U1.1 kinematics text.pdf - DAAScience10

Figure 1.9Scalar or vector?For all subsequent problems in this book, you will be using plus andminus signs to indicate direction. This method is more flexible forproblem solving and easier to use.Like distance and displacement, speed and velocity is anotherscalar-vector pair. Speed is the rate at which an object moves. It is ascalar quantity, so it has magnitude only; for example, v 50 km/h(Figure 1.9). Velocity is a vector quantity, so it has both magnitude(speed) and direction. If you are travelling south from Fort McMurrayto Lethbridge at 50 km/h, your velocity is written as v 50 km/h [S].If you designate south as negative, then v 50 km/h. Acceleration isa vector quantity that represents the rate of change of velocity. Youwill study aspects of displacement, velocity, and acceleration, andtheir interrelationships, in the sections that follow.1.1 Check and ReflectKnowledge1. What two categories of terms are used todescribe motion? Give an example of each.2. Compare and contrast distance anddisplacement.3. What is the significance of a reference point?Applications4. Draw a seating plan using the statementsbelow.(a) Chad is 2.0 m [left] of Dolores.(b) Ed is 4.5 m [right] of Chad.(c) Greg is 7.5 m [left] of Chad.(d) Hannah is 1.0 m [right] of Ed.(e) What is the displacement of a teacherwho walks from Greg to Hannah?5. A person’s displacement is 50.0 km [W].What is his final position if he started at5.0 km [E]?6. Using an autuk (a type of sealskin racquet),two children play catch. Standing 3.0 mapart, the child on the right tosses the ballto the child on the left, and then moves5.0 m [right] to catch the ball again.Determine the horizontal distance anddisplacement the ball travels from itsinitial position (ignore any vertical motion).7. Below is a seating plan for the head tableat a wedding reception. Relative to thebride, describe the positions of the groom,best man, maid of honour, and flower girl.Flowergirl3.0 m 5.0 m0.75 m 0.75 m 0.50 m 0.75 mBestmanBride Groom Maid ofHonour0.75 mRingboye TESTTo check your understanding of scalar and vectorquantities, follow the eTest links atwww.pearsoned.ca/school/physicssource.10 Unit I Kinematics

1.2 Position-time Graphs andUniform MotionYou are competing to win the Masters Golf Tournament. Thehole is 5.0 m away (Figure 1.10). You gently hit the ball withyour club and hold your breath. Time seems to stop. Then, 5.0 slater, it rolls into the hole. You have won the tournament!From section 1.1, you know that displacement is the changein an object’s position. If you replay the sequence of motions ofyour winning putt in 1.0-s intervals, you can measure thedisplacements of the golf ball from you, the putter, to thehole (Figure 1.11). Figure 1.10 You can represent motion insports using vectors and graphs.0.0 m0.0 sorigin1.0 m1.0 s2.0 m2.0 s3.0 m3.0 s4.0 m4.0 s5.0 m5.0 s Figure 1.11What is the golf ball’s displacement after each second?Table 1.1 displays the data from Figure 1.11 for the golf ball’s positionfrom you at 1.0-s intervals. By graphing the data, you can visualize themotion of the golf ball more clearly (Figure 1.12).▼ Table 1.1 Position-time dataTime (s) Position (m [right])t 0 0.0 0.0t 1 1.0 1.0t 2 2.0 2.0t 3 3.0 3.0t 4 4.0 4.0t 5 5.0 5.0Position (m [right])Position vs. Time6.05.04.03.02.01.00.00.0 1.0 2.0 3.0 4.0 5.0Time (s) Figure 1.12A position-time graph of the golf ballVelocityNotice that the graph in Figure 1.12 is a straight line. A straight line hasa constant slope. What does constant slope tell you about the ball’s motion?To answer this question, calculate the slope and keep track of theunits. Designate toward the hole, to the right, as the positive direction.Chapter 1 Graphs and equations describe motion in one dimension. 11

e SIMPractise calculatingaverage speed andaverage velocity.Go to www.pearsoned.ca/school/physicssource.riseRecall that slope . For position–time graphs, this equationrunchange in positionbecomes slope change in timeA change in position is displacement. So, the equation for slope becomesd slope td f d itf t iPHYSICS INSIGHTSpeed has magnitudeonly. Velocity has bothmagnitude and direction.velocity: rate of change in position5.0 m 0.0 m5.0 s 0.0 s1.0 m/sThe answer is positive, so the golf ball moves at a rate of 1.0 m/s [right].Notice that the units are m/s (read metres per second). These unitsindicate speed or velocity. Since displacement is a vector quantity, theslope of the position-time graph in Figure 1.12 gives you the velocity,v, of the ball: the change in position per unit time. Because you havecalculated velocity over a time interval rather than at an instant intime, it is the average velocity.v d tSpeed and Velocity4 m/s 4 m/s Figure 1.13 Objects withthe same speed can have differentvelocities.Objects travelling at the same speed can have different velocities. Forexample, a tram carries passengers across a ravine at a constant speed.A passenger going to the observation deck has a velocity of 4 m/s [right]and a passenger leaving the deck has a velocity of 4 m/s [left] (Figure 1.13).Their speeds are the same, but because they are travelling in oppositedirections, their velocities are different.1-2 Decision-Making AnalysisTraffic Safety Is Everyone’s BusinessThe IssueIn an average year in Alberta, traffic accidents claim six times more lives thanhomicide, eight times more lives than AIDS, and 100 times more lives than meningitis.Collisions represent one of the greatest threats to public safety.Background InformationIn the Canadian 2002 Nerves of Steel: Aggressive Driving Study, speeding wasidentified as one of two common aggressive behaviours that contribute to asignificant percentage of all crashes. The Alberta Motor Association’s Alberta TrafficSafety Progress Report has suggested that a province-wide speed managementprogram could significantly improve levels of road safety, decreasing both speed andcasualties. One suggested program is the implementation of the vehicle tachograph,a device required in Europe to improve road safety.e WEBTo learn more about howspeeding is a key contributingfactor in casualty collisions inAlberta, follow the links atwww.pearsoned.ca/school/physicssource.12 Unit I Kinematics

AnalysisYour group has been asked to research different traffic safety initiatives. The governmentwill use the results of your research to make the most appropriate decision.1. Research(a) how state- or province-wide speed management programs have influenceddriver behaviour(b) the societal cost of vehicle crashes(c) driver attitudes toward enforcement of and education about traffic safety issues2. Analyze your research and decide which management program should be used.3. Once your group has completed a written report recommending a particularprogram, present the report to the rest of the class, who will act asrepresentatives of the government and the community.So far, you have learned that the slope of a position-time graph representsa rate of change in position, or velocity. If an object moves atconstant velocity (constant magnitude and direction), the object isundergoing uniform motion.A position-time graph for an object at rest is a horizontal line(Figure 1.14). An object at rest is still said to be undergoing uniformmotion because its change in position remains constant over equaltime intervals.uniform motion: constant velocity(motion or rest)at rest: not moving; stationaryConcept Check(a) Describe the position of dots on a ticker tape at rest. What isthe slope of the graph in Figure 1.14?(b) Describe the shape of a position-time graph for an objecttravelling at a constant velocity. List three possibilities.Frame of ReferenceIf you were to designate the hole, rather than the putter, as the origin (startingpoint) in the golf tournament (Figure 1.15(a)), your data table wouldstart at 5.0 m [left] at time 0.0 s, instead of at 0.0 m and 0.0 s (Table 1.2).The values to the left of the hole are positive.▼ Table 1.2Position (m [right])Position vs. Time6.05.04.03.02.01.00.00.0 1.0 2.0 3.0 4.0 5.0Time (min) Figure 1.14 A position-timegraph for a stationary objectPosition-time dataTime (s)Position (m [left])t 0 0.0 5.0t 1 1.0 4.0t 2 2.0 3.0t 3 3.0 2.05.0 m0.0 s4.0 m1.0 s3.0 m2.0 s2.0 m3.0 s1.0 m4.0 s0.0 m5.0 sorigint 4 4.0 1.0t 5 5.0 0.0 Figure 1.15(a) Designating an origin is arbitrary. In this example,the hole is the origin and all positions are measured relative to it.Chapter 1 Graphs and equations describe motion in one dimension. 13

info BITOn May 31, 2004 in Moscow,Ashrita Furman of the USA walked1.6 km while continuously hulahoopingin 14 min 25 s. He alsoholds the world record for thefastest time for pushing an orangewith his nose. On August 12, 2004,he pushed an orange 1.6 km in24 min 36 s. What was his speed,in km/h and m/s, for each case?(See Unit Conversions onpage 878.)The corresponding position-time graph is shown in Figure 1.15(b).Position (m [left])Position vs. Time6.05.04.03.02.01.00.00.0 1.0 2.0 3.0 4.0 5.0Time (s) Figure 1.15(b) If you change your reference frame, the position-time graph also changes.Compare this graph with the graph in Figure 1.12.From the graph,slope vd td f d itf t i0.0 m (5.0 m)5.0 s 0.0 s1.0 m/sThe velocity of the golf ball is 1.0 m/s. What does the negative signmean? It means the ball is travelling opposite to the direction to whichthe positions of the ball are measured. It does not mean that the golf ballis slowing down. Since positions, now measured to the left of the hole(the new origin) are designated positive, any motion directed to the rightis described as being negative. In this case, you can also see that the ballis decreasing its position from the origin with increasing time. The balltravels to the right toward the hole, decreasing its position each secondby 1.0 m — it travels at 1.0 m/s to the right or 1.0 m/s [left] as indicatedby the downward slope on the graph.Concept CheckDetermine how the velocity of the golf ball can be positive if thehole is at the origin.e WEBIn November 2004, at analtitude of 33 000 m, theX-43A recorded a speed of Mach 9.Use the Internet or your local libraryto research the term “Mach” asused to describe the speed of anobject. How did this term originate?What is the difference between Machand ultrasonic? Write a brief summaryof your findings. To learn moreabout Mach, follow the links atwww.pearsoned.ca/school/physicssource.Below is a summary of what you have learned:• The slope of a position-time graph represents velocity.• The velocity is the average velocity for the time interval.• Your choice of reference frame affects the direction (sign) of your answer.• A straight line on a position-time graph represents uniform motion.Comparing the Motion of Two or More Objectson a Position-time GraphYou can represent the motions of two objects on one graph, as long as theorigin is the same for both objects. You can then use the graph to comparetheir motions, as in the next example.14 Unit I Kinematics

Example 1.2At the end of the school day, student A and student B say goodbyeand head in opposite directions, walking at constant rates. Student Bheads west to the bus stop while student A walks east to her house.After 3.0 min, student A is 300 m [E] and student B is 450 m [W](Figure 1.16).(a) Graph the position of each student on one graph after 3.0 min.(b) Determine the velocity in m/s of each student algebraically.GivenChoose east to be positive.d A 300 m [E] 300 md B 450 m [W] 450 mt 3.0 minRequired(a) position–time graph(b) velocity (v A and v B )BUSSTOPbusstopStudent B’s position:450 m [W]schoolLakeview SchooloriginWNSE +Student A’s position:300 m [E] Figure 1.16Analysis and Solution(a) Since east is the positive direction, plot student A’s position(3.0 min, 300 m) above the time axis and student B’sposition (3.0 min, 450 m) below the time axis(Figure 1.17).Position (m [E])6004002000200400600Position vs. Time1.0Time (min)2.0 3.0 Figure 1.17(b) Convert time in minutes to time in seconds.dThen use the equation v to find the velocitytof each student.60 st 3.0 min 1 min 180 s300 mv A 180 s1.7 m/sThe sign is positive, so the direction is east.450 mv B 180 s2.5 m/sThe sign is negative, so the direction is west.Practice Problems1. A wildlife biologist measures howlong it takes four animals to cover adisplacement of 200 m [forward].(a) Graph the data from the tablebelow.(b) Determine each animal’s averagevelocity.Answers1. (a)AnimalTime taken (s)Elk 10.0Coyote 10.4Grizzly bear 18.0Moose 12.9Position (m [forward])250.0200.0150.0100.050.0Position vs. Time0.00 2 4 6 8 10 12 14 16 18 20Time (s)Paraphrase(b) Student A’s velocity is 1.7 m/s [E] and student B’svelocity is 2.5 m/s [W].(b) Elk: 20.0 m/s [forward]Coyote: 19.2 m/s [forward]Grizzly bear: 11.1 m/s [forward]Moose: 15.5 m/s [forward]Chapter 1 Graphs and equations describe motion in one dimension. 15

So far, you have learned that the slope of a position–time graph representsvelocity. By comparing the slopes of two graphs, you can determinewhich object is moving faster. From the slopes of the graphs in Figure 1.17,which student is moving faster? When you represent the motions oftwo objects on the same graph, you can also tell whether the objectsare approaching or moving apart by checking if the lines are convergingor diverging. An important event occurs at the point where the twolines intersect. Both objects have the same position, so the objectsmeet at this point.Concept CheckDescribe the shape of a graph showing the motion of two objectsapproaching each other.In the next example, two objects start at different times and have differentspeeds. You will graphically find their meeting point.Example 1.3Two rollerbladers, A and B, are having a race. B gives A a head startof 5.0 s (Figure 1.18). Each rollerblader moves with a constant velocity.Assume that the time taken to reach constant velocity is negligible. IfA travels 100.0 m [right] in 20.0 s and B travels 112.5 m [right] in 15.0 s,(a) graph the motions of both rollerbladers on the same graph.(b) find the time, position, and displacement at which B catchesup with A.BAPractice Problems1. The two rollerbladers in Example 1.3have a second race in which theyeach travel the original time anddistance. In this race, they start atthe same time, but B’s initialposition is 10.0 m left of A. Takethe starting position of A as thereference.(a) Graph the motions of therollerbladers.(b) Find the time, position, and B’sdisplacement at which Bcatches up with A.Answers1. (b) t 4.0 sd 20.0 m [right]d 30.0 m [right]distance travelledby A in 5.0 sGivenChoose right to be positive.d A 100.0 m [right] 100.0 mt A 20.0 sd B 112.5 m [right] 112.5 mt B 15.0 s, started 5.0 s later Figure 1.18Required(a) position-time graph(b) time (t), position (d ), and displacement (d ) when Bcatches up with A16 Unit I Kinematics

Analysis and Solution(a) Assume that t 0.0 s at the start of A’s motion. Thus, theposition-time graph of A’s motion starts at the origin. A’s finalposition is 100.0 m at 20.0 s.The position-time graph for B’s motion starts at 0.0 m and 5.0 s(because B started 5.0 s after A). B starts moving after 5.0 s for15.0 s. Thus, at 20.0 s (5.0 s 15.0 s), B’s position is 112.5 m.Each rollerblader travels with a constant velocity, so the linesconnecting their initial and final positions are straight (Figure 1.19(a)).Position (m [right])Position vs. Time120.0B100.080.0A60.040.020.00.00.0 5.0 10.0 15.0 20.0 25.0Time (s) Figure 1.19(a)(b) On the graph in Figure 1.19(a), look for a point of intersection.At this point, both rollerbladers have the same final position.From the graph, you can see that this point occurs at t 15.0 s.The corresponding position is 75.0 m (Figure 1.19(b)).Position (m [right])Position vs. Time120.0B100.080.0A60.040.020.00.00.0 5.0 10.0 15.0 20.0 25.0 Figure 1.19(b)Time (s)To find B’s displacement, find the change in position: d d f d i.Both A and B started from the same position, d i 0. Since theyboth have the same final position at the point of intersection,d f 75.0 m.d 75.0 m 0.0 m 75.0 mThe answer is positive, so the direction is to the right.Paraphrase(b) B catches up with A 15.0 s after A started. B’s position anddisplacement are 75.0 m [right] of the origin.Chapter 1 Graphs and equations describe motion in one dimension. 17

Example 1.4From the graph in Example 1.3, find the velocities of the tworollerbladers.GivenChoose right to be positive. At the point of intersection (Figure 1.19(b)),d A 75.0 m [right] 75.0 mt A 15.0 sd B 75.0 m [right] 75.0 mt B 15.0 s 5.0 s 10.0 sRequiredvelocities of A and B (v A , v B )Analysis and SolutionTo find the velocity of each rollerblader, remember that the slope of aposition–time graph is velocity. Because the motions are uniform, theslopes will be constant for each rollerblader.d v t75.0 m 0.0 mv A 15.0 s 0.0 s5.0 m/s75.0 m 0.0 mv B 15.0 s 5.0 s75.0 m 0.0 m 10.0 s7.5 m/sThe answers are both positive, so the direction is to the right. You cansee that, in order for B to cover the same distance as A, B must movefaster because B started later.Practice Problems1. Suppose rollerblader B gives A ahead start of 5.0 s and takes 10.0 sto catch up with A at 100.0 m[right]. Determine the velocities ofrollerbladers A and B.ParaphraseA’s velocity is 5.0 m/s [right] and B’s velocity is7.5 m/s [right].Answers1. A: 6.67 m/s [right]B: 10.0 m/s [right]18 Unit I Kinematics

1.2 Check and ReflectKnowledge1. For an object at rest, what quantities ofmotion remain the same over equal timeintervals?2. For an object travelling at a constantvelocity, what quantity of motion remainsthe same over equal time intervals?3. Match each ticker tape below with thecorrect position-time graph.(i)(ii)(iii)(iv)Position (m [forward])8.07.06.05.04.03.02.01.0Position vs. Time0.00.0 1.0 2.0 3.0 4.0 5.0Time (s)4. Two friends start walking on a football fieldin the same direction. Person A walks twiceas fast as person B. However, person B has ahead start of 20.0 m. If person A walks at3.0 m/s, find the distance between the twofriends after walking for 20.0 s anddetermine who is ahead at this time. Sketcha position-time graph for both people.5. A camper kayaks 16 km [E] from a campingsite, stops, and then paddles 23 km [W].What is the camper’s final position withrespect to the campsite?6. Sketch a position-time graph for a bearstarting 1.2 m from a reference point, walkingslowly away at constant velocity for 3.0 s,stopping for 5.0 s, backing up at half thespeed for 2.0 s, and finally stopping.7. Sketch a position-time graph for a student(a) walking east to school with a constantvelocity(b) stopping at the school, which is 5 kmeast of home(c) cycling home with a constant velocityABCDApplications8. Two children on racing bikes start fromthe same reference point. Child A travels5.0 m/s [right] and child B travels 4.5 m/s[right]. How much farther from the point oforigin is child A than child B after 5.0 s?9. Insect A moves 5.0 m/min and insect Bmoves 9.0 cm/s. Determine which insect isahead and by how much after 3.0 min.Assume both insects are moving in thesame direction.10. Describe the motion in each lettered stagefor the object depicted by the positiontimegraph below.Position (m [W])Position vs. Time2016B128 AC400 4 8 12 16 20Time (s)11. A mosquito flies toward you with a velocityof 2.4 km/h [E]. If a distance of 35.0 mseparates you and the mosquito initially,at what point (distance and time) will themosquito hit your sunglasses if you aretravelling toward the mosquito with aspeed of 2.0 m/s and the mosquito istravelling in a straight path?12. Spotting a friend 5.0 m directly in front ofyou, walking 2.0 m/s [N], you start walking2.25 m/s [N] to catch up. How long will ittake for you to intercept your friend andwhat will be your displacement?13. Two vehicles, separated by a distance of450 m, travel in opposite directions towarda traffic light. When will the vehicles passone another if vehicle A is travelling 35 km/hand is 300 m [E] of the traffic light whilevehicle B is travelling 40 km/h? When willeach vehicle pass the traffic light, assumingthe light remains green the entire time?e TESTTo check your understanding of uniformmotion, follow the eTest links atwww.pearsoned.ca/school/physicssource.20 Unit I Kinematics

1.3 Velocity-time Graphs:Uniform and NonuniformMotionRecently installed video screens in aircraft provide passengerswith information about the aircraft’s velocity during the flight(Figure 1.21).x km0.0 hx km1.0 hx km2.0 hx km3.0 hx km4.0 hx km5.0 h Figure 1.22 A plane flies at a constant speed, so the distances within eachtime interval are equal. Break the plane’s motion into a series of snapshots.Record your data in a data table and then graph it.Figure 1.22 shows the data of the plane’s path. Like position-timegraphs, velocity-time graphs provide useful information aboutthe motion of an object. The shape of the velocity-time graphreveals whether the object is at rest, moving at constant speed,speeding up, or slowing down. Suppose an airplane has a cruisingaltitude of 10 600 m and travels at a constant velocity of 900 km/h [E]for 5.0 h. Table 1.3 shows the velocity-time data for the airplane.If you graph the data, you can determine the relationship betweenthe two variables, velocity and time (Figure 1.23).▼ Table 1.3Time (h) Velocity (km/h) [E]0.0 9001.0 9002.0 9003.0 9004.0 9005.0 900 Figure 1.21 Video screens are anexample of an application of velocitytimegraphs.Velocity (km/h [E])Velocity vs. Time for an Airplane90080070060050040030020010000.0 1.0 2.0 3.0 4.0 5.0Time (h) Figure 1.23A velocity-time graph for an airflightChapter 1 Graphs and equations describe motion in one dimension. 21

Designating east as the positive direction, the slope of the velocity-timegraph is:riseslope runv tv f v itf t i900 k m h 900 k m h 5.0 h 1.0 h 0 km/h 2e TECHDetermine the velocity of anobject based on the shapeof its position-time graph.Go to www.pearsoned.ca/school/physicssource.From the graph in Figure 1.23, there is no change in the plane’s velocity,so the slope of the velocity-time graph is zero.Notice the units of the slope of the velocity-time graph: km/h 2 .These units are units of acceleration. Because the plane is moving at aconstant velocity, its acceleration is zero.In general, you can recognize acceleration values by their units, whichare always distance divided by time squared. In physics, the standardunits for acceleration are metres per second per second, which is generallyabbreviated to m/s 2 (read metres per second squared).Concept Check(a) What does the slope of a position-time graph represent?(b) What does the slope of a velocity-time graph represent?Non-uniform MotionAlthough objects may experience constant velocity over shorttime intervals, even a car operating on cruise control has fluctuationsin speed or direction (Figure 1.24). How can youdescribe and illustrate a change of velocity using the conceptsof kinematics?Recall from section 1.2 that an object moving at a constantvelocity is undergoing uniform motion. But is uniform motionthe only type of motion? Perform the next QuickLab to find out. Figure 1.24 Consider the kinds of changes in velocity this carexperiences during the trip.22 Unit I Kinematics

1-4 QuickLabMatch a GraphProblemWhat type of motion does each graph describe?MaterialsLM 1-1 (provided by your teacher)rulermotion sensormasking tapeProcedure1 Study the different position-time graphs on LM 1-1.With a partner, decide what type of motion eachgraph illustrates.2 Set up the motion sensor to plot position vs. time.3 Label a starting position with masking tapeapproximately 1 m in front of the motion sensor.Move away from the motion sensor in such a waythat the graph of the motion captured approximatesthe one on the LM.4 Switch roles with your partner and repeat steps 1–3.5 Print out the graphs from your experiment. For eachgraph, construct a table of values for position and time.Questions1. Describe your motion when a horizontal line wasbeing produced on the position–time graph.2. What relationship exists between the type of motionand change in position?3. Suggest two different ways in which you couldclassify the motion described by the four graphs.4. What would the graph look like if you moved awayfrom and then back toward the motion sensor?5. What happens to the graph when you move awayfrom your initial position and then move backtoward and then beyond your initial position?e LABFor a probeware activity, go towww.pearsoned.ca/school/physicssource.Concept CheckWhich ticker tape in Figure 1.25 representsaccelerated motion? Explain. Figure 1.25Consider an object, such as a drag racer (Figure 1.26), starting from restand reaching a constant velocity over a time interval (Figure 1.27).During this time interval, the vehicle has to change its velocity from avalue of zero to a final non-zero value. An object whose velocity changes(increases or decreases) over a time interval is undergoing acceleration,represented by the variable a. Acceleration is a vector quantity. It isalso called non-uniform motion because the object’s speed or directionis changing. Figure 1.26 A drag raceraccelerates from rest.acceleration: a vector quantityrepresenting the change in velocity(magnitude or direction) per unittimescale1.0 m0.0 m0.0 s2.0 m1.0 s8.0 m2.0 s18.0 m3.0 s Figure 1.27 This sequenceillustrates a car undergoingnon-uniform motion.Chapter 1 Graphs and equations describe motion in one dimension. 23

PHYSICS INSIGHTAn object is acceleratingif it is speeding up,slowing down, orchanging direction.The following scenario illustrates acceleration.A drag race is a 402-m (quarter-mile) contest between two vehicles.Starting from rest, the vehicles leave the starting line at the same time,and the first vehicle to cross the finish line is the winner. A fan recordsthe position of her favourite vehicle during the drag race. Her resultsare recorded in Table 1.4.The position-time graph for this data is shown in Figure 1.28. Fromthe graph, note that the object is speeding up because the displacementbetween data points increases for each successive time interval. Whichticker tape in Figure 1.25 matches the graph in Figure 1.28?▼ Table 1.4Position (mTime (s) [forward])Position vs. Time for a Dragster50.00.0 0.01.0 2.02.0 8.03.0 18.04.0 32.05.0 50.0Position (m [forward])40.030.020.010.00.00.01.0 2.0 3.0 4.0 5.0Time (s) Figure 1.28What does the slopeof the graph indicateabout the speed ofthe car?Instantaneous VelocityPHYSICS INSIGHTWhen you calculate theslope of a line or curveat a single point, you arefinding an instantaneousvalue. When you calculatethe slope between twopoints, you are findingan average value.tangent: a straight line thattouches a curved-line graphat only one pointInstantaneous velocity is the moment-to-moment measure of an object’svelocity. Imagine recording the speed of your car once every secondwhile driving north. These data form a series of instantaneous velocitiesthat describe your trip in detail.Earlier in this section, you learned that determining the velocity ofan object from a position-time graph requires calculating the slope ofthe position-time graph. But how can you obtain the slope of a curve?Remember that each point on the curve indicates the position of theobject (in this case, the dragster) at an instant in time. To determine thevelocity of an object at any instant, physicists use tangents. A tangentis a straight line that touches a curve at only one point (Figure 1.29(a)).Each tangent on a curve has a unique slope, which represents thevelocity at that instant. In order for the object to be at that position, atthat time, it must have an instantaneous velocity equal to the slope ofthe tangent at that point. Determining the slopes of the tangents at differentpoints on a position-time curve gives the instantaneous velocitiesat different times. Consider forward to be the positive direction.24 Unit I Kinematics

Position vs. Time for a DragsterPosition vs. Time for a DragsterPosition vs. Time for a Dragster50.050.050.0Position (m [forward])40.030.020.010.00.00.0t1.0 2.0 3.0 4.0 5.0Time (s)dPosition (m [forward])40.030.020.010.00.00.0t1.0 2.0 3.0 4.0 5.0Time (s)dPosition (m [forward])40.030.020.010.00.00.0t1.0 2.0 3.0 4.0 5.0Time (s)d Figure 1.29(a) Figure 1.29(b) Figure 1.29(c)The slope of the tangent at2.0 s isdslope t14.0 m 0.0 m3.0 s 1.0 s14.0 m 2.0 s7.0 m/sThe slope of the tangent at3.0 s is30.0 m 0.0 mslope 4.0 s 1.75 s30.0 m 2.25 s13 m/sThe slope of the tangent at4.0 s is47.0 m (15.0 m)slope 5.0 s 3.0 s32.0 m 2.0 s16 m/sThe sign is positive, so at 2.0 s,the velocity of the dragster is7.0 m/s [forward].At 3.0 s, the velocity of thedragster is 13 m/s [forward].At 4.0 s, the velocity of thedragster is 16 m/s [forward].Using Slopes of Position-time Graphs to DrawVelocity-time GraphsYou can now create a new table using the slopes of the position-timegraphs in Figures 1.29(a), (b), and (c). See Table 1.5. Remember thatthe slope of a position-time graph is velocity. These slope values areactually instantaneous velocities at the given times. You can use thesethree velocities to draw a velocity-time graph (Figure 1.30). Theresulting velocity-time graph is a straight line that goes through theorigin when extended. This means that the dragster has started from rest(0 velocity). The graph has a positive slope. To find the meaning ofslope, check the units of the slope of a velocity-time graph. They are(m/s)/s, which simplify to m/s 2 . These units are the units of acceleration.Since the velocity-time graph in this example is a straight linewith non-zero slope, the acceleration of the object is constant, so theobject must be undergoing uniformly accelerated motion.info BITWhen jet fighters come in to landon an aircraft carrier, they stop soquickly that pilots sometimes loseconsciousness for a few seconds.The same thing can happen whena pilot ejects from an aircraft, dueto enormous acceleration.uniformly accelerated motion:constant change in velocity perunit timeChapter 1 Graphs and equations describe motion in one dimension. 25

▼ Table 1.5Velocity (m/sTime (s) [forward])2.0 7.03.0 134.0 16Velocity (m/s [forward])20151050Velocity vs. Time1.0 2.0 3.0 4.0 5.0Time (s) Figure 1.30 This velocity-time graph represents anobject undergoing uniformly accelerated motion.Acceleration (m/s 2 [forward])Acceleration vs. Time4.54.03.53.02.52.01.51.00.50.00.0 1.0 2.0 3.0 4.0Time (s) Figure 1.31 An accelerationtimegraph for an object undergoinguniformly accelerated motion is astraight line with zero slope.PHYSICS INSIGHTIf the acceleration-timegraph has a non-zeroslope, the accelerationis changing (is nonuniform).The slopeof an acceleration-timegraph is called jerk,with units m/s 3 .Just as the slope of a position-time graph reveals the rate at whichposition changes (velocity), the slope of a velocity-time graph revealsthe rate at which velocity changes (acceleration). Calculate the slopeof the line in Figure 1.30 as follows, designating forward as positive:riseslope runa v t v f vitf ti10 m/s (4 m/s) 2.5 s 1.0 s4 m/s 2The answer is positive, so the car is accelerating at 4 m/s 2 [forward].The resulting acceleration-time graph is shown in Figure 1.31. Youknow that the velocity-time graph for an object undergoing uniformmotion is a horizontal line (with zero slope, as in Figure 1.23).Similarly, a horizontal line on an acceleration-time graph indicatesuniform acceleration.Concept CheckIf the position-time graph for an object undergoing positive accelerationis a parabola, such as the one in Figure 1.28, what is theshape of the position-time graph for an object undergoing negativeacceleration? What would a ticker tape of the motion of an objectthat is slowing down look like?After driving your all-terrain vehicle (ATV, Figure 1.32) through afield, you see a wide river just ahead, so you quickly bring the vehicleto a complete stop. Notice in Figure 1.33 that, as your ATV slowsdown, the displacement in each time interval decreases. Figure 1.32ATVs can undergo a wide variety of motions.26 Unit I Kinematics

scale1.0 m0.0 m0.0 s13.5 m1.0 s24.0 m2.0 s31.5 m3.0 s36.0 m4.0 s37.5 m5.0 s Figure 1.33This ATV is undergoing non-uniform motion. It is accelerating, in this case, slowing down.Example 1.5 shows the calculations and resulting velocity-time graphfor an object that is slowing down uniformly.Example 1.5▼ Table 1.6Position (mThe position-time data for an ATVTime (s) [forward])approaching a river are given inTable 1.6. Using these data,0.0 0.0(a) draw a position-time graph1.0 13.5(b) draw a velocity-time graph(c) calculate acceleration2.0 24.03.0 31.5Analysis and Solution4.0 36.0Designate the forward directionas positive.(a) For the position-time graph,5.0 37.5plot the data in Table 1.6 (Figure 1.34).Position (m [forward])40.035.030.025.020.015.010.05.0Position vs. Time0.00.0 1.0 2.0 3.0 4.0 5.0 Figure 1.34Time (s)(b) Since the position-time graph is non-linear, find theslope of the tangent at 2.0 s, 3.0 s, and 5.0 s(Figures 1.35(a), (b), and (c)).Position vs. Timed Position(m [forward])40.030.020.010.0td0.00.0 1.0 2.0 3.0 4.0 5.0Time (s) Figure 1.35(a)17.0 m 2.0 s 8.5 m/sPractice Problems1. Draw a position-time graph fromthe velocity-time graph given below.Velocity (m/s [right])1412108642002. Calculate the acceleration usingthe graph below.Velocity (m/s [N])Answers1.2502. 1.0 m/s 2 [N]Velocity vs. Time20 30 40Time (s)Chapter 1 Graphs and equations describe motion in one dimension. 271210864200210Velocity vs. Time4 6 8 10 12Time (s)slope 50t032.5 m15.5 m0 10 20 30 40 3.0 s1.0 sTime (s)Position (m [right])200150100Position vs. Time

e MATHFor an alternative methodto create a velocity-timegraph from the positiontimedata points, visitwww.pearsoned.ca/school/physicssource.Position(m [forward])Position vs. Time40.030.0dt20.010.00.00.0 1.0 2.0 3.0 4.0 5.0Time (s) Figure 1.35(b)Position(m [forward])40.0Position vs. Time30.020.010.00.00.0 1.0 2.0 3.0 4.0 5.0Time (s) Figure 1.35(c)slope d t37.0 m(26.0 m)4.0 s2.0 s 1 1.0m2.0s 5.5 m/sThis tangent is a horizontalline, so its slope is zero.PHYSICS INSIGHTvavaspeedingupspeedingupThe slopes of the tangents give the instantaneous velocities (Table 1.7).Positive signs mean that the direction is forward. Plot the data on avelocity-time graph (Figure 1.36).15.014.013.0▼ Table 1.712.011.0 Velocity vs. Time10.0Velocity (m/s9.0Time (s) [forward])8.07.06.02.0 8.55.04.03.0 5.53.02.01.05.0 00.00.0 1.0 2.0 3.0 4.0 5.0 6.0Time (s) Figure 1.36(c) Find acceleration by calculating the slope of thevelocity-time graph.a v t0.0 m/s (8.5 m/s)5.0 s 2.0 s2.8 m/s 2The acceleration of the ATV is 2.8 m/s 2 . Because the forward directionwas designated as positive, the negative sign means that the directionof acceleration is backward.Velocity (m/s [forward])vavaslowingdownslowingdownNegative Acceleration Does Not Necessarily Mean Slowing DownIn Example 1.5, the value for acceleration is negative. What is the meaningof negative acceleration? When interpreting the sign of acceleration,you need to compare it to the sign of velocity. For example, for the dragracer that is speeding up, the direction of its velocity is the same as thedirection of its acceleration (see the calculation of the slope of the velocitytimegraph for Figure 1.30). When the directions (signs) of velocity andacceleration are the same (positive or negative), the object is speeding up.28 Unit I Kinematics