PHYSICS 20 Unit 1 Kinematics Block 1 PDF
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This document appears to be a set of physics notes, specifically focusing on kinematics. It covers key concepts like scalar and vector quantities, uniform and uniformly accelerated motion and 2-dimensional motion, alongside relevant equations and examples. There are both theoretical explanations and practical problem-solving exercises for students.
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Name: ____________________________ PHYSICS 20 UNIT 1: KINEMATICS 1 Key Concepts scalar quantities uniformly accelerated motion vector quantities...
Name: ____________________________ PHYSICS 20 UNIT 1: KINEMATICS 1 Key Concepts scalar quantities uniformly accelerated motion vector quantities two-dimensional motion uniform motion Specific Knowledge Outcomes □ 20–A1.1k define, qualitatively and quantitatively, displacement, velocity and acceleration □ 20–A1.2k define, operationally, and compare and contrast scalar and vector quantities □ 20–A1.3k explain, qualitatively and quantitatively, uniform and uniformly accelerated motion when provided with written descriptions and numerical and graphical data □ 20–A1.4k interpret, quantitatively, the motion of one object relative to another, using displacement and velocity vectors □ 20–A1.5k explain, quantitatively, two-dimensional motion in a horizontal or vertical plane, using vector components. Specific Skill Outcomes 20–A1.3s analyze data and apply mathematical and conceptual models to develop and assess possible solutions □ construct graphs to demonstrate the relationships among displacement, velocity, acceleration and time for uniform and uniformly accelerated motion □ analyze a graph of empirical data to infer the mathematical relationships among displacement, velocity, acceleration and time for uniform and uniformly accelerated motion □ solve, quantitatively, projectile motion problems near Earth’s surface, ignoring air resistance □ relate acceleration to the slope of, and displacement to the area under, a velocity-time graph 20–A1.4s apply conventions of science in communicating information and ideas and in assessing results □ use appropriate Système international (SI) units, fundamental and derived units and significant digits □ use appropriate numeric, symbolic, graphical and linguistic modes of representation to communicate ideas, plans and results □ use delta notation correctly when describing changes in quantities 2 Lesson 1: Position and Displacement Vector Quantities A vector is a quantity that has both a magnitude (amount) and a direction. E.g. position, displacement, velocity, acceleration A vector can be represented in many different ways, but we will use these three ways in physics 20. Description Geometrically Algebraically R 15 m, left R 15 m R 15 m Leflet Right Position ( d ) The position of an object is simply the location of an object, relative to an origin. The origin is where you measure from. oss.me e.g. Object 1 0 A B 35 cm 55 cm If the origin is A, If the origin is B, d 1 = 35cm d1 = 55cm 35 am Right 55cm left chdange podsition Displacement ( d ) a in Displacement is the change in position of an object. i.e. d d f d i It is the straight line distance from the initial position to the final position i.e. the “start to finish” vector Example: An object starts at a location 60 m east of the origin. It then moves to a location 40 m west of the origin, and then it finally moves to a location 25 m east of the origin. Express each position as an algebraic vector. What is the overall displacement of the object? 35m.West sdtotif.fi am sd df di60m _35 3 35m 25m Displacement can also be the total of any number of individual displacement vectors. i.e. dtotal d1 d 2 ... Example: A bike travels 47 km South, and then it travels 111 km North. Express each displacement vector as an algebraic vector. Determine the bike’s total displacement. sdtsd do 4 ii sdsdtot 111km 158km NKmf.ir 47km f 47km Scalar Quantities 64km 64kmENorth.TL A scalar is a quantity that shows only magnitude (how much), but it does not indicate direction e.g. time, distance, speed Distance ( d ) Distance is the length of the path travelled by an object. Distance and displacement can have the same magnitude when the motion is all in one direction. dtotal d1 d2 ... Example: Determine the distance travelled by the objects in the previous two examples. 4 HOMEWORK (Speed, Position, and Displacement) 1. A rock, which if originally located 72 m above a cliff, is dropped. If its final position is 18 m below the cliff, then what is its displacement? 2. A train is located 14 km South of a city. If it then experiences a displacement of 30 km North, then what is the final position of the train? sat de di now df set di 3. 30km A motion is illustrated below: 14km 164nA 14.83 cm 0 12.74 cm 16.19 cm Determine the overall displacement and distance travelled. 5 4. An object experiences a displacement of 1.13 m towards the right. If the object’s final position is 25 cm right of the origin, then what was the object’s original position? 5. An object starts 37 m E of a rock. It then moves to a position 171 m E of rock, and finally, to a position 84 m W of rock. Determine the overall displacement and distance travelled. 6. An object moves 16 km towards the North, it then moves 26 km towards the South, and finally, it moves 40 km towards the North. Determine its overall distance and displacement. SOLUTIONS 1. 90 m downward 2. 16 km North of the city 3. d = 16.19 cm right ; d = 71.33 cm 4. 88 cm left of the origin 5. d = 121 m West ; d = 389 m 6. d = 82 km ; d = 30 km North 6 Lesson 2: Speed/Velocity Velocity ( v ) is the rate of change of position (i.e. how quickly the displacement takes place). The average velocity of an object can be determined by: d There are two methods for calculating displacement: Equation: vave t Change in position d d f di when given “locations”. m Total displacement dtotal d1 d 2 ... when given separate intervals Units: s of motion. where d is the displacement of the object (in m) Δt is the time it takes for the object to make the displacement (in s) Velocity always indicates the direction the object is travelling. 51.0m 27.0m's Example: A toy plane is located 51.0 m north of a city. If it then travels at 27.0 m/s south for 3.90 s, determine its final location. ii sf i dEfi54.3msathD 1 fi it df di df 54.3m Speed vs Velocity If it di 27.0ms 3.90s 51.0m Speed is a scalar quantity (total distance over total time), which means it does not have direction. Velocity is a vector quantity (total displacement over total time) having a direction. Example: A car travels 77 km/h north for 3.8 h, and then travels 49 km/h south for 2.6 h. Determine: the average speed? the average velocity? so it sd 77km h 3.8h 292.6km IN Ñ if Adi 49km h 2.6h 127.4 km s Sott Sd site 292.6km 127.4km 165.2kmCN e i 7 Vane 66km t 6 1h Graphical Analysis (Uniform Motion) The slope of a position-time graph is velocity. o When an object is moving in uniform motion, it has a constant (average) velocity. o On a position-time graph, this would be indicated by constant slope. o Any change in the slope indicates a change in velocity. Example Position as a Funciton of Time 20.0 15.0 9m Slope veto Com 10.0 Position (m) 20m 5.0 6m 0.0 Time (s) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 are Slope Em -5.0 0.53m s 0.53ms south a) When is the object stationary? d) During what interval is the speed the greatest? 3 5s Steepslope fastest 8 12s b) During what interval is the object travelling e) What is the average velocity over the whole south? 15.0 s? 8 15s c) What is the velocity at 7.0 s? Tae E f) What is the average speed over 15.0 s? slope 3.0m t.IS 9iii fi 2.5n 8 HOMEWORK (speed/velocity) A. 1. A satellite can travel 3.4 105 m in 1.1 minutes. Determine its speed, in km/s. d 3.4 10 m 340km 5.2k 1 V h t I 1min 66s 2. A pitcher could routinely throw a fastball at a horizontal speed of 160 km/h, as verified by a radar gun. How long would the ball take to reach home plate 18.4 m away? 3. Fv It An electron moves from a position that is 4.20 mm below an origin to a position that is 11.5 mm above the origin in 4.80 ns (nanoseconds). Determine its velocity, in m/s. 4. A sailboat, originally located at a position that is 5.00 km East of a pier, then moves at a average velocity of 4.20 m/s West for 37.0 minutes. Determine the sailboat’s final position. off it di t 3 2220s III df Tave minx f 4.20ms 2220s 5000m 4324m 5. Given the position-time graph shown, calculate the average velocity. (2 sig digs) 4.32kmWest Note: Right is positive. 6. Starting from home, you bicycle 24 km North in 2.5 hours, then turn and pedal straight home in 1.5 hours. What is your average speed and average velocity over the (a) first 2.5 hours, and (b) entire trip? 9 B. 7. A plane is flying at an average velocity of 650 km/h North. How long does it take (in seconds) to travel from a position 1450 m South of a city to 925 m North of the city? 8. On average, a blink lasts about 100.0 milliseconds. How far does a MIG-25 “Foxbat” fighter travel during a pilot’s blink if the plane’s average speed is 2110 mi/h (3376 km/h)? 9. For the graph, assume right is positive. a) Determine the object’s position at: (i) 0 min (ii) 12 min (iii) 20 min b) Determine the overall distance and displacement for the intervals: (i) 2 - 12 min (ii) 13 - 18 min (iii) 6 - 22 min c) When is the object (i) furthest left of the origin? (ii) furthest right? d) When is the object moving (i) to the left? (ii) to the right? e) When is the object at rest? f) Calculate the average velocity (in m/min) of the object at: (i) 8.0 min (ii) 15 min g) For the overall trip, what is the: (i) average velocity? (ii) speed? Answers in m/min. 10 10. An object moves at a average velocity of 7.40 km/h South for 3.00 hours. If its final position is 8.50 km South of the origin, then determine its original position (relative to the origin). 11. During a road trip, a car travels at a average velocity of 85 km/h towards the East for 4.0 hours. After resting at this location for 1.0 hour, the car then travels at 75.0 km/h towards the West for 3.0 hours. For the entire trip, including the rest time, determine: a) the average speed b) the average velocity SOLUTIONS 1. 5.2 103 m/s = 5.2 km/s 2. v = 44.44 m/s ; t = 0.414 s 3. 3.27 10 m/s upward 6 4. t = 2220 s ; d = 9.324 km West ; d f = 4.32 km West of the pier 5. Slope 3.3 103 m/s Left 6. a) vavg = 9.6 km/h ; v avg = 9.6 km/h North b) vavg = 12 km/h ; v avg = 0 7. d = 2375 m North ; v = 180.556 m/s North ; t = 13.2 s 8. v = 937.778 m/s ; d = 93.78 m 9. a) (i) 15 m right of origin (ii) 40 m right of origin (iii) 25 m left of origin b) (i) 25 m ; 25 m right (ii) 65 m ; 65 m left (iii) 83 m ; 47 m left c) Furthest left: 18 - 21 min Furthest right: 11 - 13 min d) Moving left: 13 - 18 min Moving right: 4 - 11, 21 - 24 min e) 0 - 4, 11 - 13, 18 - 21 min f) (i) 2.0 m/min right (ii) 13 m/min left g) Average velocity: 0.63 m/min left Average speed: 4.8 m/min 10. d = 22.2 km South ; d i = 13.7 km North of origin 11. d1 = 340 km (E) ; d2 = 225 km (W) ; tT = 8.0 h ; vavg = 71 km/h ; v avg = 14 km/h East 11 Lesson 3: Acceleration Acceleration ( a ) is the rate of change of velocity (i.e. how quickly the velocity changes). m v v f vi s or m a Units: t t s s2 Acceleration and Velocity If the acceleration is positive (i.e. a > 0), then the velocity is increasing. If the acceleration is negative (i.e. a < 0), then the velocity is decreasing. If the acceleration is zero, there is no change in velocity (i.e. constant velocity). Acceleration and Speed When the velocity and the acceleration are in the same direction, it is speeding up. When the velocity and the acceleration are in opposite directions, it is slowing down. Example. While braking, a car experiences a backward acceleration of 7.40 m/s2 for 2.88 s. At the end, it is moving forward at 31.5 km/h. What is its initial velocity? 2 8 7.40 a 2.885 If 31.5km h s 8.75ms at If Ji Ji Ti Te at 8.75ms 7.40ms 2.88s Ji 30.062 m s VERSE sforwardf e30 im 12 HOMEWORK (Acceleration) A. 1. If it takes a vehicle 3.40 seconds to speed up from rest to 60.0 km/h, then what is the magnitude of its acceleration (in m/s2) ? 2. When a vehicle has its brakes applied, it slows down from 90.0 km/h to rest. If the acceleration is 7.50 m/s2 backwards, then how much time did this motion take? 3. An object is moving at a velocity of 8.00 m/s towards the North. If it experiences an acceleration of 1.30 m/s2 South for 2.60 seconds, then what is the object’s final velocity? B. 4. An object experiences an acceleration of 4.50 m/s2 towards the West for 8.10 minutes. What was the object’s change in velocity? 13 5. An electron, initially moving at a velocity of 7.40 106 m/s towards the left, experiences an acceleration for 50.0 s (microseconds). If its final velocity is 3.30 106 m/s towards the right, then what is the object’s acceleration? 6. An object, originally moving at 35 cm/s towards the right, experiences an acceleration of 1.8 m/s2 towards the left. If the final velocity of the object is 74 cm/s towards the left, then determine how much time this motion took. 7. An object experienced an acceleration of 0.350 m/s2 towards the East for 2.80 minutes. If its final velocity was 45.0 m/s towards the West, then what was its initial velocity? SOLUTIONS (For the solutions below, North, East, and Right are considered positive directions) 1. vi = 0 ; vf = 16.6667 m/s ; t = 3.4 s ; a = 4.90 m/s2 2. vi = 25 m/s ; v f = 0 ; a = 7.50 m/s2 ; t = 3.33 s 3. vi = 8.0 m/s ; a = 1.30 m/s2 ; t = 2.60 s ; v f = 4.62 m/s North 4. t = 486 s ; a = 4.50 m/s2 ; v = 2.19 103 m/s West 5. vi = 7.4 106 m/s ; v f = 3.3 106 m/s ; t = 50 106 s ; a = 2.14 1011 m/s2 Right 6. vi = 0.35 m/s ; v f = 0.74 m/s ; a = 1.80 m/s2 ; t = 0.61 s 7. v f = 45 m/s ; a = 0.35 m/s2 ; t = 168 s ; vi = 104 m/s West 14 Lesson 4: Graphical Analysis (Uniform Acceleration) Position-Time Graphs There are two ways to calculate velocity when the velocity is changing throughout the motion: 1. Average Velocity To find the average velocity from t = t1 to t = t2 : secant v avg = Slope of secant line from the point at t = t1 to the point at t = t2 t1 t2 t 2. Instantaneous Velocity To find the velocity at t = t1: tangent v1 = Slope of tangent line at t = t1 Use two points on the tangent line. t1 t Example. Determine: a) the average velocity from 0.25 s to 5.00 s Ref: North + b) the instantaneous velocity at 1.00 s South I 50 48 46 44 42 40 slope 38 36 slope Vare Vinst 34 32 30 3 19 Position (m) 28 26 24 15 59 22 20 18 16 14 12 10 I 8.21ms 8 6 4 2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time (s) 15 Velocity-Time Graphs - there are two major calculations you make: Acceleration - acceleration is the slope of a velocity-time graph - if the velocity-time graph has a constant slope (i.e. a straight line), then the object has a constant acceleration Displacement - displacement is the area under a velocity-time graph a i.e. Area between the line and the t-axis Example. Velocity as a function of time 3 East + West 2.5 Velocity ( 102 m/s) 2 a_slope 9 im's 1 1.5 A 1 5.0 2 0.5 141 Az 1 ms 0 0 5 10 15 20 25 30 35 40 Time (s) Determine the object’s : a) acceleration b) displacement from 9.0 s to 28.0 s Slope 2xio'mls 2280m A lxw.la A 1 Items 1045m AT 3325m sd 3.33km 16 skip Example: Draw the position-time graph for each of the following velocity-time graphs. a. b. skip Example: Given the following parabolic position-time graph, draw both the velocity-time graph and the acceleration-time graph for the same object. a. 17 HOMEWRK (Graphical Analysis) Ref: Up +, Down Both A and B 1. a) Determine the average velocity from: (i) t = 0 to t = 4.4 s (ii) t = 1.6 s to t = 5.4 s b) Determine the instantaneous velocity at: (i) t = 1.6 s (ii) t = 2.8 s (iii) t = 4.4 s A. 2. Assume North is positive. a) Determine the acceleration of the object. b) Determine the displacement of the object for the first 3.1 s 18 B. 3. Assume North is positive. a) What is the object’s velocity at: (i) 35 s? (ii) 75 s? b) When is the object at rest? c) When is it (i) moving North? (ii) moving South? d) Determine its acceleration at (i) 20 s (ii) 35 s (iii) 65 s e) When is it moving North and slowing down? f) When is it moving South and speeding up? g) Calculate the object’s displacement for: (i) the first 60 s (ii) the first 120 s 4. Skip Sketch position-time and velocity-time graphs for each. (Ref: Forward is positive) a) a constant backward b) moving forward and c) moving backward and velocity speeding up slowing down SOLUTIONS 1. a) (i) 6.4 m/s up (ii) 6.3 m/s down b) (i) 11 m/s up (ii) 0 (iii) 15 m/s down 2. a) 5.5 103 m/s2 North b) 3.9 102 m North 3. a) (i) 4.0 m/s N (ii) 6.0 m/s S b) 0, 60, 120 s c) Moving North: 0 - 60, 120 - 140 s Moving South: 60 - 120 s d) (i) 0.13 m/s2 N (ii) 0 (iii) 0.60 m/s2 S e) 45 - 60 s f) 60 - 70 s g) (i) 1.5 102 m N (ii) 60 m S 4. a) b) c) 19 9 slope experiment ñ m s error 100 Lesson 5: Kinematic Equations In general, if an object has a constant acceleration, you can use the following equations: v f vi vi v f vi v f a d t where v avg t 2 2 1 1 d v i t a t 2 2 2 v f vi 2 a d d v f t a t 2 2 2 Note: If an object reaches a maximum displacement, then it has come to rest. d If the object has a constant velocity, the acceleration is zero. Use v . t Example. A ball is rolled up an incline at 24.1 km/h. If its maximum displacement is 9.6 m, determine the time of the motion. V 24.1km h 6.694mi sd t 8.6m e ᵗ Example. A particle accelerates at 470 m/s2 North for 810 ms. If its final location is 123 m South of its original location, determine its initial velocity. a 470m s d it t 0.810s d 123m Ift It SEE vi v vi t ffomso.es Vi 342 m s south 20 HOMEWORK (Kinematics Equations) A. 1. A car, starting at rest, accelerates forward at an average rate of 5.8 m/s2. How much time would it take for the car to travel a displacement of 70 m? V O s Iat a 5.8m s sd sd 70m fat t 2. 215 An airplane must take off from an airfield over a distance of 370 m. If it starts from rest, and its take-off speed is 240 km/h, then what is the magnitude of its average acceleration? d 370m Vi O Vf Vi 2ad f VF 240km h a a 3. An object accelerates from rest to a speed of 75.0 km/h. If it travels a distance of 150 m, then how much time did it take? d Eft 4. An object accelerates towards the left at 13.0 m/s2 over a displacement of 44.0 m to the right. If the object’s final velocity is 15.0 m/s to the right, then what is its initial velocity? 21 5. An object, having an initial velocity of 21 m/s upward, moves for a total time of 5.0 seconds. If its final position is 48 m below its original position, then what is its average acceleration? B. 6. An object, having an initial velocity of 11.6 m/s North, travels for a total time of 1.90 s. If its overall displacement is 7.20 m South, then what is its final velocity? 7. A car, travelling at an initial speed of 110 km/h, has its brakes applied and it slows down to a speed of 40.0 km/h. If its braking distance is 38.0 m, what is its average acceleration? 8. An object accelerates at a rate of 420 m/s2 West for 3.90 seconds. If its overall displacement is 720 m East, then what is the object’s final velocity? 22 9. An object, initial moving upward with a speed of 9.50 m/s, experiences a constant acceleration of 1.60 m/s2 downwards. How much time would it take for it move up to a maximum displacement, come to rest, and then come back down to a position that is 4.00 m above its original position? Hint: You may wish to find the final velocity first. 10. While driving her car at a speed of 90.0 km/h, Mrs. Jones sees an obstruction in the road. It takes her 0.80 s to react and put her foot on the brake. If, when the brake is applied, the car “decelerates” at a uniform rate of 9.3 m/s2 to rest, what is the total displacement of the car (from when she first sees the obstruction)? SOLUTIONS 1. 4.9 s 2. 6.01 m/s2 3. 14.4 s 2 4. 37 m/s right 5. 12 m/s down 6. 19.2 m/s South 2 7. 10.7 m/s backwards 8. 634 m/s West 9. vf = 8.8006 m/s downwards ; t = 11.4 s 10. Reaction distance: 20 m ; Braking distance: 33.6 m ; Total distance: 54 m 23 Quizanthusday Distance total Displacement Average velocity Motion Graphs constant velocity Accelerating dL Velocity Time Graph acceleration displacements 3 Questions using 5 king L S III t Velocity To Lesson 6: Projectiles (1-D) A projectile is any object that is thrown or dropped into the air. Assuming no other forces, all projectiles near the surface of the Earth experience the same acceleration caused by the gravitational force exerted by the Earth: V30 m Acceleration due to gravity: a g = 9.81 downwards s2 t.fi When a projectile is thrown or dropped vertically, there is no horizontal motion and: its vertical velocity will be zero for an instant at maximum height. (but it is still accelerating downward due to gravity) at any equal position (height) on its path, the vdown = - vup Ld tup at any equal position on its path, then tup = tdown Note: If an object is dropped, then vi = 0 (relative to the person) If an object is rolled up or down a slope, a 9.81 m/s2 Y tdow Vi Example: A ball is thrown straight upward at 47.0 m/s. How much time does it take to return to its original (release) height? t47.0m s t am t f4 a 9.81m s ffmd 9.58D at Example. On the Moon, gravity is 1/6 that of Earth. When a rock is thrown straight up, it takes 18.4 s for it to return to its release height. What is its max height? amoon f 9 1ms 1.635mg top E 18.4s 9.2s Vf 0 692nd sd sd f fat 24 sat 1.635m5 92 69.19m t HOMEWORK (1-D Projectiles) Assume 1-D motion and no air resistance. A. 1. A rock is thrown upward at a speed of 15.0 m/s. a) What would its maximum height be (measured from its release height)? b) How much time would it take to go up and then return to its original height? t.FI L L f a a.am 2. d A t t.ve On Planet X, an object is launched upward with a speed of 28.0 m/s. If it takes a total time of 7.50 seconds for it to go up and return to its original (release height), a) what is the acceleration due to gravity on this planet? b) what is the maximum height of this object? Theoretical 3. An object is launched upward at a speed of 26.0 m/s. What is this object’s velocity when it reaches a height of 30.0 m above its release height? Why are there two answers? 4. An object is launched upward at a speed v. If it is in the air for 3.80 s and its final position is 11.0 m above the original position, then what is the speed v ? 25 B. 5. An object is dropped from a height of 85.0 m. What is the object’s average velocity? 6. An object is thrown upward at a speed of 17 m/s. How much time would it take for this object to reach a position that is 10 m below its original position? Hint: Find the final velocity first. 7. On Planet G, an object is thrown upward and it flies through the air for 16.0 seconds. If the object’s velocity is 13.0 m/s downward when it is located 24.0 m below its original position, then what is the acceleration due to gravity on this planet? 8. On Planet Y, an object that is dropped from a height of 12.0 m takes 3.20 s to reach the ground. If another object on this planet is thrown up in the air and it takes 8.50 s to reach its original position, then to what maximum height does it reach? SOLUTIONS 1. a) 11.5 m b) 3.06 s 2. a) 7.47 m/s2 down b) 52.5 m 3. It reaches this height going up and going down. v = 9.35 m/s up and 9.35 m/s down 4. 21.5 m/s 5. vf = 40.837 m/s downwards ; vavg = 20.4 m/s downward 6. vf = 22.0273 m/s downwards ; t = 4.0 s 7. 1.44 m/s2 downwards 8. a = 2.3438 m/s2 downwards ; d = 21.2 m 26 Lesson 7: Two Dimensional Vectors Direction of 2-D Vectors y For 1-D motion, you can simply use a reference system (i.e. a positive and negative direction) since there are only two directions. 20 Note However, for 2-D motion, there are an infinite number of possible 9200 x directions. There are two common methods to show the direction of 2-D 3050 vectors: #1. NEWS (i.e. North, South, East, West) #2. PCS (Polar Coordinate System) 615 - angles measured from N, S, E, or W axis - angle measured from the positive x-axis - angle is always acute (less than 90) Positive angles = counterclockwise Negative angles = clockwise N y W 60 E 60 60 120 4 x S = 60 N of W = + 120 (or 240) axis it touches Example 1. Example 2. Express the vector’s angle Sketch a vector with an angle that is in NEWS and PCS. 50 E of N. 90490 00 Eta 20 20 Wofs NEWS: _____________ 70 Sofw PCS: ______________ 250 or 110 27 Adding 2-D Vectors (Tail-to-Tip) Consider the two vectors shown: i V2 V1 If we wish to add these two vectors tail-to-tip (i.e. V1 V2 ), do the following: Place the tail of the second vector on the tip of the first vector The resultant (or sum) vector R is drawn from the origin to the tip of the second vector Resultant = “Start to finish” vector R V1 V2 R V2 V2 V1 V1 If the vectors are at right angles, then a right triangle is formed. tan V2 tan 2 2 Pythagorean Formula: R V1 V2 Soh Cah Toa: V1 Example. A hiker walks the following path from a cabin: 40 m West, then 70 m North. What is her overall displacement (mag and dir)? op sati.it sd 80.62m 81m 70m tan 60.255 40mF 60 Nofw sd 8lm 60No 28 HOMEWORK (2-D vectors; tail-to-tip addition) A. 1. For the following vectors, a) 14 b) express the angle in NEWS and in PCS. 68 2. Sketch the resultant vector by adding each set of vectors. V1 a) V2 b) V2 V3 V1 3. Determine the resultant displacement (magnitude and direction) for the following: a) A person walked 8.50 km North, and then 5.70 km East N r Note b) A boat sailed 12.0 km West, and then 19.0 km South 4. A person walked 1.8 km West, and then an unknown distance South. If the magnitude of the resultant displacement is 3.0 km, then determine: a) the distance the person walked South b) the angle of the resultant displacement 29 5. Sketch the resultant displacement vector, using a tail-to-tip diagram. No calculations. 14 m at 20 N of W, and then 25 m at 10 W of S B. 6. it Determine the resultant displacement (magnitude and direction) for the following: a) A person walked 8.0 blocks East, 5.0 blocks North, and then 14.0 blocks West C 85T is 8 b) A plane flew 140 km North, then 290 km East, and then 300 km South 7. A boat travelled an unknown distance North, and then 19.0 km East. If the magnitude of the resultant displacement is 23.0 km, then determine: a) the distance the boat travelled North b) the angle of the resultant displacement 30 8. Three vectors are added together: V1 = 72.0 units South ; V 2 = 55.0 units East ; V 3 If the resultant vector is zero, then determine the magnitude and direction of V 3. 9. Sketch the resultant displacement vector, using a tail-to-tip diagram. No calculations. 5 km at 30 E of S, and then 11 km at 70 N of E SOLUTIONS 1. a) 14 W of N ; 104 (256) b) 68 W of S ; 158 (202) 2. a) R V1 V2 b) R V1 V2 V3 (see diagram on right) 3. a) 10.2 km at 33.8 E of N (56.2 N of E) b) 22.5 km at 57.7 S of W (32.3 W of S) 4. a) 2.4 km b) 53 S of W (37 W of S) 5. See diagram on right. 6. a) 7.8 blocks at 40 N of W (50 W of N) b) 331 km at 28.9 S of E (61.1 E of S) 7. a) 13.0 km b) 55.7 E of N (34.3 N of E) 8. 90.6 units at 52.6 N of W (37.4 W of N) 9. See diagram on right. 31 Lesson 8: Adding By Components Vector Components To determine the magnitude of each component: Opp Vx y y from Hp Horizontal V Vy Fiscal W Wy OPP Hyp Adj x V x W Wx ay sin x Vx V sin cos x Wx W cos V W Vy Wy cos V y V cos sin W y W sin V W Example. Determine the x- and y-components of the vector V = 240 m at 30 W of S. i.si I Adding 2-D Vectors (Component Method) Draw and determine Components. CARPO Find the x- and y-components of all vectors Determine the total in each direction. Add up the x, add up the y R x x1 x2 ... and R y y1 y2 ... Make sure to incorporate direction (i.e. north +, south -) Draw a vector addition diagram with the totals tip-to-tail. Add R x and R y tail-to-tip Use Pythagorus to determine the resultant magnitude, use tan θ to determine direction Ry R Rx 2 Ry 2 and tan Rx 32 Example. A boat completes 3 legs of a journey: 300 m West, 500 m at 20.0 E of S, and 600 m at 10.0 N of E, determine its overall displacement (mag and dir). if 300m I it fffids d3y 600cos100 t591m dzx 3 104m day 600sin10 dz Sinf 500sin20 171m _of day dz cost 500 0520 470m dxto 300m 171m 591m 461m 366m dytor o 470m 104m 366m d 46T 589m tan tail 38 sdt 589m 38 Sof HOMEWORK (Components) A. 1. Determine the x- and y-components for the following vectors: a) 2.50 km at 180 b) 3.90 m South 2.50km off 0 9s.am Y O SOFE c) 27.0 km at 340 d) 928 m at 43.0 N of W de É fa dy E 33 2. A vector has an x-component of 160 m East and a direction of 71.0 N of E. a) What is the y-component of the vector? b) The magnitude of the vector? n COS 71 Vy ii H E m 3. You wish to add three vectors. Their components are shown below: Vector 1: x1 = 15.0 km West y1 = 46.0 km North Vector 2: x2 = 67.0 km East y2 = 93.0 km South Vector 3: x3 = 80.0 km West y3 = 12.0 km South What are the magnitude and direction of the resultant (sum) vector? 4. Add the following vectors, using components. a) 45.0 m North, 60.0 m at 210 b) 3.90 km at 62.0 S of E, 5.80 km at 24.0 S of W 34 B. 5. A vector has a y-component of 3.10 km South. The x-component is directed East and the magnitude of the resultant vector is 5.00 km. a) What is the x-component? b) What is the angle of the resultant vector? 6. Add 7.00 km at 18.0 W of N and 11.0 km at 69.0 E of N using components. 7. Two vectors are added: V1 = 460 units at 81.0 N of W and V 2. If the resultant vector is 270 units at 25.0 N of E, then determine V 2. SOLUTIONS 1. a) x = 2.50 km W ; y = 0 b) x = 0 ; y = 3.90 m S c) x = 25.4 km E ; y = 9.23 km S d) = 137 ; x = 679 m W ; y = 633 m N 2. a) y = 465 m N b) R = 491 m 3. xT = 28 km W ; yT = 59 km S ; R = 65.3 km at 64.6 S of W (25.3 W of S) 4. a) Rx = 51.9615 m , Ry = 75 m ; R = 91.2 m at 55.3 N of W (34.7 W of N) b) Rx = 3.4677 km ; Ry = 5.8026 km; R = 6.76 km at 59.1 S of W (30.9 W of S) 5. a) x = 3.92 km E b) 38.3 S of E (51.7 E of S) 6. Rx = 8.1063 km , Ry = 10.5994 km ; R = 13.3 km at 52.6 N of E (37.4 E of N) 7. x2 = 316.66 , y2 = 340.23 ; V2 = 465 units at 47.1 S of E (42.9 E of S) 35 Lesson 9: Relative Velocity A reference frame is a set of coordinates that allows for measurements of position, speed, etc. An observational frame of reference is one where the observer is at rest in that frame. To measure the speed of car relative to the ground means we measure the speed of the car as seen by an observer at rest on the ground. The velocity of an object is dependent upon the frame of reference of the observer. E.g. A person holding a ball in their hand as they drive at 60 km/h would measure the speed of the ball as zero, relative to them or the car. A person standing on the side of the road would measure the speed of the ball to be 60 km/h, relative to the ground. Add or subtract velocities to determine relative velocity in other frames of reference E.g. A person walking along a moving walkway (in the same direction) will have a greater velocity relative to the ground then a person walking at the same speed along the ground. This is determined by adding the velocity of the person to the velocity of the walkway. Objects Moving in a Current For an object moving relative to a current (wind or water) the vector equation is: it vobject relative to ground vobject relative to current vcurrent relative to gound vactual vobject no current vcurrent E.g. A plane can travel at 100 km/h in still air. If it aims east into a wind that blows at 20 km/h west, then determine the ground distance that it will travel in 5.0 h. plane Iwind VACTUAL nd 100km h f 20km h 80km h V d vt 80km h 5.0h 4.041rad I 2-D Uniform Motion The addition of velocities is not restricted to one-dimensional situations. The vectors can be added even when they are at angles to each other, they just need to be added tip to tail. 1 Remember that constant velocities (a = 0) can be calculated using d vi t (0) t 2 d v t 2 36 Example: A pilot sets the airspeed (velocity relative to air) of a plane at 300 km/h north, but there is a wind blowing east at 50 km/h relative to the ground. Draw a vector diagram to show the velocity relative to the ground, and the displacement of the plane relative to the ground. Velocity Diagram: 1 d vt Displacement Diagram: Wind 50 km/h Air speed 300 km/h Actual d Velocity 304km h Same direction as the actual tan 9.46 9 E velocity 90 EoFN Example: A boat can travel at a max speed of 11.0 m/s relative to the water. It sets a heading south across the 90.0-m width of a river flowing east at 6.00 m/s relative to the shore. Determine: a) the boat’s velocity relative to the shore (w.r.t ground) b) the total distance it travels I dB 90m Actual 8 ACTUAL t.it E s I dtctVAc.tt ndActuac 103mD Example: A pilot sets the airspeed (velocity relative to the air) of a plane at 240 km/h, but there is a wind blowing west at 60 km/h relative to the ground. Draw a vector diagram to show the direction the plane must aim in order to have an actual velocity that is due south. START sine C a b 2404km h w̅ wind sin 4,2 LEGO Jacture a sin 232k 14 E Vwind 60km h 37 Example: A boat can travel at a max speed of 14 m/s in still water. It must travel due East (wrt the ground) across a 120 m wide river, which has a current flowing at 8.0 m/s South. Determine: a) the direction the boat must aim b) the time to get across since Eider dart Fidid in sin 0 350 4 10.450 VACTUAL HOMEWORK (Relative Velocity) Ém v t A. 1. A boat can travel at 7.0 m/s in still water. If it is placed in a river that has a current of 3.0 m/s South, what would be the boat’s velocity with respect to the shore if the boat headed: a) upstream (opposite to current) b) downstream (same as current) 2. A boat that can travel at 11.0 m/s in still water is placed into a 135 m wide river that has a current of 8.0 m/s North. If it is aimed directly West, then calculate: a) the velocity of the boat with respect to the ground b) the time it takes to travel to the other side of the river c) how far downstream the boat travels ne s Vact day Visit it 114 1 38 3. A river that is 57.0 m wide has a current that is 4.90 m/s West. A boat can travel at 6.50 m/s in still water. If the boat must travel directly North across the stream, a) at what angle must the boat be aimed? b) what would be the magnitude of boat’s velocity (with respect to the shore)? c) how much time would it take to reach the opposite shore? B. 4. A plane that can travel at 400 km/h in still air heads directly into a 50.0 km/h wind that blows directly West. How long would it take this plane to reach its destination, which is 1900 km due East? 5. A plane that can travel at 200 km/h in still air heads directly South. However, a wind comes from the West (towards the East) and blows it off course. If the angle of the plane’s final velocity is 6.50 E of S, then determine: a) the wind speed b) how far the plane travels in 3.00 hours 39 6. A plane must travel due East, but there is a wind that is blowing from the North (towards the South) at 45.0 km/h. If the plane’s actual speed is 170 km/h, then determine: a) the heading (angle) of the plane b) the speed of the plane if it was in still air 7. A plane that can travel at 120 km/h in still air heads towards the West. However, there is a 55.0 km/h wind that is directed at 70.0 N of E. What is the actual velocity of the plane (with respect to the ground)? SOLUTIONS 1. a) 4.0 m/s upstream b) 10.0 m/s downstream 2. a) 14 m/s at 36 N of W (54 W of N) b) 12 s c) 98 m 3. a) 48.9 E of N (41.1 N of E) b) 4.27 m/s c) 13.3 s 4. v = 350 km/h East ; t = 5.43 h 5. a) 22.8 km/h b) v = 201.29 km/h ; d = 604 km 6. a) 14.8 N of E (75.2 E of N) b) 176 km/h 7. xT = 101.1889 km/h ; yT = 51.6831 km/h ; v = 114 km/h at 27.1 N of W (62.9 W of N) 40 Lesson 10: Horizontal Projectiles Horizontal components: gravity-free path (constant velocity) - moves at a constant velocity ( a x 0 ) 1 d x vi t (0) t 2 d x vxt 2 V ᵈ dropped ball (a = g) Vertical components: - behaves like a dropped object i.e. v i y 0 (starts at rest) a y g = 9.81 m/s2 downward 9.81m s Example. A rock is thrown horizontally at a speed of 27.0 m/s from a building that is 19.0 m high. Determine the horizontal distance travelled. IT lsd vii t.at g ÉÉÉ d fat t.EE t.i Vx d dx vx dx 53.1mD 41 HOMEWORK (Horizontal Projectiles) Assume no AR and a level surface. A. 1. A rock is thrown horizontally off of a 10 m high building with a speed of 14 m/s. How far away from the base of the building will the rock hit? 2. A crate is dropped from a plane that is moving forward at a speed v at a height of 85.0 m. If the crate lands 120 m horizontally from the release position, then what is the speed v ? 3. A bullet is projected horizontally from a height H at a speed of 350 m/s. If it lands 225 m horizontally from the release position, than what is the initial height H ? 4. A rock is thrown horizontally off of a cliff that is 35.0 m high with a speed of 18.0 m/s. What is the velocity of the rock just before it hits? Hint: Find both components of the final velocity first. titlad fan to VFI Tad 26.2ms f v ftp t ff8 3t 42 tan 55.5 FromHorizontal 34.5 fromvertical B. 5. On Planet X, a rock is thrown horizontally at a speed of 29.0 m/s and at a height of 3.20 m. If the rock travels a horizontal distance of 52.0 m, then what is the acceleration due to gravity? 6. A rock is thrown horizontally from a height of 2.50 m. When it lands, the angle of its velocity vector is 36.0 below the horizontal. What was the initial speed? 7. A car driver, travelling at 90.0 km/h, notices that he is heading directly for a cliff when he is 20.0 m away from the edge. He applies the brakes and experiences a backward acceleration of 8.00 m/s2, but it is not enough. The car goes over the cliff. If the cliff is 12.0 m high, then calculate how far the car travels horizontally while in the air (until the car lands). SOLUTIONS 1. t = 1.4278 s ; d = 20 m 2. t = 4.1628 s ; v = 28.8 m/s 3. t = 0.6429 s ; h = 2.03 m 4. vfx = 18 m/s forward ; vfy = 26.2 m/s down ; v = 31.8 m/s at 55.5 below horizontal 5. t = 1.7931 s ; a = 1.99 m/s2 down 6. vfy = 7.0036 m/s ; vi = vfx = = 9.64 m/s 7. Moving at 17.4642 m/s when it reaches the cliff’s edge. While in air: t = 1.5641 s ; d = 27.3 m 43 Lesson 11: Diagonal Projectiles 1st Step: Draw the components of the initial velocity (vy = 0) down vi viy dmax θ vix Horizontal motion: moves at a constant velocity ( a x 0 ) Vertical motion: vi = vix viy ≠ 0 so it behaves like an object thrown 1 vertically d x vi t (0) t 2 d x vxt 2 vi viy a y g = 9.81 m/s2, downward Note: At any given height, vup = vdown From the same height, the time up to the top is the same as the time down to that height. The overall path of the projectile is a parabola. Example. A golf ball is hit with a speed of 78.0 m/s at 23.0 above the horizontal. Determine the horizontal distance the ball travels until it lands. (No AR) Components ofVelocity mom's iii 1 i dx a V IFom.at i 9.81m s V t dy.TV 30.48ms 30.48m's f dy Vx't t f 6.21 44 446m ᵗdy 0 HOMEWORK (Diagonal Projectiles) Assume no AR and a level surface. A. 1. A golf ball is struck with an initial velocity of 47.0 m/s at an angle of 28.0 above the horizontal. If this ball lands and rolls and additional 15.0 m, then determine the maximum height that this ball achieves. iiiiiiiinini.no h Vx 47cos28o 4 sm 2 Vf V Zad s a dmax 24.8 2. A football is kicked with an initial velocity of 25 m/s at a projection angle of 34. Will this ball go over the 9.2 m uprights, if they are located 40 m away? Prove by calculation. Geek 1 Y Miss www.iii.fi dx 40m theft t dy t 9 45 It 1.9299 s t 1.9299 s d vit fat 13,9711.929,9 0.5 9.81m5 1.9299s 8.71m 45 B. 3. An object is thrown with an initial speed of 14.0 m/s. If the maximum height of this object is 6.00 m, then what was the projection angle? a.Ya.ams y fi 6.00m 4. A soccer ball is projected at an angle of 31.0. If it stays in the air for a total time of 3.00 seconds, then what was its initial speed? 5. For the world record shot put in 1990, the shot was projected at a velocity of 14.5 m/s at 37 above the horizontal. If the release height was 2.0 m, then how far (horizontally) was it thrown? Hint: Find the vertical component of the final velocity first. SOLUTIONS 1. a) t = 4.4985 s ; dair = 186.68 m ; dT = 202 m b) hmax = 24.8 m 2. t = 1.93 s ; h = 8.71 m Not high enough (h < 9.2 m) 3. viy = 10.8499 m/s ; = 50.8 above the horizontal 4. viy = 14.715 m/s ; vi = 28.6 m/s 5. vfy = 10.7419 m/s downward ; t = 1.9845 s ; d = 23 m 46 b.at f E TAct Veuurent 10.0m s Act Five Eofs Distance Travelled case H f.se 112m Vx 150m s X 9 a 9.81m s dx Vig dy 4 2.025 Ufy 1829 dx vxt d Kitkat 303m t 2 2.02s V COS 122.2mg Vig v sine 44.5m I v.in 4 9.075 iii's Vfy 44 fm vtt dy Vf 1109m9.07 dx damax 111 Vig 40.5in 40 25.7ms 30.6m 40 cos 40 my 0 d IQ 33 50m w 80m 60 Nofw 80m 0 som Y 0 X 80 cos 60 40m y 80 sin 60 69m to 50 40 90m g at 69m 695 902 113m tan 37 Nofw PHYSICS 20 Unit 1 Kinematics Review 47 One Dimensional Kinematics Practice Exam: A. Conceptual Questions 1. Identify the following as a scalar (S) or a vector (V). Time ____ Position ____ Distance ____ Displacement ____ Speed ____ Velocity ____ Acceleration ____ 2. Position is the location of an object with respect to ____________________ 3. Displacement is the _______________ to ____________________ vector. 4. a) An object moves 10 km North, and then it moves 10 km South Overall distance: __________ Overall displacement: ____________ b) An object moves 40 m West, and then it moves 50 m East Overall distance: __________ Overall displacement: ____________ 5. What is the slope of a position-time graph? ________________________ 6. What is uniform motion? __________________________________________ How does it look on a position-time graph? __________________________________ 7. When does it mean when an object accelerates? ________________________________________ 8. An object has a negative velocity and a positive acceleration (North +, South ). Which direction is it moving? _________ Is it speeding up or slowing down? _____________ 9. What does it mean to have: a) zero position? _______________________________ b) zero velocity? _________________________________________ c) zero acceleration? _______________________________________________ 10. On a position-time graph, how do you find: a) average velocity? ________________________________________________ b) instantaneous velocity? _____________________________________________ 11. On a velocity-time graph, how do you find: a) acceleration? ___________________________________________________ b) displacement? _________________________________________________ 12. If an object has a constant velocity: a) what is its acceleration? __________ b) which equation? ____________________ 13. What is the one condition required to use the acceleration equations on the formula sheet? _____________________________________________________________________ 14. What does it mean when an object undergoing uniform acceleration has reached a maximum displacement? ________________________________________________________ 48 One Dimensional Kinematics Practice Exam: B. MULTIPLE CHOICE 1. What is the measurement shown, 20.3 20.4 kg to proper significant digits? A. 20.3 kg B. 20.38 kg C. 20.382 kg D. 20.3820 kg 2. How many significant digits does the number 0.004 500 m have? A. 2 B. 4 C. 6 D. 7 3. What is the quantity 0.000 000 000 000 005 83 seconds in scientific notation? A. 5.83 1015 s B. 5.83 1017 s C. 5.83 1015 s D. 5.83 1017 s 4. If you subtract the quantities 140.3 m 15.76 m, then what is the answer (to proper significant digits)? A. 124.54 m B. 124.5 m C. 125 m D. 1.2 102 m 5. If you multiply the quantities 20.400 m 0.0720 m, then what is the answer (to proper significant digits)? A. 1.4688 m2 B. 1.469 m2 C. 1.47 m2 D. 1.5 m2 6. If the quantity 746 Ms is converted to seconds, then what is the proper conversion? A. 7.46 106 s B. 7.46 104 s C. 7.46 106 s D. 7.46 108 s 7. If the quantity 4.92 107 m is converted to nanometres (nm), then what would be the correct conversion? A. 4.92 1016 nm B. 4.92 nm C. 492 nm D. 4.92 1016 nm 8. F has an inverse-square relationship with G. If G gets multiplied by 4, what happens to F? A. Multiplied by 4 B. Multiplied by 16 C. Divided by 4 D. Divided by 16 49 y 9. What is the equation of the straight line shown? 9.5 A. y = 1.1 x + 4.0 B. y = 1.1 x + 9.5 C. y = 0.94 x + 4.0 2.0 D. y = 0.94 x + 9.5 8.0 x 10. Which of the following is NOT true about the manipulated variable? A. It has values that are set by the experimenter (not measured). B. It tends to change regularly (i.e. in regular increments). C. Its values are plotted on the y-axis on the graph. D. All of the above are true about the manipulated variable. 11. A vehicle travels at a constant speed of 50.0 km/h. How far can this vehicle travel in 0.217 minutes? A. 181 m B. 234 m C. 385 m D. 650 m 12. An object moves 110 m East, and then moves 270 m West. What is the overall displacement? A. 160 m West B. 380 m West C. 160 m East D. 380 m East 13. An object starts at a position that is 14.0 m South of the origin. If the object then moves at a velocity of 2.80 m/s North for 9.10 seconds, what is its final position? A. 11.5 m South of the origin B. 39.5 m South of the origin C. 11.5 m North of the origin D. 39.5 m North of the origin 14. An object has a positive velocity but a negative acceleration. If forward is considered positive, then which is the best interpretation of the motion? A. moving backward and speeding up B. moving forward and speeding up C. moving backward and slowing down D. moving forward and slowing down 15. An object is initially travelling at 4.0 m/s towards the left. If it is then accelerated for 1.9 s and its final velocity is 6.0 m/s towards the right, what is its average acceleration? A. 1.1 m/s2 left B. 1.1 m/s2 right C. 5.3 m/s2 left D. 5.3 m/s2 right 50 16. For the position-time graph shown, up is positive. 14 13 The instantaneous velocity 12 at t = 2.2 s is 11 10 A. 4.1 m/s down 9 Position (m) B. 4.1 m/s up 8 7 C. 7.5 m/s down 6 D. 7.5 m/s up 5 4 3 17. What is the average velocity 2 for the first 1.4 seconds? 1 0 A. 7.1 m/s down 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 B. 7.1 m/s up Time (s) C. 8.6 m/s down D. 8.6 m/s up 18. A car has its brakes applied and it experiences a backward acceleration of 8.2 m/s2. If it comes to rest over a distance of 17 m, then how much time did this take? A. 2.0 s B. 3.4 s C. 4.1 s D. 5.6 s 19. An object, moving North at a speed of 14.0 m/s, experiences a constant acceleration of 6.50 m/s2 towards the South. If its final position is 11.0 m South of its original position, then what is its final velocity? A. 7.28 m/s South B. 18.4 m/s South C. 7.28 m/s North D. 18.4 m/s North 20. An object, starting at rest, accelerates to a speed of 70.0 km/h in 5.40 seconds. How far did it travel in this time? A. 52.5 m B. 105 m C. 189 m D. 378 m 21. Which of the following velocity-time graphs shows an object moving West and speeding up a constant acceleration? East is positive. A. B. v v t t C. D. v v t t 51 For the position-time graph shown, right is positive: 10 8 22. In which of the following intervals is 6 the object moving right? 4 Position (km) A. 7 - 9 h 2 B. 9 - 12 h 0 C. 16 - 20 h 0 2 4 6 8 10 12 14 16 18 20 -2 D. None of the above. -4 -6 -8 23. What is the overall displacement of Time (h) the object for the entire trip? A. 0 km B. 3.0 km left C. 3.0 km right D. None of the above. 24. What is the overall distance travelled by the object for the entire trip? A. 0 B. 3.0 km C. 26.0 km D. 33.0 km 25. In which of the following intervals is the object speeding up? A. 0 - 4 h B. 16 - 20 h C. Both A and B D. Neither A nor B 26. What is the velocity of the object at 11 h? A. 0.55 km/h left B. 0.55 km/h right C. 2.7 km/h left D. 2.7 km/h right 27. When is the object at rest? A. 9 h, 20 h B. 4 - 7 h, 12 - 16 h C. Both A and B D. Neither A nor B 28. What is the magnitude of the average velocity for the first 7 hours? A. 0 B. 1.0 km/h C. 1.4 km/h D. 1.9 km/h 52 For the velocity-time graph, North is positive. 30 20 29. On which intervals below is the object 10 moving North? Velocity (m/s) A. 5 - 8 s 0 0 2 4 6 8 10 12 14 16 18 20 B. 10 - 14 s -10 C. Both A and B -20 D. Neither A nor B -30 -40 30. On which intervals is the object at rest? Time (s) A. 8 and 20 s B. 0 - 5, 14 - 16 s C. Both A and B D. Neither A nor B 31. On which interval is the acceleration towards the South? A. 5 - 8 s B. 16 - 20 s C. Both A and B D. Neither A nor B 32. Calculate the magnitude of the acceleration at 12 s. A. 0.71 m/s2 B. 2.1 m/s2 C. 2.5 m/s2 D. 7.5 m/s2 33. On which interval is the object speeding up? A. 5 - 8 s B. 8 - 10 s C. Both A and B D. Neither A nor B 34. What is the displacement of the object in the first 8.0 seconds? A. 2.6 102 m South B. 2.6 102 m North C. 3.2 102 m South D. 3.2 102 m North 53 C. WRITTEN RESPONSE 1. A car travels at 110 km/h East for 3.8 hours, and then it travels at 95 km/h West for 6.7 hours. Calculate the car’s average speed and average velocity. Answers in km/h. 2. An object has an initial velocity of 576 km/h North. It then experiences an average acceleration of 15.0 m/s2 South for a time t, and in the end, its final position is 70.0 m South of its original location. Calculate the time t. Hint: You may wish to find the final velocity first. 54 3. The position of a cart was measured using a ticker tape apparatus. The results are as shown: t (s) 1 1.5 2.0 2.5 3.0 3.5 4.0 d (cm) 56 77 95 118 134 152 173 Graph the data on the grid given. Title: ________________________________________________ Slope (and units): y-intercept (and units): What is the significance of the slope and the y-intercept? What is the equation of the line of best-fit? (using the original variables) 55 One Dimensional Kinematics Practice Exam: (SOLUTIONS) A. Conceptual Questions (SOLUTIONS) 1. Time (S) ; Position (V) ; Dist. (S) ; Displacement (V) ; Speed (S) ; Velocity (V) ; Accel (V) 2. Origin 3. Start to finish vector 4. a) 20 km ; 0 b) 90 m ; 10 m East 5. slope = velocity 6. constant velocity (a = 0) 7. change in velocity 8. moving S, slowing down 9. a) at origin b) at rest c) no change in velocity 10. a) secant slope b) tangent slope 11. a) slope b) area btw line and t-axis 12. a) 0 b) v = d / t 13. constant acceleration 14. come to rest (v = 0) B. Multiple Choice (SOLUTIONS) 1. C 9. D 17. B 25. D 33. B 2. B 10. C 18. A 26. C 34. A 3. A 11. A 19. B 27. B 4. B 12. A 20. A 28. B 5. C 13. C 21. D 29. B 6. D 14. D 22. C 30. A 7. C 15. D 23. B 31. B 8. D 16. C 24. D 32. C C. Written Response (SOLUTIONS) 1. Leg 1: d1 = v1 t1 = (110 km/h) (3.8 h) = 418 km (East) Ref: East +, West Leg 2: d2 = v2 t2 = (95 km/h) (5.7 h) = 636.5 km (West) - d T 418 km 636.5 km Average speed: v 2 = 1.0x10 km/h tT 3.8 h 6.7 h r r d T 418 km 636.5 km Average velocity: v = 20 km/h = 20 km/h West tT 3. 8 h 6.7 h 2. vi = +160 m/s ; a = 15.0 m/s2 ; d = 70.0 m ; t ? Ref: North +, South - 2 2 2 v f vi 2ad v f vi 2ad = 166.433 m/s (negative, since moving down) v f vi v f vi 166.433 160 a t = 21.8 s t a 15.0 173 cm 56 cm 3. Avg speed: Slope = = 39 cm/s 4.0 s 1.0 s y-intercept: 16 cm (initial position) Equation: y = mx + b d = 39 t + 16 56 Two Dimensional Kinematics Practice Exam: A. Conceptual Questions 1. Sketch a vector with the following angles: a)
