Kinematics

1. Kinematics Kinematics is the study of the motion of objects which includes distance, displacement, speed, velocity, acceleration, and time.

2. Vocabulary Used In Kinematics Frame of Reference • A set of axes or a point from which you make measurements. These measurements can be position, distance, displacement, speed, velocity, etc. Position • Separation between a frame of reference and an object measured in cm, m, km.

3. Distance • Length measurement of separation. Displacement • A change in position of an object with respect to starting point. Do not consider distance and displacement synonymous!

4. If you walk across a room, turn around and return to your starting point, you have traveled the 2 × width of the room. However, your displacement is zero because you ended up where you started.

5. Vector or Scalar? In everyday life, when you measure things like mass, volume, length, temperature, etc. you often forget the unit and never provide direction. • Quantities that only have magnitude (size) are called scalar quantities. In physics, we often make measurements that have both magnitude and direction and never forget the units!

6. Examples of vector quantities are displacement, velocity, and acceleration. For example, when you say that you are traveling at 35 km/h N, you are stating your direction as well as your rate. When you weigh something and find it to be 25 N (newtons), you are giving the magnitude and direction (understood to be straight down).

7. distance s = = time Vector vs Scalar Two very important measured quantities are speed and velocity. Average speed is a scalar quantity because only the magnitude is given with no specified direction. • average speed • 100 m/s

8. displacement v = = Average velocity is a vector quantity because in addition to magnitude, direction is always given. • average velocity • 100 m/s N Velocity can either be positive or negative. • If the object is moving in the direction that has been defined as positive, then the velocity is positive, otherwise it is negative. time

9. N S This graph shows the displacement of an object that started next to you at the origin. The displacement is positive throughout the time interval (t = 0-16 s) which means the object moved in a positive direction with respect to you. When the object comes to a stop at 16 s, its maximum displacement is 35 m N.

10. From 0-4 s, the object travels with a constant velocity as indicated by the constant slope. From 4-6 s, the object has stopped (the slope is 0), from 6-10 s, the object travels at a higher constant velocity than 0-4 s, and from 10-16 s, it moves the slowest of its three velocities.

11. Δs Δs Δs Δs 10. m 18. m v v v = Δt Δt Δt Δt 4.0 s 3.0 s = = 0-4 s: v = = 3.3 m/s N v = 4-6 s: 0 m/s 6-10 s: = = = 4.5 m/s N

12. Instantaneous vs Average Velocity Instantaneous velocity is the velocity at one particular instant of time. • When driving a car, each time you look at the speedometer, you are reading the instantaneous velocity provided you are not lost. To determine your average velocity, you would need to use your odometer and a watch.

13. How fast are you going at 2.0 s? Ans. 3.3 m/s N How fast are you going at 5.0 s? Ans. 0 m/s How fast are you going at 13.0 s? Ans. Find the slope of the line between 10.0 and 16.0 s. The slope is constant so the velocity is constant. If the velocity is constant, then the object must still be moving north.

14. Δs vave = Δt Constant Velocity Problems A train leaves Providence traveling at a constant velocity of 36.0 m/s due west. • How long does it take the train to travel 1620.0 m? vave = 36.0 m/s Δs = 1620.0 m

15. 1620.0 m 36.0 m/s = Δt Δt = 45.0 s . (b) What is the velocity of the train in km/h? 1 km 3600 s m vave = 36.0 × × s 103 m 1 h vave = 130. km/h W

16. mi 1760 yd 36 in 2.54 cm × × × 55 1 mi 1 yd 1 in h 1 m 1 km × × 102 cm 103 m You plan on driving from the west coast to the east coast of the United States (4600 km) at 55 mi/h and you plan to stop every 2.0 h to rest and refuel which will take 30. min. (a) How long will the trip take? vave = 88 km/h =

17. vave = km h Δs Δt 4600 km = 88 Δt . Δt = 52.27 h 1 stop 30 min ΔtT = 52.27 h + 52.27 h × × 2 h 1 stop 1 h = × 65.3 h 60 min

18. 4600 km 65.3 h Δs vave = 70.4 km/h = = Δt (b) What is the vave for the entire trip?

19. Acceleration Acceleration is defined as the rate at which the velocity changes. • A change in velocity can be either in magnitude, direction, or both. • A rate always implies a time interval - /s, /h, /fortnight.

20. vf - vi m/s m Δv aave = = = = tf - ti Δt s s2 • This means acceleration always has two units of time and they do not have to be the same units although many times they are. • An example of acceleration would be 9.80 m/s/s = 9.80 m/s2 or -9.80 m/s2. The average acceleration is given by

21. Acceleration is a vector quantity meaning it has both magnitude and direction. • If the acceleration is positive and in the same direction of the velocity, the object speeds up. • If the acceleration is negative and in the opposite direction of the velocity, slows down.

22. As with velocity, average acceleration must always have two points to be calculated and instantaneous acceleration has only one point.

23. Interpreting Acceleration Problems If the problem says: • “… starts from rest …”, vi = 0 m/s • “… moves at a constant velocity …“, aave = 0. • “… comes to rest …”, vf = 0.

24. This graph shows the velocity of an from 0-16.0 s. From 0-4.0 s, the object is accelerating positively. From 4.0-6.0 s, the object is accelerating at a greater rate. From 6.0-9.0 s, the object is moving at a constant velocity.

25. From 9.0-11.0 s, the object is decelerating. From 11.0-13.0 s, the object is again speeding up. From 13.0-16.0 s, the object is accelerating at a lower rate than from 11.0-13.0 s. All of the above descriptions were determined by looking qualitatively at the slope of the graph in each time interval.

26. Where is the object in relation to you? • You don’t know! • Think about it. The object could have started next to you or started from rest 100 m in front of you. • Would the same graph apply in either scenario? • You do know that the object continues to get further from you.

27. Δv Δv Δv Δ a a Δt Δt Δt Δt 6.25 m/s2 N 1.25 m/s2 N = = = = 0-4 s: 5.0 m/s a = = = 4.0 s a 4-6 s: 12.5 m/s 2.0 s 6-8.5 s: = = 0

28. Acceleration Due To Gravity All objects accelerate at the same rate when air resistance is ignored. • This is true no matter how large or small an object is, no matter the height from which it is dropped. a = g = 9.80 m/s2 = 980 cm/s2= 32 ft/s2 • g changes from location to location.

29. g is a vector quantity meaning it has both magnitude and direction. • g changes from location to location. • The sign convention can be very confusing and usually there is not a lot of uniformity from resource to resource.

30. vi > 0 g = -9.80 m/s2 Symmetry Considerations B C Point A: Point C: vi = 0 g = 9.80 m/s2 Point B: Point D: vi > 0 vi = vf D A g = -9.80 m/s2 g = 9.80 m/s2

31. Symmetry Considerations B C Point A to Point B: Velocity is decreasing but is positive. g = -9.80 m/s2 Point C to Point D: Velocity is increasing but is negative. D A g = 9.80 m/s2

32. vf+vi v = 2 Kinematics Equations The following equations are for constant acceleration. vf = vi + aΔt x = xi + vit + ½aΔt2 vf2 = vi2 + 2a(x-xi)

33. Some thoughts about the previous equations: • The symbols i and f stand for initial and final respectively. Sometimes, a subscript 0 is used for initial and no subscript is used for final. It really does not matter, it is all what you get used to.

34. x and xi represent the position and sometimes s is used in place of x. Many times xi = 0 which simplifies the equation. x – xi represents the displacement. Many times x is used to represent horizontal displacement and y is substituted for x when the displacement is vertical.

35. Some g Problems You throw a ball straight up to the top of a flagpole and it takes 4.0 s for the ball to return. • What is the height of the flagpole in meters? ΔtT = Δtup + Δtdown = 4.0 s Δtup = Δtdown = 2.0 s g = ±9.80 m/s2

36. x = xi + vit + ½aΔt2 Δx = 0 + ½ × 9.80 m/s2 × (2.0 s)2 = 20. m Some final thoughts: • From symmetry considerations, you may use the upward or downward path. • The downward path was chosen which meant both xi and vi = 0.

37. Many times there may be an alternate formula(s) which could be used to solve the problem which is fine. (b) What is the velocity of the ball if you catch it from the same height you threw it? vf2 = vi2 + 2a(x-xi) = vi2 + 2gΔx vf2 = 0 + 2 × 9.80 m/s2 × 20. m vf =20. m/s

38. Another g Problem A skydiver jumps from an airplane and his parachute opens after 1.8 s. (a) What is his speed when his parachute opens? vi = 0 g = 9.80 m/s2 vf = vi + gΔt = 0 + 9.80 m/s2 × 1.8 s vf = 18 m/s

39. vi + vf 0 + 18 m/s vave 9.0 m/s = = = 2 2 (b) What was the average speed during the first 1.8 s? (c) How far does the skydiver fall in 1.8 s? Δs vave = Δt Δs = 9.0 m/s × 1.8 s = 16 m

40. (d) Immediately after the parachute is opened, what happens to the skydiver’s acceleration? Justify your answer. As soon as the parachute is opened, the skydiver is no longer in free fall. The parachute creates air resistance which means that the weight of the skydiver is not the only force acting. The acceleration decreases.

41. (e) Immediately after the parachute is opened, what happens to the skydiver’s velocity? Justify your answer. At the instant the parachute is opened, the acceleration decreases but it is still positive meaning the velocity still is increasing. The velocity will be increasing by less than 9.80 m/s2.

42. Graphing Uniform Acceleration When the acceleration of an object is uniform, its displacement varies with the square of the time. This results in the graph of Displacement vs Time to be a parabola. The displacement increases much faster for equal sized intervals than the time does.

43. You can perform the same type of graphical analysis as done before, but somewhat differently. As shown in the next graph, the instantaneous velocity does not equal the average velocity. • This only occurs when you have uniform motion. • Uniform motion is either moving in a straight line at constant speed or not moving at all (a = 0).

44. To find the instantaneous velocity at 17.0 s, one must draw a tangent to the curve at 17.0 s. Taking the slope of the tangent, gives: . Δs 1416 m – 920 m v 165 m/s = = = 3.0 s Δt

45. Δs = Δt Don’t get confused! • We did use two points on the tangent line to determine the slope but the slope was computed for only one point on the parabola. To determine the average velocity during a time interval, you use the same calculation as before vave

46. And The Analysis Goes On … Another graph used extensively with accelerated motion is Velocity vs Time.

47. One thing to note is that the slope is constant which is indicative of uniform or constant velocity. The slope of the straight line is given by: . Δv 98.0 m/s – 0 m/s a 9.80 m/s2 = = = 10.0 s Δt

48. One thing to note is that the slope is constant which is indicative of uniform or constant velocity. The slope of the straight line is given by: . Δv 98.0 m/s – 0 m/s a 9.80 m/s2 = = = 10.0 s Δt

49. The Speed Trap A motorist is speeding on Rt. 95N at a constant velocity of 30.0 m/s. He zips by a speed trap and it takes the state police officer 5.00 s to finish his coffee. He then pursues the motorist accelerating at a constant 5.20 m/s2. • How long does it take for the police officer to catch the speeder?

50. Δtm = t + 5.00 s Δtp = Δtp vave = 30.0 m/s vi = 0 Δxm = Δxp ap = 5.20 m/s2 Δxp = Δxm, xi = 0 Δxm = vave Δtm = 30.0(t + 5.00 s) x = xi + vit + ½aΔt2 Δxp = 0 + ½ × 5.20 m/s2 × Δt2 30.0(t + 5.00) = ½ × 5.20 × Δt2