 0 The SI unit is the meter, m. –Motion is relative displacement. –Speed is the time rate of motion. Speed has no direction.

5. 2.1 Displacement and Velocity Displacement –the change in position of an object and the direction of change –  x = x f –x i or  x = x f –x 0 What is the skater’s displacement? What is the skater’s distance traveled?  x = + 140 m d = 420 m

6. 2.1 Displacement and Velocity Displacement has direction and sign. Motion to the right or east is positive (+). Motion to the left or west is negative (-). Motion to the north or up is positive (+). Motion to the south or down is negative (-).

7. Chapter 2 Positive and Negative Displacements 2 nd 3 rd 4 th

8. 2.1 Displacement and Velocity Velocity –Average velocity is the total displacement divided by the time interval during which the displacement occurred –SI Unit: m/s –can be positive or negative depending on the sign of the displacement –Average velocity is equal to the constant velocity needed to cover a certain displacement in a given time interval; it is not necessarily the average of the initial and final velocities. (It is equal to the average of the initial and final velocities when acceleration is constant.) v avg == = xtxt x f - x i  t x f – x 0  t

9. 2.1 Displacement and Velocity A car travels from A to B which is 100 km. If the first half of the distance is driven at 50 km/h and the second half is driven at 100 km/h, what is the average velocity? Know: v avg = xtxt  x = 100 km What is  t?  t =  x v avg  t 0-50km = = = 1.0 h  x v avg 50  km 50 km/h  t 50km – 100 km = = = 0.5 h  x v avg 50  km 100 km/h v avg = = = 67 km/h xtxt 100 km 1.5 h

10. 2.1 Velocity and Speed Velocity describes motion with both a direction and a numerical value (a magnitude). Speed has no direction, only magnitude. Average speed is equal to the total distance traveled divided by the time interval.

11. 2.1 Displacement and Velocity p. 44 #1 v avg = + 0.98 m/s  t = 34 min  x = ?  t = 34 min x 60 s/min = 2040 s v avg = xtxt  x = v avg  x  t  x = 0.98 m/s x  2040 s  x = 1999.2 m  x = 2.0 x 10 3 m East

12. 2.1 Displacement and Velocity p. 44 #2, #4, and #6

13. 2.1 Interpreting Motion Graphically For any position-time graph, we can determine the average velocity by drawing a straight line between any two points on the graph. If the velocity is constant, the graph of position versus time is a straight line. The slope indicates the velocity. Time (s) B A C Position (m)

14. Object 1: positive slope = positive velocity Object 2: zero slope= zero velocity Object 3: negative slope = negative velocity 2.1 Interpreting Motion Graphically (m) (s)

15. 2.1 Displacement and Velocity Position-Time Graphs (Distance-Time Graphs) –Describe the motion for each graph (1, 2, and 3) 1 2 3

16. 2.1 Displacement and Velocity Position – Time graphs illustrate velocity. A B C D E What is the total displacement of the object? the total distance traveled? Calculate v avg for AB, BC, CD, DC, AC, AD, and AE. v avg, AB = +6 m/s v avg, BC = 0 m/s v avg, AD = -1 m/s v avg, CD = -4 m/s v avg, DE = 2.7 m/s v avg, AC = 4 m/s v avg, AE = 0 m/s

17. 2.1 Displacement and Velocity Instantaneous Velocity –is the velocity of an object at some instant or at a specific point in the object’s path. –is equal to the slope of the tangent line at a specific point (instant in time).

18. 2.2 Acceleration = Change in Velocity Acceleration is the change in velocity. An object accelerates if its speed, direction, or both change. Acceleration is the rate at which velocity changes over time. Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.

19. 2.2 Acceleration SI units: m/s 2 Acceleration has direction and magnitude. m/s s ( )

20. 2.2 Acceleration Consider a train moving to the right, so that the displacement and the velocity are positive. When the velocity in the positive direction is increasing, the acceleration is positive, as at A. When the velocity is constant, there is no acceleration, as at B. When the velocity in the positive direction is decreasing ((-) slope), the acceleration is negative, as at C. The slope of the velocity-time graph is the average acceleration.

21. 2.2 Positive or Negative Acceleration VelocityAccelerationMotion ++speeding up in + direction --speeding up in - direction +-slowing down in + direction -+slowing down in - direction - or +0constant velocity 0- or +speeding up from rest 00at rest

22. 2.2 Constant Acceleration

23. 2.2 Motion Graphs Time (s) Acceleration(m/s 2 ) Displacement -Time Graph slope = m/s velocity area = m x s meaningless Velocity -Time Graph slope = m/s/s acceleration area = m/s x s = m displacement Acceleration-Time Graph slope = m/s 2 /s meaningless area = m/s 2 x s = m/s velocity

24. 2.2 Constant Acceleration Displacement depends on acceleration, initial velocity, and time. The time interval between each image is the same. Strobe photograph with 0.1s time intervals. Why does the ball get farther and farther apart in the same amount of time? (The distance between the images increases while the time interval remains constant.) The ball is accelerating with constant acceleration due to gravity.

25. 2.2 Constant Acceleration Velocity-Time Graph The relationships between displacement (  x), velocity (v), and constant acceleration (a) are expressed by equations that apply to any object moving with constant acceleration. v avg = xtxt v i + v f 2 v avg = vivi vfvf * * * v avg Constant acceleration

26. At constant acceleration 2.2 Displacement with Constant Acceleration v avg = and v avg = xtxt v i + v f 2 Using these two equations derive an expression for displacement. xtxt = ½(v i + v f )  x = ½(v i + v f )  t = ½(v i + v f )

27. 2.2 Final Velocity and Acceleration From the definition of acceleration calculate final velocity, v f, from initial velocity, acceleration, and time. a = v f – v i  t Solve for v f. a  t = v f – v i a  t + v i = v f v f = a  t + v i

28. 2.2 Displacement with Constant Acceleration Displacement with Constant Acceleration when v f is not known.  x = ½(v i + v f )  t v f = a  t + v i  x = ½(v i + a  t + v i )  t  x = ½(2v i  t + a  t 2 )  x = v i  t + ½a  t 2 v f = a  t + v i This equation is useful for finding the displacement required for an object to reach a certain speed. Use v f = a  t + v i to determine t and then this formula for  x.

29. 2.2 Final Velocity w/o  t at Constant Acceleration If  t is unknown, the final velocity depends on initial velocity, acceleration, and displacement v f = a  t + v i Solve  x = ½(v i + v f )  t for  t and substitute it into the equation above.  x = ½(v i + v f )  t 2  x = (v i + v f )  t 2  x (v i + v f ) =  t v f = a  t + v i v f = a  + v i 2  x (v i + v f ) (v f - v i ) (v i + v f ) =2a  x v f 2 = v i 2 + 2a  x v f - v i = a 2  x (v i + v f )

30. 2.2 Formulas when Acceleration is Constant  x = ½(v i + v f )  t  x = ½(v f )  t v f = v i + a  t v f = a  t  x = v i  t + ½a  t 2  x = ½a  t 2 v f 2 = v i 2 + 2a  x v f 2 = 2a  x Object initially at rest, v i = 0 Yea let’s PRACTICE!!

31. 2.3 Falling Objects

32. Free fall is the motion of any object under the influence of gravity only, i.e. no air resistance (friction) of any kind. Any falling object's motion is at least approximately free fall as long as: 1.The object is relatively heavy compared to its size. 2.The object it falls for a relatively short time. 3.The object it is moving relatively slowly. (This means air resistance depends on speed, size, and shape.) An object doesn't have to be falling to be in free fall. If you throw a ball upward its motion is still considered to be free fall, since it is moving under the influence of gravity.

33. 2.3 Falling Objects Free fall is motion at constant acceleration. a g = acceleration due to gravity g = acceleration due to Earth’s gravity g = 9.81m/s 2 = 32 ft/s 2 at Earth’s surface (about 22 mi/hr/s) a g = - g = - 9.81m/s 2 = - 32 ft/s 2 (choose up (+) down is (-) For objects thrown up into the air, acceleration is constant during upward and downward motion. Acceleration is -9.81 m/s 2 when the object is going up and when the object is coming down.

34. 2.3 Falling Objects Acceleration of gravity, "g" in the Solar System At the surface ofg (m/s 2 ) isMass (kg) * Radius (m) * Sun 1 2751.99 x 10 30 6.95 x 10 8 Mercury3.73.30 x 10 23 2.44 x 10 6 Venus8.94.87 x 19 24 6.05 x 10 6 Earth9.85.97 x 10 24 6.38 x 10 6 Moon (of Earth)1.67.35 x 10 22 1.73 x 10 6 Mars3.76.42 x 10 23 3.40 x 10 6 Jupiter 1 251.90 x 10 27 7.15 x 10 7 Ganymede (moon of Jupiter) 1.41.48 x 10 23 2.63 x 10 6 Europa (moon of Jupiter)1.34.80 x 10 22 1.57 x 10 6 Saturn 1 10.45.68 x 10 26 6.03 x 10 7 Uranus 1 8.98.68 x 10 25 2.56 x 10 7 Neptune 1 111.02 x 10 26 2.48 x 10 7 Pluto0.71.27 x 10 22 1.14 x 10 6 1 Assuming that these objects actually had a surface... * Mass and radius data taken from The Nine PlanetsThe Nine Planets http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Kinematics/ff_velocity_acc.htm

35. 2.3 Falling Objects As the object moves up it decelerates, so its velocity decreases. At the peak of the object’s path it stops and turns around. The object’s velocity then begins to increase in the negative direction. Velocity – Time Graph What does it tell us? What does the slope indicate? What type of motion is represented?

36. 2.3 Falling Objects Falling Object Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0 m above the floor, how long will it be in the air before it strikes the floor? Sample Problem

37. 2.3 Falling Objects Terminal Velocity –constant speed an object reaches when falling through a resisting medium (air) –depends on the the object’s mass, shape, and size

38. 2.3 Falling Objects p.64, #1  x = 239 m a g = - 3.7 m/s 2 239 m v i = 0 v f = ?  t = ? a. v f 2 = v i 2 + 2a  x v f 2 = 2a  x v f 2 = 2(-3.7m/s 2 )(-239m) = 1768.6 m 2 /s 2 v f = - 42 m/s b. a = v f - v i  t  t == vfavfa -42 m/s -3.7m/s 2 = 11s

39. 2.3 Falling Objects p.64, #3

40. Chpt. 2 Reading Guide Key p. 47 #3 20.0 s 22 s a. v avg = 50.0 m/20.0 s = 2.50 m/s v avg = xtxt b. v avg = 50.0 m/22.0 s = 2.27 m/s c. v avg = 0.0 m/s

41. Chpt. 2 Reading Guide Key p. 47 #4 v 1 = 0.90 m/s v 2 = 1.90 m/s 780 m v avg = xtxt a.  t 1 = = 780 m / 0.90m/s = 866.67 s  t 2 = = 780 m / 1.90m/s = 410.53 s 866.67 s - 410.53 s = 456 s = 460 s xv1xv1 xv2xv2

42. Chpt. 2 Reading Guide Key p. 47 #4 v 1 = 0.90 m/s v 2 = 1.90 m/s ? m v avg = xtxt b. 866.67 s - 410.53 s = 456 s = 460 s xv1xv1 xv2xv2 - = 330 s t 1 – t 2 = 5.50 min = 330 s v1v1 v2v2 - = 330 s xx  0.90  1.90 - = 330 s xx  x = 564.3 m = 560 m

43. Chpt. 2 Reading Guide Key p. 47 #6 Bear B It went farther in the same amount of time. Bear A At 8 min, the slope for Bear A is steeper. No