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Unit B Changes in Motion

Unit B Changes in Motion. Chapter 1 Describing Motion. Average Speed. Average speed is equal to the total distance traveled divided by the total time. Average speed = total distance elapsed time v = d t Pg 169 # 1-3. Uniform Motion.

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Unit B Changes in Motion

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  1. Unit B Changes in Motion Chapter 1 Describing Motion Science 20 Unit 2

  2. Average Speed • Average speed is equal to the total distance traveled divided by the total time. • Average speed = total distance • elapsed time • v = d • t • Pg 169 # 1-3 Science 20 Unit 2

  3. Uniform Motion • Uniform motion is motion in a straight line at a constant speed. • Uniform motion is rare • Non-uniform motion is when there is a change in speed (speeding up or slowing down) or a change in direction. • Instantaneous speed is the speed at any one point in time Science 20 Unit 2

  4. Scalar Quantity • Scalar quantities consist of magnitude only and no indication of direction. • Speed, time and volume are scalar quantities • Pg 172 #6-8 • Pg 173 # 2-4 Science 20 Unit 2

  5. Velocity • Position is a vector quantity describing the location of a point relative to a reference point • Vector quantity is a quantity consisting of magnitude and direction • Sign convention – north is positive, to the right is positive – south is negative and to the left is negative Science 20 Unit 2

  6. Displacement is a vector quantity describing the length and direction in a straight line from the starting position to the final position. • Average velocity is a vector quantity describing the change in position over a specified time Science 20 Unit 2

  7. Example • A high school athlete runs 100 m south in 12.20 s. What is the velocity in m/s and km/hr? Science 20 Unit 2

  8. Scale Diagrams • Resultant displacement is the vector sum of individual displacements. • Head-to-tail method: a method where the tail of a succeeding vector arrow begins at the head of the preceding vector arrow. Science 20 Unit 2

  9. Draw a vector diagram to represent the following. A person walks 300 m south and then turns around and walks 150 m north. What is the persons displacement? Science 20 Unit 2

  10. Pg 181 #15 • Pg 184 # 17 • Pg 185 # 2-5 Science 20 Unit 2

  11. Lab Activity • Using a ticker tape timer pull a dynamics cart at constant speed. • Mark ‘0’time when uniform motion starts – dots are equally spaced. • Count every five dots and mark ticker tape • Measure and chart data from the ticker tape Science 20 Unit 2

  12. Graph a Position vs Time Graph, calculate the slope of the line. • Graph a Distance vs Time Graph • Graph a Velocity vs Time Graph, calculate displacement at time = to 5 tocks • Hand in Chart and three graphs Science 20 Unit 2

  13. Work • Pg 193 #3&4 Science 20 Unit 2

  14. Acceleration • Acceleration is a change in velocity during a time interval(speeding up or slowing down) • Acceleration is a vector quantity • A force is required to change motion in some way Science 20 Unit 2

  15. Acceleration cont’d • Units for acceleration – m/s2 • Formula is Science 20 Unit 2

  16. v = vf – vi • vf = final speed (m/s) • vi = initial speed (m/s) Science 20 Unit 2

  17. Example • A car traveling a 50 m/s speeds up to 95 m/s over 6 s. What is the car’s acceleration? • vi = 50 m/s • vf = 95 m/s • t = 6 s Science 20 Unit 2

  18. Examples • The velocity of a car increases from 2 m/s at 1.0s to 16 m/s at 4.5 s. What is the car’s average acceleration? Science 20 Unit 2

  19. Rearrange Formula • vf = vi + at • Use this formula to find final speed when an object is accelerating • Example – If a car with a velocity of 2.0 m/s at time zero, accelerates at a rate of +4.0 m/s for 2.5 s, what is its velocity at the end of its acceleration? Science 20 Unit 2

  20. Work • Pg 200 #25 • Pg 203 #26 & 28 Science 20 Unit 2

  21. Acceleration Lab • Using the ticker tape timer drop an object from the top of the stair well. • Chart data from ticker tape • 1. Time • 2. Position • 3. Velocity • 4. Acceleration Science 20 Unit 2

  22. Lab Continued • Graphs to be completed • 1. Position vs Time graph • 2. Velocity vs Time graph – calculate slope • 3. Acceleration vs Time graph Science 20 Unit 2

  23. Displacement Equation Science 20 Unit 2

  24. Acceleration Due to Gravity • Acceleration due to gravity is 9.81 m/s2 • Gravitational acceleration will act on any object moving up or down in the atmosphere • Example jumping, throwing a ball up, falling off a building etc Science 20 Unit 2

  25. Example • A rock is thrown straight up in the air. It reaches a height of 18.6 m in 2.1 s. Calculate the initial velocity t = 2.1 s d = 18.6 m a = - 9.81 Vf = 0 Vi = ? Vi = vf - at = 0 – (-9.81)(2.1) Vi = 20.6 m/s Science 20 Unit 2

  26. Work • Pg 199 # 24 • Pg 203 # 27 • Pg 208 # 32 • Pg 209 # 33 Science 20 Unit 2

  27. Another Distance(Displacement) Equation • When final velocity is not given in the original data and acceleration is given use the following formula Science 20 Unit 2

  28. Example 1 • A boy leaves the surface of a trampoline with an initial velocity of 11.8 m/s, straight up. Determine the displacement after 0.8 s. • vi = 11.8 m/s • t = 0.8 s • a = -9.81 m/s2 Science 20 Unit 2

  29. Example 2 • A diver steps off the ledge of a platform and enters the water 5.0 m below. If the initial velocity of the diver was zero, determine the time it took the diver to reach the water. • d = -5.0 m • a = -9.81 m/s2 • vi = 0 Science 20 Unit 2

  30. Reaction Distance • Reaction time is critical in the stopping of a vehicle. • Includes – the time it takes the drivers brain to recognize there is a need to stop and the time it takes the driver’s foot to move from the gas pedal to the brake pedal. • Reaction time varies from person to person Science 20 Unit 2

  31. Reaction distance is the distance the vehicle travels while the driver is reacting. • Braking distance is the distance a vehicle travels from the moment the brakes are first applied to the time the vehicle stops. • Stopping Distance = Reaction Distance +Braking Distance Science 20 Unit 2

  32. Example • The typical reaction time for most drivers is considered to be about 1.50 s. This includes the time required to identify the danger (0.75 s) and the time required to react to the danger (0.75 s) The ability of vehicles to decelerate varies greatly however. Traffic safety engineers often use a deceleration value of 5.85m/s2 to calculate the minimum stopping distance for a vehicle on smooth, dry pavement. Science 20 Unit 2

  33. Determine the distance traveled while reacting, the distance traveled while braking and the minimum stopping distance of a vehicle traveling 110 km/h. • While reacting d = vt = (31)(1.50) = 46.5 m Science 20 Unit 2

  34. While stopping (braking): • Vi = 31 m/s • Vf = 0 • a = - 5.85 m/s Science 20 Unit 2

  35. Science 20 Unit 2

  36. Work • P. 213 #3, 4 • Pg 216 # 38 • Pg 218 # 39-40 • Pg 220 #1 & 4 Science 20 Unit 2

  37. Braking • Force of friction is contact between two surfaces that acts to oppose the motion of one surface past the other. • Friction is a force • All forces are a push or pull. • Forces are measured in Newtons – N. • Brakes – particularly brake pads and rotors – are designed to produce additional friction between the rotating wheels and the fixed frame of the vehicle Science 20 Unit 2

  38. Net force – is the vector sum of all forces acting on an object. • In the case of braking the net force includes • 1. Force of air resistance • 2. Force of road resistance • 3. Force applied by the braking system Science 20 Unit 2

  39. Another factor that affects the rate of deceleration is the mass of the vehicle. • Larger trucks require a much greater stopping distance Science 20 Unit 2

  40. Newton’s Second Law of Motion • Newton’s Second Law of Motion states that an object will accelerate in the direction of the net force applied. Fnet = ma Units are N = kg m/s2 Science 20 Unit 2

  41. Example • A vehicle with a mass of 1250 kg is traveling 45 km/h east, when the driver engages the brakes to stop at an intersection.If the net force on the vehicle is 7000 N west, determine the magnitude and direction of the deceleration of the vehicle while the net force is applied. Science 20 Unit 2

  42. Determine the length of time the net force must be applied to stop the vehicle. Science 20 Unit 2

  43. Work • Pg. 222 #43 • Pg 226 # 45 • Pg 227 # 1-4 (copy questions) Science 20 Unit 2

  44. Speeding Up • Newton’s second law also explains what happens when a vehicle increases its velocity. • The additional force required to make the vehicle move faster is called the applied force. • The net force results from the vector sum of the applied force and the force of friction Science 20 Unit 2

  45. Science 20 Unit 2

  46. Newton’s First Law of Motion (Inertia) • This law states that in the absence of a net force, an object in motion will tend to maintain its velocity, and an object at rest will tend to remain at rest. Science 20 Unit 2

  47. Example - One • The engine of a motorcycle supplies an applied force of 1880 N, west, to overcome frictional forces of 520 N, east. The motorcycle and rider have a combined mass of 245 kg. Determine the acceleration. Science 20 Unit 2

  48. Science 20 Unit 2

  49. Example - Two • A car with a mass of 1075 kg is traveling on a highway, The engine of the supplies an applied force of 4800 N, west, to overcome frictional forces of 4800 N, east. Determine the acceleration. Science 20 Unit 2

  50. Science 20 Unit 2

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