Physics 11 Distance, Position, and Displacement PDF

1.1 Distance, Position, and Displacement You see and interact with moving objects every day. Whether you are racing down a ski hill or running for a s...

1.1 Distance, Position, and Displacement You see and interact with moving objects every day. Whether you are racing down a ski hill or running for a school bus, motion is part of your everyday life. Long jump athletes are very aware of distance, position, and displacement. Long jumpers run down a stretch of track to a foul line and then jump as far as possible into a sand pit (Figure 1). Their goal is to increase the distance of their jumps. To do this, they focus on their speed, strength, and technique. Successful long jumpers master this goal by applying the physics of motion. Learning Tip SI Metric Units The SI system (le Système international d’unités) provides base units for measuring fundamental physical quantities. Examples of base units are metres (m) for distance and displacement and seconds (s) for time. Some physical quantities use derived units that are defined in terms of a combination of base units. An example of a derived unit is metres per second (m/s), which is the SI unit for speed and velocity. Figure 1 Long jumpers attempt to maximize the horizontal distance of their jumps. Describing the Motion of Objects To understand the motion of objects, we must first be able to describe motion. Physicists use a number of specific terms and units to describe motion. You are likely kinematics the study of motion familiar with many of these terms and units. motion a change in an object’s location Kinematics is the term used by physicists and engineers to describe the study of as measured by a particular observer how objects move. What exactly is motion? Motion is a change in the location of an object, as measured by an observer. Distance, in physics terms, means the total length distance (d ) the total length of the path travelled by an object in motion of the path travelled by an object in motion. The SI metric base unit for distance is the metre (m). To help you understand the terms that describe motion, imagine that direction the line an object moves along you are at your home in Figure 2. You are at the location marked “0 m.” If you walk from a particular starting point directly from home to your school in a straight line, you will travel a distance of 500 m. If you walk from your school to the library and then return home, you will travel an additional distance of 700 m 1 1200 m 5 1900 m. Career Link If your friend wants to know how to get to the library from your home, telling Many branches of engineering him to walk for 1200 m is not very helpful. You also need to tell your friend which use principles of kinematics. To direction to go. Direction is the line an object moves along from a particular starting learn more about becoming an point, expressed in degrees on a compass or in terms of the compass points (north, engineer, west, east, and south). Directions can also be expressed as up, down, left, right, for- ward, and backwards. Directions are often expressed in brackets after the distance (or g o t o nelso n sc ience other value). For example, 500 m [E] indicates that the object is 500 m to the east. 8   Chapter 1 Motion in a Straight Line NEL Direction is important when describing motion. If the school in Figure 2 is your starting point, the library is in a different direction from your school than your home is. If the library is your starting point, then your school and home are in the same direction. W E library school home 0m 500 m Figure 2 Distance and direction along a 1200 m straight line Scalar and vector Quantities A scalar quantity is a quantity that has magnitude (size) only. Distance is an example scalar a quantity that has only of a scalar quantity. Since direction is so important in describing motion, physicists magnitude (size) frequently use terms that include direction in their definitions. A vector is a quantity vector a quantity that has magnitude that has magnitude (size) and also direction. An arrow is placed above the symbol for (size) and direction a variable when it represents a vector quantity. Position and displacement WEB LInK Position is the distance and direction of an object from > a particular reference point. To review scalar and vector Position is a vector quantity represented by the symbol d. Notice the vector arrow above quantities, the symbol d. This arrow indicates that position is a vector: it has a direction as well as a magnitude. For example, if home is your reference point, the position of the school go T o N ELSoN S C i EN C E in Figure 2 is 500 m [E]. Note that the magnitude of the position is the same as the straight-line distance (500 m) from home to school, but the position also includes the > position (d ) the distance and direction of direction (due east [E]). The position of the school from point 0 m can be described an object from a reference point by the equation > d school 5 500 m 3 E 4 Now assume that the library is your reference point, or the point 0 m. The position of the school from the reference point (library) can be described by the equation > d school 5 700 m 3 W 4 Once the position of an object has been described, you can describe what hap- > pens to the object when it moves from that position. This is displacement—the> displacement (Dd ) the change in change in an object’s position. Displacement is represented by the symbol Dd. position of an object Notice the vector arrow indicating that displacement is a vector quantity. The tri-> angle symbol ∆ is the Greek letter delta. Delta is always read as “change in,” so Dd is read as “change in position.” As with any change, displacement can be calculated by subtracting the initial position vector from the final position vector: > > > Dd 5 d final 2 d initial When an object changes its position more > than once (experiences two or more displacements), the total displacement Dd T of the object can be calculated by adding the displacements using the following equation: > > > Dd T 5 Dd 1 1 Dd 2 NEL 1.1 Distance, Position, and Displacement 9  Tutorial 1 Calculating Displacement for Motion in a Straight Line When you walk from one place to another, your position changes. This change in your position is displacement. The displacement can be calculated > using > your >position at the beginning and the end of your journey with the equation Dd 5 d final 2 d initial. Remember that position is a vector quantity, so you have to take direction into account. In the following Sample Problems, we will calculate displacements using a range of techniques. Refer to Figure 3 for the first three Sample Problems. In Sample Problem 1, we will calculate the displacement of an object with an initial position of 0 m. W E mall library school home 0m 1000 m 500 m 1200 m Figure 3 Sample Problem 1: Calculating Displacement from a Zero Starting Point by Vector Subtraction > Imagine that you walk from home to school in a straight-line Required: Dd route. What is your displacement? > > > Analysis: Dd 5 d school 2 d home Solution > > > Figure 3 shows that home is the starting point for your journey. Solution: Dd 5 d school 2 d home When you are at home, your position has not changed. Therefore, 5 500 m 3 E 4 2 0 m your initial position is zero. Your school has a position of 500 m [E] > Dd 5 500 m 3 E 4 relative to your home. > > Statement: Your displacement when walking from your home to Given: d school 5 500 m 3 E 4 ; d home 5 0 m school is 500 m [E]. Sample Problem 2: Calculating Displacement by Vector Subtraction What is your displacement if you walk from your school to the Statement: Your displacement when walking from school to the library? Note that all positions are measured relative to your home. library is 700 m [E]. > > Defining the initial starting position of your motion as 0 m will Given: d school 5 500 m 3 E 4 ; d library 5 1200 m 3 E 4 > often make displacement problems simpler. In Sample Problem 2, Required: Dd if we had defined 0 m as being the location of the school, it would > > > Analysis: Dd 5 d library 2 d school have been obvious from the diagram that the displacement from > > > the school to the library is 700 m [E]. Solution: Dd 5 d library 2 d school 5 1200 m 3 E 4 2 500 m 3 E 4 > Dd 5 700 m 3 E 4 Sample Problem 3: Calculating Total Displacement by Vector Addition One night after working at the library, you decide to go to the If the displacement > from the library to your home is mall. What is your total displacement when walking from the represented by Dd 1 and the displacement > from your home library to the mall? to the > mall is represented by Dd 2 , then the total displacement, > > Dd T, is given by vector addition of these two displacements: Given: Dd 1 5 1200 m 3 W 4 ; Dd 2 5 1000 m 3 W 4 (from Figure 3) > > > > Dd T 5 Dd 1 1 Dd 2 Required: Dd T > > > Analysis: In this problem, we are not simply calculating a Solution: Dd T 5 Dd 1 1 Dd 2 change in position, we are finding the sum of two different 5 1200 m 3 W 4 1 1000 m 3 W 4 > displacements. The displacements are given in Figure 3. Dd T 5 2200 m 3 W 4 To calculate the total displacement, we will need to use vector Statement: When walking from the library to the mall, you addition. This is simple if both vectors have the same direction. experience a displacement of 2200 m [W]. 10 Chapter 1 Motion in a Straight Line NEL  Sample Problem 4: Calculating Total Displacement by Adding Displacements in Opposite Directions A dog is practising for her agility competition. She leaves her At this point, it appears that we have a problem. We need to add trainer and runs 80 m due west to pick up a ball. She then a vector with a direction [W] to a vector with a direction [E]. We carries the ball 27 m due east and drops it into a bucket. can transform this problem so that both vectors point in the same What is the dog’s total displacement? direction. To do so, consider the direction [E] to be the same as Solution “negative” [W]. The vector 27 m [E] is the same as −27 m [W]. We can therefore rewrite the equation as follows. In this problem, the given values are displacements. To calculate the total displacement, we will add these two displacement vectors. 5 80 m 3 W 4 2 27 m 3 W 4 > > > Given: Dd 1 5 80 m 3 W 4 ; Dd 2 5 27 m 3 E 4 Dd T 5 53 m 3 W 4 > Statement: The dog’s total displacement is 53 m [W]. Required: Dd T > > > Analysis: Dd T 5 Dd 1 1 Dd 2 > > > Solution: Dd T 5 Dd 1 1 Dd 2 5 80 m 3 W 4 1 27 m 3 E 4 Practice 1. A golfer hits a ball from a golf tee at a position of 16.4 m [W] relative to the clubhouse. The ball comes to rest at a position of 64.9 m [W] relative to the clubhouse. Determine the displacement of the golf ball. T/I [ans: 48.5 m [W]] 2. A rabbit runs 3.8 m [N] and stops to nibble on some grass. The rabbit then hops 6.3 m [N] to scratch against a small tree. What is the rabbit’s total displacement? T/I [ans: 10.1 m [N]] 3. A skateboarder slides 4.2 m up a ramp, stops, and then slides 2.7 m down the ramp before jumping off. What is his total displacement up the ramp? T/I [ans: 1.5 m [up]] Vector Scale Diagrams In Tutorial 1, you used algebra to determine the displacement of an object in a straight line. However, there is another method you can use to solve displacement problems: vector scale diagrams. Vector scale diagrams show the vectors associated with a dis- vector scale diagram a vector diagram placement drawn to a particular scale. A vector can be represented by a directed line drawn using a specific scale segment, which is a straight line between two points with a specific direction. Line seg- directed line segment a straight line ments have magnitude only (Figure 4(a)). A directed line segment is a line segment > between two points with a specific with an arrowhead pointing in a particular direction (Figure 4(b)). For example, > AB is direction a line segment in the direction from point A to point B. Line segment BA is the same line segment but in the direction from point B to point A (Figure 4(c)). A directed line segment that represents a vector always has two ends. The end with the arrowhead is referred to as the tip. The other end is the tail. A vector scale diagram is a representa- Career Link tion of motion using directed line segments drawn to scale with arrowheads to show their specific directions. Vector scale diagrams are very useful in measuring the total Geomatics technicians collect data displacement of an object from its original position. using GPS, surveying, and remote sensing. They also record data using B B B geographic information systems tip tail (GIS). GIS analysts capture and manage spatial information for use in government and industry. To learn tail tip more about becoming a geomatics A A A technician or GIS analyst, (a) (b) (c) > > go to nels on sc i ence Figure 4 (a) A line segment (b) Directed line segment AB (c) Directed line segment BA NEL 1.1 Distance, Position, and Displacement    11  > > Consider two displacements, Dd 1 = 700 m [W] and Dd 2 = 500 m [W]. We can determine the total displacement that results from adding these vectors together by drawing a vector scale diagram. In general, when drawing a vector scale diagram, you should choose scales that produce diagrams approximately one-half to one full page in size. The larger the diagram, the more precise your results will be. Figure 5 shows a vector diagram drawn to a scale where 1 cm in the diagram rep- resents 100 m in the real world. Note that each vector in Figure 5 has a tip (the end with an arrowhead) and a tail (the other end). Vectors can be added by joining them tip to tail. This is similar to using a number > line in mathematics. Thus, after applying our chosen scale, Figure > 5 shows Dd 1 drawn as > a vector 7.0 cm in length pointing due> west. The tip of Dd 1.is joined to the tail of Dd 2. In other words, the displacement Dd 2 is drawn as a directed line >segment that is 5.0 cm long pointing > due west, starting where the displacement Dd 1 ends. The total displacement, Dd T, is the displacement from the> tail, or start, of the first vector to the tip, or end, of the second vector. In this case, Dd T points due west and has a length of 12 cm. Converting this measurement by applying our scale gives a total displacement of 1200 m [W]. For straight-line motion, vector scale diagrams are not very complex. We will look at more advanced vector scale diagrams in Chapter 2 when we consider motion in two dimensions. N scale 1 cm : 100 m ∆dT  1200 m [W] Figure 5 Vector scale diagram ∆d2  500 m [W] tip ∆d1  700 m [W] tail  Determining Total Displacement for Two Motions in Opposite Directions Tutorial 2 Using Vector Scale Diagrams In the following Sample Problem, we will determine displacement by using vector scale diagrams. Consider an example in which motion occurs in two opposite directions. Sample Problem 1: Using a Vector Scale Diagram to Determine the Total Displacement for Two Motions in Opposite Directions Imagine that you are going to visit your friend. Before you get N there, you decide to stop at the variety store. If you walk 200 m [N] scale 1 cm : 100 m from your home to the store, and then travel 600 m [S] to your store friend’s house, what is your total displacement? Solution ∆d1  200 m [N] Let> your initial displacement from your home to the store be Dd 1 and > your displacement from the store to your friend’s house home be Dd 2. > Ontario Physics > 11 U Given: Dd 1 = 0176504338 200 m [N]; Dd 2 = 600 m [S] ∆d2  600 m [S] > Required: Dd TFN C01-F04-OP11USB > > > ∆dT Analysis: Dd TCO 5 Dd 1 1 Dd 2 NGI > Solution: Figure 6 shows > the given vectors, with> the tip of Dd 1 joined to the tailPass of> Dd 2. The resultant 6th pass > vector Dd T is drawn > in red, from the tail of Approvedto the tip of. The direction of T is [S]. > Dd 1 Dd 2 Dd friend’s Dd T measuresNot 4 cmApproved in length in Figure 6,> so using the scale of house 1 cm : 100 m, the actual magnitude of Dd T is 400 m. Figure 6 Solution scale diagram for adding vectors with a change Statement: Relative to your starting point at your home, your in direction total displacement is 400 m [S]. 12   Chapter 1 Motion in a Straight Line Art is drawn to scale. NEL  Practice 1. A car drives 73 m [W] to a stop sign. It then continues on for a displacement of 46 m [W]. Use a vector scale diagram to determine the car’s total displacement. T/I C [ans: 120 m [W]] 2. A robin flies 32 m [S] to catch a worm and then flies 59 m [N] back to its nest. Use a vector scale diagram to determine the robin’s total displacement. T/I C [ans: 27 m [N]] 1.1 Summary Motion involves a change in the position of an object. Motion can be described using mathematical relationships. A scalar is a quantity that has magnitude (size) only. A vector is a quantity that has magnitude (size) and direction. You can determine the displacement of an object by subtracting the start position from the end position. You can determine total displacement by adding two or more displacements together algebraically or by using a vector scale diagram. Vectors can be added by joining them tip to tail. 1.1 Questions 1. Which of the following quantities are vectors, and which 5. Determine the total displacement for each of the following are scalars? Be sure to explain the reasoning for your motions by algebraic methods and by using scale answer. K/U C diagrams. T/I C > > (a) A bird flies a distance of 20 m. (a) Dd 1 5 10 m 3 W 4 ; Dd 2 5 3.0 m 3 W 4 (b) A train is travelling at 100 km/h due north. > > (b) Dd 1 5 10 m 3 W 4 ; Dd 2 5 3.0 m 3 E 4 (c) It takes an athlete 10.37 s to run 100 m. > > (c) Dd 1 5 28 m 3 N 4 ; Dd 2 5 7.0 m 3 S 4 2. Explain the following in your own words: K/U C > > (d) Dd 1 5 7.0 km 3 W 4 ; Dd 2 5 12 km 3 E 4 ; (a) the difference between position and displacement > (b) the difference between distance and displacement Dd 3 5 5.0 km 3 W 4 3. What is the displacement of a locomotive that changes its 6. A person walks 10 paces forward followed by 3 paces position from 25 m [W] to 76 m [W]? T/I forward, and finally 8 paces backwards. T/I C (a) Draw a vector scale diagram representing this person’s 4. A car changes its position from 52 km [W] to 139 km [E]. motion. Use a scale of 1 cm 5 1 pace. What is the car’s displacement? T/I (b) Check your answer by pacing out this motion yourself. How close is your experimental result to that predicted by your vector scale diagram? NEL 1.1 Distance, Position, and Displacement    13  1.2 Speed and Velocity If you have been a passenger in a car or are taking driving lessons, speed is something you have thought about. Knowing the speed at which a vehicle is moving is important for safety. Excessive speed is a contributing factor in many collisions. Speed can be measured in different ways. Police use laser speed devices to detect the speed of moving vehicles (Figure 1). In the laboratory, scientists and engineers can use electronic devices such as motion sensors to measure speed. Average Speed Figure 1 A laser speed device can The average speed of a moving object is the total distance travelled divided by the total accurately measure the speed of an time elapsed. You are probably familiar with the speedometer of a passenger vehicle, oncoming vehicle. which tells the speed of the vehicle in kilometres per hour (km/h). However, the SI unit for speed is metres per second (m/s). You do not need a special device like a police speed device to measure speed. If you average speed (vav ) the total distance know the distance travelled and the time it took an object to travel that distance, you travelled divided by the total time taken can calculate the average speed of the object using the equation to travel that distance Dd vav 5 Dt where vav is the average speed, Δd is the distance travelled, and Δt is the change in time. Like distance, speed is a scalar quantity. In the following Tutorial, we will deter- mine the average speed of an object using this equation. Investigation 1.2.1 Tutorial 1 Calculating Average Speed Watch Your Speed (p. 46) The following Sample Problems will demonstrate how to use the equation for In this investigation you will use the average speed. average speed equation to determine average speed in a study of vehicles Sample Problem 1: Determining Average Speed passing an observation point. Your dog runs in a straight line for a distance of 43 m in 28 s. What is your dog’s average speed? Given: ∆d = 43 m; Δt = 28 s Required: vav Dd Analysis: vav 5 Dt Dd Solution: vav 5 Dt 43 m 5 28 s vav 5 1.5 m/s Statement: Your dog’s average speed is 1.5 m/s. Sample Problem 2: Determining the Distance Travelled by a Ball Moving at Constant Speed A baseball rolls along a flat parking lot in a straight line at a constant speed of 3.8 m/s. How far will the baseball roll in 15 s? Given: vav = 3.8 m/s; Δt = 15 s Required: ∆d 14   Chapter 1 Motion in a Straight Line NEL  Dd Analysis: vav 5 Dt Dd 5 vavDt Solution: Dd 5 vavDt m 5 a3.8 b 115 s2 s Dd 5 57 m LEARNING TIP Statement: The ball will roll 57 m in 15 s. Rounding in Calculations As a general rule, round final answers Practice to the same number of significant digits 1. A paper airplane flies 3.7 m in 1.8 s. What is the airplane’s average speed? T/I as the given value with the fewest [ans: 2.1 m/s] significant digits. Take extra care when rounding digits with multiple parts. You 2. A cheetah can run at a maximum speed of 30.0 km/h, or 8.33 m/s. How far can a will see in this book that extra digits cheetah run in 3.27 s? T/I [ans: 27.2 m] are carried in intermediate calculations. 3. How long does it take a rock to fall through 2.8 m of water if it falls at a constant For more help with rounding, refer to speed of 1.2 m/s? T/I [ans: 2.3 s] the Skills Handbook. Research This Searching for Speeders Skills: Researching, Analyzing, Communicating, SKILLS Identifying Alternatives, Defending a Decision HANDBOOK A5 A laser speed device is used by police officers to measure the speed of moving vehicles. This device sends a pulse of infrared laser light at the speed of light (3.0 × 108 m/s) toward a moving vehicle. The laser pulse reflects off the vehicle and returns to a sensor on the speed device. A computer chip in the speed device determines the time it took for the pulse to travel to and from the moving vehicle. The speed device uses one half of this very short time and the speed of light to calculate the distance to the moving vehicle. The speed device’s computer uses multiple distance readings to determine how the vehicle’s distance is changing with time and then calculates the vehicle’s speed. Modern speed devices send thousands of pulses of light each second, providing a high level of accuracy. 1. Conduct research to investigate how common laser speed devices are in the region where you live. 2. Investigate how speed affects the number of automobile collisions and fatalities in Canada. 3. Investigate alternative methods the police could use to determine the speed of a vehicle. A. Does the use of laser speed devices have an impact on the number of automobile collisions and fatalities in Canada? K/U B. Do you feel that the use of laser speed devices is the preferred way for police to monitor the speed of automobiles? C C. Laser speed devices and video recorders can now be used to capture the speed of a moving vehicle, the vehicle’s licence plate number, the date, and the time in the same image. If these devices are set in a fixed position, they can operate without the need for a police officer to be present. Data can be collected electronically and speeders can be sent a ticket through the mail. Do you support the use of devices like these in Ontario? Justify your decision. C A GO T O N ELSON SC IEN C E NEL 1.2 Speed and Velocity 15  Average Velocity > average velocity (v av ) the total The average velocity of an object in motion is its total displacement, or change in displacement, or change in position, position, divided by the total time taken for the motion. Velocity describes change in divided by the total time for that position over time. For instance, a cyclist travelling east at a constant speed of 11 m/s displacement has a velocity of 11 m/s [E]. Since it has direction and magnitude, average velocity is a vector quantity. The SI unit for velocity is metres per second (m/s). The symbol for > average velocity is vav. position–time graph a graph describing A position–time graph is a graph that describes the motion of an object, with position the motion of an object, with position on the vertical axis and time on the horizontal axis. Figure 2 shows a position–time on the vertical axis and time on the graph for the motion of a rolling ball measured by students during an experiment. horizontal axis Notice that the points on the graph form a straight line that moves upward from left to right. Whenever an object is moving at a constant velocity, the position–time graph of that motion is a straight line. slope (m) a measure of the steepness You may recall from your mathematics studies that the slope of a line describes its of a line steepness. The symbol for slope is m. Slope is determined, as shown in Figure 2, by rise vertical change between two points comparing the magnitude of the rise (the change between points on the y-axis) and on a line the magnitude of the run (the change between the same points on the x-axis). You can run horizontal change between two points use this technique whether the graph passes through the origin or not; for example, on a line in Figure 3 the motion begins at a position of 10 m [E] when t 5 0 s. For an object moving at a constant velocity, so that its position–time graph is a straight line, the key relationship is this: The slope of a position–time graph gives the velocity of the object. Learning Tip The steeper the graph, the greater is the object’s displacement in a given time Rates of Change Average speed and average velocity interval, and the higher is its velocity. This can be confirmed > using the information are examples of rates of change— in Figure 2. Since the y-axis shows change in position, Dd , and the x-axis shows change an important concept in science that in time, ∆t, the formula for the slope of this graph can be rewritten as follows: describes how quickly a quantity rise is changing. Velocity is the rate of slope 5 change of position, which means run that the more rapidly an object’s > > d2 2 d1 position is changing, the greater is m5 the magnitude of its velocity. t2 2 t1 > Dd or m 5 Position v. Time for a Rolling Ball Dt 60 55 Position v. Time for a Rolling Ball 50 45 45 40 40 35 35 d (m [E]) 30 30 d (m [E]) 25 25 20 20 rise  40 m [E] 15 15 10 10 5 5 run  10 s 0 0 0 2 4 6 8 10 12 0 2 4 6 8 10 12 t (s) t (s) Figure 2 Calculating the slope of a position– Figure 3 A position–time graph with non-zero time graph initial position 16   Chapter 1 Motion in a Straight Line NEL The (average) velocity of a moving object is given by the equation > > Dd v av 5 Dt To determine the slope—the average velocity—from the zero point to the final data point for the x-axis and y-axis for the motion shown in Figure 2, substitute the initial and final displacement and time values into the equation we just derived: > > Dd v av 5 Dt > > d2 2 d1 m5 t2 2 t1 40 m 3 E 4 2 0 m 5 10 s 2 0 s m 5 4 m/s 3 E 4 The velocity of the rolling ball is 4 m/s [E]. Note that the slopes of the graphs shown in Figure 2 and Figure 3 are the same, 4 m/s [E]. The two motions are different in that the motion described by Figure 2 started 0 m away from the observer, whereas the motion graphed in Figure 3 had an initial position of 10 m [E] from the observer. Calculating average velocity from the slope of a position–time graph is a very useful technique because average velocity is often difficult to measure directly. However, position can be easily measured with equipment such as tape measures, motion sensors, and laser speed devices. Velocity (a vector quantity) is to speed (a scalar quantity) as displacement (a vector quantity) is to distance (a scalar quantity). The equation for average velocity should therefore look similar to the equation for average speed, except that velocity and dis- placement are vectors: > > Dd Learning Tip v av 5 Dt Calculations with Vectors > In general, you cannot divide one where Dd is the change in position and ∆t is the change in time during the given vector by another, as dividing a time interval. This is the same equation as the one we just derived using the slope of direction has no meaning. However, if a position–time graph. the directions of both vectors are the Note that Δt can also be described by the equation Δt 5 t2 2 t1. Often, we can sim- same, you can disregard the direction plify this equation by considering t1, the start time, to be 0 s. In the following Tutorial, and divide one magnitude by the other. we will use the average velocity equation to determine unknown values. Tutorial 2 Solving Problems Using the Equation for Average Velocity The equation for average velocity can be used to solve for any of the three variables in the average velocity equation when the other two are known. In the following Sample Problems, we will review solving equations for an unknown variable using the equation for average velocity. Sample Problem 1: Calculating the Average Velocity of an Object > On a windy day, the position of a balloon changes as it is blown > Dd 82 m [N] away from a child in 15 s. What is the average velocity Analysis: v av 5 Dt of the balloon? > > Dd Solution Solution: v av 5 Dt We are given the change in time and the change in position of 82 m 3 N 4 the balloon, so we can solve for average velocity. 5 15 s > > Given: Dd 5 82 m 3 N 4 ; Dt 5 15 s         v av 5 5.5 m/s 3 N 4 > Statement: The average velocity of the balloon is 5.5 m/s [N]. Required: vav NEL 1.2 Speed and Velocity   17  Sample Problem 2: Calculating the Time for a Displacement to Occur A subway train travels at an average velocity of 22.3 km/h [W]. > km 1h 1 min 1000 m vav 5 a22.3 3W 4 b a ba ba b How long will it take for the subway train to undergo a h 60 min 60 s 1 km displacement of 241 m [W]? > > 22.3 3 W 4 3 1000 m Given: v av 5 22.3 km/h 3 W 4 ; Dd 5 241 m 3 W 4 5a b 60 3 60 s > Required: ∆t vav 5 6.1944 m/s 3 W 4 1two extra digits carried2 > > Dd Now that we have converted units, we can use the average Analysis: v av 5 Dt velocity equation to determine the change in time. > Since we are given the average velocity and the displacement of > Dd v av 5 the subway train, we can rearrange the average velocity equation Dt to solve for the change in time. > Dd Solution: Before we solve this problem, we must first make sure Dt 5 > v av that all of the given values are converted to SI metric units. This will require us to convert 22.3 km/h to metres per second. We will 241 m 3 W 4 5 do this by multiplying 22.3 km/h by a series of ratios equal to 1. 6.1944 m/s 3 W 4 We will use these ratios so that the units that we do not want Dt 5 38.9 s (kilometres and hours) will cancel, and we will be left with the Statement: It takes the subway train 38.9 s to be displaced units we do want (metres and seconds). 241 m [W] from its starting point. Practice 1. What is the average velocity of a soccer ball that is kicked and travels 2.17 m [E] in 1.36 s? T/I [ans: 1.60 m/s [E]] 2. How long will it take a cat to run 8.2 m [N] at an average velocity of 3.7 m/s [N]? T/I [ans: 2.2 s] Motion with Uniform and Non-uniform Velocity motion with uniform or constant Motion with uniform or constant velocity is motion at a constant speed in a straight velocity motion of an object at a constant speed in a straight line line. It is the simplest type of motion that an object can undergo, except for being at rest. Note that both requirements (constant speed and straight line) must be met motion with non-uniform velocity for an object’s velocity to be uniform. In contrast, motion with non-uniform velocity (accelerated motion) motion in which is motion that is not at a constant speed or not in a straight line. Motion with non-uniform the object’s speed changes or the object velocity may also be called accelerated motion. Table 1 shows some examples of motion does not travel in a straight line with uniform velocity. You will learn more about accelerated motion in Section 1.3. Table 1 Examples of Uniform and Non-uniform Velocity Example Uniform velocity Non-uniform velocity Explanation A car travels down a straight highway at a ✓ The car is travelling at a constant steady 100 km/h. speed in a straight line. A passenger on an amusement park ride ✓ The passenger is travelling at a travels in a circle at a constant speed of constant speed but not in a straight 1.2 m/s. line. She is travelling in a circle. A parachutist jumps out of an aircraft. ✓ ✓ Before he opens the parachute, the (after parachute opens) (before parachute opens) speed of the parachutist will increase due to gravity. Once the parachute is opened, his speed will become constant due to air resistance. He will then fall at a constant speed in the same direction (downwards). 18   Chapter 1 Motion in a Straight Line NEL  Determining Types of Motion from Position–Time Graphs Recall that the slope of a position–time graph gives the velocity of an object. A position– time graph that describes constant velocity must be a straight line because motion with constant velocity is motion at a constant speed. Therefore, the slope of the position–time graph must also be constant. Table 2 shows five position–time graphs that represent commonly occurring types of motion. You will see these types of motion frequently in investigations. By the end of this unit, you should be able to identify the type of motion from the characteristics of its position–time graph. Table 2 Interpreting Position–Time Graphs Position–time graph Type of motion Example Graph A graph is a horizontal straight line the slope of a horizontal straight line is zero the object has a velocity of zero d(m [E]) the object is at rest the object is at a constant positive position relative to the reference position the object is stationary at a location to the east of the t (s) reference position Graph B graph is a horizontal straight line the slope of a horizontal straight line is zero the object has a velocity of zero d (m [E]) the object is at rest 0 the object is at a constant negative position relative to t (s) the reference position the object is stationary at a location to the west of the reference position Graph C graph is a straight line with positive slope straight lines with non-zero slopes always represent constant (non-zero) velocity from the y-axis, we know the object is moving d (m [E]) eastward the object’s velocity can be determined from the slope of the graph (rise divided by run) t (s) Graph D g raph is a straight line with positive slope, which represents constant (positive) velocity from the y-axis, we know the object is moving d(m [E]) eastward the object’s velocity can be determined from the slope of the graph since graph D has the steeper slope, we can conclude t (s) that this object has a greater velocity than the object described by graph C Graph E g raph is a straight line, which represents constant velocity the slope of the graph is negative the object’s velocity can be determined from the slope d (m [E]) of the graph note that the direction for position on the y-axis is 11USB given by a vector with direction [E] the negative slope indicates that the object is moving westward t (s) NEL 1.2 Speed and Velocity   19  Mini Investigation Bodies in Motion SKILLS Skills: Performing, Observing, Analyzing HANDBOOK A6.2 Motion sensors are devices that send out ultrasonic wave pulses. position–time graphs to show the overall shape and features When some of these waves reflect off an object, the waves return to of the graphs that you predict the motion sensor will the motion sensor. A computer can then analyze the data from the generate for each of the following scenarios: returning waves and generate real-time position–time graphs. Using slow constant speed speeding up motion sensors can help you understand position–time graphs. fast constant speed slowing down Equipment and Materials: motion sensor; computer or 4. Using the motion sensor, generate real-time position–time graphs computer interface; graph paper; pencil for each scenario in Step 3 using body motions. 1. Connect the motion sensor to a computer or computer interface. A. Compare your predictions with the results. Explain any 2. Place the motion sensor on a lab bench. Make sure that you differences. T/I C have about 4 m of free space in front of the motion sensor. B. If there is time, use the motion sensor to generate graphs that 3. You will be using your body to generate position–time resemble as many letters of the alphabet as you can. T/I C data with the motion sensor. Before you begin, sketch 1.2 Summary Average speed is equal to the total distance travelled divided by the time taken for the motion. A position–time graph describes motion graphically. The slope of the position–time graph gives the velocity of an object. Average velocity is equal to the total displacement divided by the time taken for the motion. In other words, velocity describes change in position over time. Motion with uniform or constant velocity is motion at a constant speed in a straight line. Objects that are undergoing constant velocity have a position–time graph that is a straight line. 1.2 Questions 1. When you are solving a problem, how do you know if you 5. Copy and complete Table 3 in your notebook. T/I are given a speed value or a velocity value? K/U Table 3 2. Define motion with uniform velocity in your own words. K/U C > > 3. Give two real-life examples each of motion with uniform v av Dd Dt velocity and motion with non-uniform velocity. A 12.6 m [S] 16.3 s 4. Determine the velocity for the motion described by the graph in Figure 4. T/I 2.0 × 103 m/s [E] 25 m [E] 16 40 m/s [N] 0.25 s 14 12 6. What is the displacement of a horse that runs at a velocity of 3.2 m/s [S] for 12 s? T/I d7 (m [W]) 10 8 7. How many seconds would it take a car travelling at 6 100.0 km/h to travel a distance of 16 m? T/I 4 8. What is the velocity (in metres per second) of a Canadian 2 Forces CF-18 fighter jet that travels 8.864 km [S] in 0 0.297 min? T/I 0 1 2 3 4 5 t (s) Figure 4 C01-P17-0P11USB 20 Chapter 1 Motion in a Straight Line NEL  Acceleration 1.3 Some theme parks have rides in which you are slowly carried up in a seat to the top of a tower, and then suddenly released (Figure 1). On the way down, your arms and hair may fly upward as the velocity of your seat increases. The thrill of this sudden change in motion can frighten and exhilarate you all at once. An even bigger thrill ride, however, is to be a pilot in a jet being launched from the deck of an aircraft carrier. These giant ships use catapults to move 35 000 kg jets from rest (0 km/h) to 250 km/h in just 2.5 s. While it is true that objects sometimes move at constant velocity in everyday life, usually the velocities we observe are not constant. Objects that experience a change in velocity are said to be undergoing acceleration. Acceleration describes how quickly an object’s velocity changes over time or the rate of change of velocity. We can study acceleration using a velocity–time graph, which, similar to the position–time graph, has time on the horizontal axis, but velocity rather than position on the vertical axis. Figure 1 Sudden changes in velocity Velocity–time graphs are particularly useful when studying objects moving with are part of the thrill of midway rides. uniform velocity (zero acceleration) or uniform acceleration (velocity changing, but > acceleration (a av) how quickly an at a constant rate). The velocity–time graphs for both uniform velocity and uniform object’s velocity changes over time acceleration are always straight lines. By contrast, the position–time graph of an accelerated motion is curved. Table 1 shows the position–time graphs of objects (rate of change of velocity) moving with accelerated motion. velocity–time graph a graph describing the motion of an object, with velocity on the vertical axis and time on the horizontal axis Table 1 Interpreting Position–Time Graphs Position–time graph Type of motion Example Graph A graph is a curve on any graph that curves, the slope or steepness of the C01-F13-0P11USB graph changes from one point on the graph to another since the slope in graph A is constantly changing, the velocity is not constant d (m [E]) since the graph lies above the x-axis and its slope is increasing, the velocity of the object is also increasing (speeding up) in a positive or eastward direction t (s) Graph B graph is a curve t (s) on any graph that curves, the slope or steepness of the graph changes from one point on the graph to another since the slope in graph B is constantly changing, the velocity is not constant since the graph lies below the x-axis and its slope is d (m [E]) negative but getting steeper, the object is speeding up as it moves in a negative or westward direction P11USB Group NEL 1.3 Acceleration   21 rowle  Table 1 (continued) Position–time graph Type of motion Example Graph C graph is a curve on any graph that curves, the slope changes from one point on the curve to another since the slope in graph C is constantly changing, the velocity is not constant d (m [E]) since the graph lies above the x-axis and its slope is decreasing, the velocity of the object is also decreasing (slowing down) in a positive or eastward direction t (s) Graph D graph is a curve t (s) on any graph that curves, the slope or steepness of the graph changes from one point on the graph to another since the slope in graph D is constantly changing, the velocity is not constant d (m [E]) since the graph lies below the x-axis and its slope is negative and getting shallower, the object is slowing down as it moves in a negative or westward direction Determining Acceleration from a Velocity–Time Graph Figure 2 shows a velocity–time graph for a skateboard rolling down a ramp. Notice that the line of the graph goes upward to the right and has x-intercept and y-intercept of zero. We can calculate the slope of the graph in Figure 2 using the equation SB rise p slope 5 e run > Dv 1m/s2 m5 Dt 1s2 Since acceleration describes change in velocity over time, this suggests that the (average) acceleration of the skateboard is given by the equation > > Dv a av 5 Dt SB Learning Tip p Slope and Area of Velocity v. Time for a Skateboard Rolling Down a Ramp e Velocity–Time Graphs 35 The slope of a velocity–time graph 30 gives the acceleration of the object. The area under a velocity–time 25 v (m/s [S]) graph gives the displacement of the 20 object. Why is displacement related 15 to the area under a velocity–time graph? One way to think about it is 10 this: the greater the velocity during 5 a given time interval, the greater 0 the area under the graph, and the 0 2 4 6 8 10 12 greater the displacement over that t (s) time interval. Figure 2 Velocity–time graph for a skateboard rolling down a ramp 22   Chapter 1 Motion in a Straight Line NEL  In other words, The slope of a velocity–time graph gives the average acceleration of an object. Acceleration over a time interval, that is, average acceleration, is given by the equation change in velocity average acceleration 5 change in time > > Dv aav 5 Dt > > > vf 2 vi or aav 5 Learning Tip Dt Recall that the SI unit for velocity is metres per second (m/s) and the SI unit for Square Seconds? time is seconds (s). The SI unit for acceleration is a derived unit. A derived SI unit is What is a square second? Good a unit created by combining SI base units. We can derive the units for acceleration by question! When we write acceleration dividing a velocity unit (m/s) by a time unit (s), as follows: units as m/s2, we are not implying units of acceleration 5 units of velocity per second that we have measured a square second. This is simply a shorter way m of expressing the derived unit. You s can also read the unit as “metres per 5 s second, per second”—describing m 1 how many metres per second of 5 3 s s velocity are gained or lost during m each second of acceleration. 5 2 s Tutorial 1 Calculating Acceleration > > > vf 2 vi When you are given values for any three variables in the equation a av 5 for Dt acceleration, you can solve for the missing variable. Sample Problem 1 What is the acceleration of the skateboard in Figure 2? Consider Statement: The skateboard is accelerating down the ramp at the motion between 0 s and 10 s. 3 m/s2 [S]. > > What would the acceleration be if the same velocity change Given: v i 5 0 m/s; vf 5 30 m/s 3 S 4 ; t i 5 0 s; t f 5 10 s > of 30 m/s [S] took place over a time interval of only 5 s? Required: aav > Analysis: To calculate the average acceleration, which is the > Dv aav 5 slope of the graph, we use the defining equation in the form that Dt > > includes v i, v f, t i, and t f: > 30 m/s 3 S 4 > Dv 5 5s aav 5 Dt > > > aav 5 6 m/s2 3 S 4 vf 2 vi 5 The time interval is shorter, so a more rapid acceleration occurs. tf 2 ti > > > vf 2 vi Solution: aav 5 tf 2 ti 30 m/s 3 S 4 2 0 m/s 5 10 s 2 0 s > aav 5 3 m/s2 3 S 4 NEL 1.3 Acceleration   23  Sample Problem 2 > > A bullet is found lodged deeply in a brick wall. As part of the > vf 2 vi investigation, a forensic scientist is experimenting with a rifle Analysis: a av 5 Dt that was found nearby. She needs to determine the acceleration > > > vf 2 vi that a bullet from the rifle can achieve as a first step in linking Solution: a av 5 the rifle to the bullet. During a test firing, she finds that the rifle Dt bullet accelerates from rest to 120 m/s [E] in 1.3 3 1022 s m m 120 3 E 4 2 0 as it travels down the rifle’s barrel. What is the bullet’s average s s 5 acceleration? 1.3 3 1022 s > > Given: v i 5 0 m/s; v f 5 120 m/s 3 E 4 ; Dt 5 1.3 3 10 22 s > > a av 5 9.2 3 103 m/s2 3 E 4 Required: a av Statement: The acceleration of the bullet is 9.2 3 103 m/s2 [E]. Sample Problem 3 When a hockey player hits a hockey puck with his stick, the To solve this dilemma, we can use a technique from Section 1.2. velocity of the puck changes from 8.0 m/s [N] to 10.0 m/s [S] We will change a vector subtraction problem into a vector addition over a time interval of 0.050 s. What is the acceleration of problem. Recall that negative [N] is the same as positive [S]. the puck? > 10.0 m/s 3 S 4 1 8.0 m/s 3 S 4 > > a av 5 Given: v i 5 8.0 m/s 3 N 4 ; v f 5 10.0 m/s 3 S 4 ; Dt 5 0.050 s 0.050 s > Required: a av 18.0 m/s 3 S 4 > > 5 0.050 s > vf 2 vi Analysis: a av 5 > Dt aav 5 3.6 3 102 m/s2 3 S 4 > > > vf 2 vi Statement: The hockey puck’s acceleration is 3.6 3 102 m/s2 [S]. Solution: a av 5 Dt Notice that the initial velocity of the puck is north, while its final > 10.0 m/s 3 S 4 2 8.0 m/s 3 N 4 velocity is south. The acceleration is in the opposite direction to a av 5 the initial motion, so the puck slows down and comes to rest. 0.050 s It then continues accelerating south, increasing its velocity to At this point, it would appear that we have a dilemma. In the 10 m/s [S]. This is why the final velocity is due south. Sample numerator of the fraction, we must subtract a vector with a Problem 3 can also be solved by using a vector scale diagram. direction [N] from a vector with a direction [S]. Practice 1. A catapult accelerates a rock from rest to a velocity of 15.0 m/s [S] over a time interval of 12.5 s. What is the rock’s average acceleration? T/I [ans: 1.20 m/s2 [S]] 2. As a car approaches a highway on-ramp, it increases its velocity from 17 m/s [N] to 25 m/s [N] over 12 s. What is the car’s average acceleration? T/I [ans: 0.67 m/s2 [N]] 3. A squash ball with an initial velocity of 25 m/s [W] is hit by a squash racket, changing its velocity to 29 m/s [E] in 0.25 s. What is the squash ball’s average acceleration? T/I [ans: 2.2 × 102 m/s2 [E]] You can use both the defining equation for average acceleration and velocity–time graphs to determine other information about the motion of an object, besides accel- eration itself. In Tutorial 2, you will use the equation in a different form to determine a different quantity. Tutorial 3 introduces the idea of determining displacement via the area under a velocity–time graph. 24   Chapter 1 Motion in a Straight Line NEL  Tutorial 2 Solving the Acceleration Equation for Other Variables In the following Sample Problem, we will explore how to solve the defining acceleration equation for other variables. Sample Problem 1: Solving the Acceleration Equation for Final Velocity > > > A racehorse takes 2.70 s to accelerate from a trot to a gallop. If a av Dt 5 v f 2 v i the horse’s initial velocity is 3.61 m/s [W] and it experiences an > > > acceleration of 2.77 m/s2 [W], what is the racehorse’s velocity v f 5 a av Dt 1 v i when it gallops? m m > > 5 a2.77 3 W 4 b 12.70 s2 1 3.61 3 W 4 s2 s Given: Dt 5 2.70 s; v i 5 3.61 m/s 3 W 4 ; a av 5 2.77 m/s2 3 W 4 > 5 7.48 m/s 3 W 4 1 3.61 m/s 3 W 4 Required: v f > > > vf 5 11.1 m/s 3 W 4 > vf 2 vi Analysis: a av 5 Statement: When it gallops, the racehorse has a velocity of Dt 11.1 m/s [W]. Solution: Rearrange the equation for acceleration to solve for the final velocity. Practice 1. How long does it take a radio-controlled car to accelerate from 3.2 m/s [W] to 5.8 m/s [W] if it experiences an average acceleration of 1.23 m/s2 [W]? T/I [ans: 2.1 s] 2. A speedboat experiences an average acceleration of 2.4 m/s2 [W]. If the boat accelerates for 6.2 s and has a final velocity of 17 m/s [W], what was the initial velocity of the speedboat? T/I [ans: 2.1 m/s [W]] Tutorial 3 Determining Displacement from a Velocity–Time Graph By further analyzing the velocity–time graph shown in Figure 2 on page 22, we can determine even more information about the motion of the skateboard. Figure 3 has the same data plotted as Figure 2, except that the area under the line is shaded. The area under a velocity– time graph gives the displacement of the object. Note how the area under the straight line in Figure 3 forms a triangle. The area (A) of a triangle is determined by the equation 1 A 5 bh 2 > 1 Dd 5 bh 2 where b is the length of the base of the triangle and h is the height of the triangle. We can determine the displacement of the skateboard from Figure 3 by calculating the area of this triangle. Velocity v. Time for a Skateboard Rolling Down a Ramp 35.0 30.0 25.0 v (m/s [S]) 20.0 15.0 height: h 30.0 m/s [S] 10.0 5.0 base: b 10.0 s 0 0 2.0 4.0 6.0 8.0 10.0 12.0 t (s) Figure 3 Velocity–time graph showing the area underneath the line NEL 1.3 Acceleration   25  Sample Problem 1: Determining Displacement from a Velocity–Time Graph What is the displacement represented by the graph in Figure 3 > 1 m on the previous page? Solution: Dd 5 110.0 s2 a30.0 3 S 4 b 2 s Given: b = 10.0 s; h = 30.0 m/s [S] > Dd 5 150 m 3 S 4 > Required: Dd Statement: The object was displaced 150 m [S] in 10 s. > 1 Analysis: Dd 5 bh 2 Sample Problem 2: Determining Displacement from a More Complex Velocity–Time Graph What is the displacement represented by the graph in Figure 4 over the time interval from 0 s to 10.0 s? Velocity v. Time for a Complex Motion 12.0 10.0 8.0 v (m/s [S]) 6.0 h  w  10.0 m/s [S] 4.0 2.0 b  5.0 s l  5.0 s 0 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 t (s) Figure 4 Velocity–time graph showing more complex motion > The graph in Figure 4 is a more complex velocity–time graph Required: Dd than we worked with in Sample Problem 1. However, the > displacement of the object can still be determined by calculating Analysis: Dd 5 Atriangle 1 Arectangle > the area under the velocity–time graph. Solution: Dd 5 Atriangle 1 Arectangle To calculate this displacement, we will need to break the area 1 under the line into a rectangle and a triangle, and add these two 5 bh 1 lw 2 areas together. We know the area (A) of a triangle is given by 1 1 m m Atriangle 5 bh. The area of a rectangle is its length (l ) multiplied 5 15.0 s2 a10.0 3 S 4 b 1 15.0 s2 a10.0 3 S 4 b 2 2 s s by its width (w ), or Arectangle = lw. 5 25 m 3 S 4 1 50.0 m 3 S 4 hysics 11 U > Dd 5 75 m 3 S 4 38 Given: b = 5.0 s; h = 10.0 m/s [S]; l = 5.0 s; w = 10.0 m/s [S] Statement: The object has travelled 75 m [S] after 10.0 s. C01-F21-OP11USB CrowleArt Group Practice Deborah Crowle 1. Determine the displacement represented by the graph in Figure 4 over the following 3rd pass time intervals: T/I (a) from 0 s to 4.0 s [ans: 16 m [S]] oved (b) from 0 s to 7.5 s [ans: 50 m [S]] 26   Chapter 1 Motion in a Straight Line NEL  Mini Investigation Motion Simulations SKILLS Skills: Predicting, Performing, Observing, Analyzing, Communicating HANDBOOK A2.1 In this investigation, you will use a computer simulation for four different motion scenarios and then analyze the scenarios by graphing. Equipment and Materials: computer access; graphing paper 1. Go to the Nelson website and find the link for this Mini Investigation. 2. Go to the simulation, and take a few minutes to familiarize yourself with how it operates. You will be asked to run this simulation for each of the following scenarios: a positive velocity value a negative velocity value a negative initial position and a positive velocity value a negative initial position and a negative velocity value A. Before you run these scenarios, write a brief statement and draw a sketch to predict how each graph will appear. T/I C B. After you have run each scenario, sketch the resulting position–time and velocity–time graphs from the actual data you collected. T/I C C. Compare your sketches from Question A to the graphs that were generated when you ran each scenario. Explain any discrepancies or misconceptions that you may have had. T/I go T o N ELSoN SC iEN C E Figure 5 Launch of NASA’s Mars Pathfinder mission Instantaneous velocity and Average velocity The velocity of any object that is accelerating is changing over time. In motion with motion with uniform acceleration uniform acceleration, the velocity of an object changes at a constant (uniform) rate. motion in which velocity changes at During a launch (Figure 5), a spacecraft accelerates upward at a rapid rate. NASA a constant rate personnel may need to determine the velocity of the spacecraft at specific points in > instantaneous velocity ( vinst ) the time. They can do this by plotting position and time data on a graph > and determining velocity of an object at a specific instant the spacecraft’s instantaneous velocity. Instantaneous velocity, or v inst, is the> velocity of in time an object at a specific instant in time. By comparison, average velocity (v av) is deter- mined over a time interval. Both types of velocity are rates of change of position, but they tell us different things about the motion of an object. Tutorial 4 Determining Instantaneous and Average Velocity Figure 6 on the next page shows a position–time graph of an object that is undergoing uniform acceleration. Moving along the curve, the slope of the curve progressively increases. From this, we know that the velocity of the object is constantly increasing. To determine the instantaneous velocity of the object at a specific time, we must calculate the slope of the tangent of the line on the position–time graph at that time. A tangent is a straight line that contacts a curve at a single point and extends in the same direction as the slope of the curve at the point. A plane mirror can be used to draw a tangent to a curved line. Place the mirror as perpendicular as possible to the line at the point desired. Adjust the angle of the mirror so that the real curve merges smoothly with its image in the mirror, which will occur when the mirror is perpendicular