2020 Tokyo Olympics Physics Matters Textbook PDF

# Chapter 2: Newtonian Mechanics ## Kinematics ### What You Will Learn - What are speed, velocity and acceleration? - How do we analyse motion graphically? - What is acceleration of free fall? ### 2.1 What Are Speed, Velocity and Acceleration? #### Disciplinary Idea Matter and energy make up the...

# Chapter 2: Newtonian Mechanics ## Kinematics ### What You Will Learn - What are speed, velocity and acceleration? - How do we analyse motion graphically? - What is acceleration of free fall? ### 2.1 What Are Speed, Velocity and Acceleration? #### Disciplinary Idea Matter and energy make up the universe. #### Helpful Note A unit of time can be a second, a minute or an hour. #### Learning Outcomes - State what is meant by speed and velocity. - Calculate average speed using distance travelled / time taken. - State what is meant by uniform acceleration and calculate the value of acceleration using change in velocity / time taken. - Interpret given examples of non-uniform acceleration. #### Speed * **If Usain Bolt were to race against a cheetah in a 100-metre sprint, who would be the winner (Figure 2.1)?** * To find out we will need to compare their speeds. Speed refers to how fast something moves. Speed is the distance travelled per unit time. Its SI unit is metre per second (m/s). * **Speed = distance travelled / time taken** * Based on Usain Bolt's 100-metre fastest record time of 9.58 s, Speed = 100 m / 9.58 s = 10.4 m/s. * Compare this with the cheetah's average running speed shown in Figure 2.2 #### Average Speed * Table 2.1 shows the results for men's running events at the 2020 Tokyo Olympics. * **Table 2.1 Results for men's running events at the 2020 Tokyo Olympics** | Athlete | Country | Event/m | Time/s | Average Speed / m/s | | ------------------------ | -------- | -------- | ------ | ------------------- | | Jacobs Lamont Marcell | Italy | 100 | 9.80 | 10.2 | | Andre de Grasse | Canada | 200 | 19.62 | 10.2 | | Steven Gardiner | Bahamas | 400 | 43.85 | 9.12 | | Emmanuel Kipkurui Korir | Kenya | 800 | 105.06 | 7.61 | * The speeds shown in Table 2.1 are average speeds. Average speed assumes that each athlete ran at the same speed throughout the entire distance. * **Average speed = total distance travelled / total time taken** * In reality, the athletes did not run at the same speed throughout their races. The speed at one instant may be different from the speed at another instant. The speed of an object at a particular instant is known as its instantaneous speed. #### Worked Example 2A * A car travels 6 km in 5 minutes. * (a) Calculate its average speed in m/s. * (b) Is the average speed of a car higher than the cheetah's speed shown in Figure 2.2? #### Answer * (a) Average speed = 6 x 1000 m / 5 x 60 s = 20 m/s. * (b) No. The car's average speed of 20 m/s is lower than the cheetah's average speed of 30 m/s. #### Differences Between Distance and Displacement * We have learnt about the differences between distance and displacement. Can you recall them? * Figure 2.4 shows the motion of an object from point A to point B and then to point C. We shall use it to illustrate the similarity and differences between distance and displacement. #### Distance * The total length covered by a moving object regardless of the direction of motion. * Has magnitude only. * SI unit: metre (m). * Distance travelled by the object from A to B, and then to C = 10 m + 2 m = 12 m. #### Displacement * The distance measured in a straight line from a fixed reference point. * Has both magnitude and direction. * SI unit: metre (m). * Displacement of the object travelling from A to B, and then to C = 10 m - 2 m = 8 m (from A to C). * Refers to the shortest distance and direction of an object from the starting point. #### Worked Example 2B * Figure 2.5 shows a car that travels 5.0 km due east from point O and makes a U-turn to travel another 3.0 km to reach the ending point E. * (a) Calculate the distance covered, and * (b) its displacement. #### Answer * (a) Distance covered = 5.0 km + 3.0 km = 8.0 km * (b) Taking the direction due east of point O as positive, Displacement = 5.0 km - 3.0 km = 2.0 km. The displacement of the car is 2.0 km due east of the starting point O. #### Velocity * We have learnt in Chapter 1 that when determining the velocity of an object, we need to know the speed of the object and the direction in which it is travelling. When calculating velocity, we use displacement instead of distance. * **Velocity = displacement / time taken** * To calculate the average velocity we use the following formula: * **Average velocity = total displacement / total time taken** #### Worked Example 2C * Figure 2.6 shows a car that travels 5.0 km due east and makes a U-turn to travel another 7.0 km. If the time taken for the whole journey is 0.20 h, calculate the: * (a) average speed; and * (b) average velocity of the car. #### Answer * (a) Average speed = (5.0 + 7.0) km / 0.20 h = 60 km/h * (b) Taking the direction due east of point O as positive, Average velocity = (5.0 - 7.0) km / 0.20 h = -10 km/h. The car is moving at an average velocity of 10 km/h westwards. ### Acceleration #### Disciplinary Idea Forces help us to understand motion. Characterising motion with position, displacement, velocity and acceleration provides the basis for formulating and appreciating Newton's laws of motion that provide the link to forces. #### Helpful Note The delta symbol, Δ, is used to represent change or difference. For example, in the equation for calculating uniform acceleration, Δv indicates the change in velocity. #### Uniform Acceleration * When the change (increase or decrease) in the velocity of an object for every unit of time is the same, the object undergoes constant or uniform acceleration (Table 2.3). * **Table 2.3: Object moving with uniform acceleration** | Time/s | Velocity/m/s | | ------ | ------------- | | 0 | 0 | | 1 | 20 | | 2 | 40 | | 3 | 60 | | 4 | 80 | | 5 | 100 | * Acceleration is the rate of change of velocity. * **Acceleration = change of velocity / time taken** * Acceleration is a vector quantity. Its SI unit is metre per second per second (m/s²). #### Worked Example 2D * A car at rest starts to travel in a straight path. It reaches a velocity of 12 m/s in 4 s (Figure 2.9). What is its acceleration, assuming that it accelerates uniformly? * **Answer** * We assign the direction to the right as positive. * Given: initial velocity u = 0 m/s (since the car starts from rest), Final velocity v = 12 m/s, time taken Δt = t - t0 = 4 s. * Since its acceleration is assumed to be uniform, a = (12 - 0) m/s / 4 s = 3.0 m/s². The acceleration is 3.0 m/s². #### Worked Example 2E * The velocity of a golf ball rolling in a straight line changes from 8 m/s to 2 m/s in 10 s (Figure 2.10). What is its deceleration, assuming that it is decelerating uniformly? * **Thought Process** * When an object is slowing down it is decelerating. Its final speed (v) is lower than its initial speed (u). Hence, the acceleration of the object (a = Δv / Δt ) is a negative value. The deceleration of an object that is slowing down is the positive value of its acceleration. * **Answer** * We assign the direction to the right as positive. * Given: initial velocity u = 8 m/s, final velocity v = 2 m/s, time taken Δt = 10 s. * Since its acceleration is assumed to be uniform, a = (2 - 8) m/s / 10 s = -0.60 m/s². The negative calculated value indicates that the ball is slowing down (i.e. decelerating). The deceleration of the ball is the positive value of its acceleration when the ball is slowing down. The deceleration is 0.60 m/s². #### Non-uniform Acceleration * An object undergoes non-uniform acceleration if the change in its velocity for every unit of time is not the same (Table 2.4). * **Table 2.4: Object moving with non-uniform acceleration** | Time/s | Velocity/m/s | | ------ | ------------- | | 0 | 0 | | 1 | 10 | | 2 | 40 | | 3 | 60 | | 4 | 70 | | 5 | 50 | * Note that the change in velocity is not the same for every second. The moving object is undergoing non-uniform acceleration (Figure 2.11). #### Let's Practise 2.1 * 1. At the start of a journey the odometer (a meter that tracks the total distance a car has travelled) showed an initial reading of 50 780 km. At the end of the journey, the odometer reading was 50 924 km. The journey took two hours. What was the average speed of the journey? * (a) km/h; and * (b) m/s? * 2. "Velocity is speed in a specific direction". Is this statement true or false? Explain your answer. * 3. (a) What is acceleration? * (b) Given that the velocity of an object moving in a straight line changes uniformly from u to v in time t, write an expression for the acceleration of the object. ### 2.2 How Do We Analyse Motion Graphically? #### Learning Outcomes - Plot and interpret a displacement-time graph and a velocity-time graph. - Deduce the motion of a body from the shape of a displacement-time graph. - Deduce the motion of a body from the shape of a velocity-time graph. - Calculate the area under a velocity-time graph to determine the displacement, for motion with uniform velocity or uniform acceleration. #### Displacement-Time Graphs * The displacement-time graph of an object gives us some information about the motion of the object. Let us look at the motion of a car as an example. In Figure 2.12, the car is travelling away from the starting point O. It travels along a straight line in one direction. * The displacement of the car is measured at every second. The displacement and time are recorded and a graph is plotted using the data. There are four possible scenarios for the motion of the car. * Figures 2.13 and 2.14 show the displacement-time graphs of the car at rest and travelling at a uniform velocity. * The gradient of a displacement-time graph of an object gives the velocity of the object. #### Car at rest * Time/s: 0, 1, 2, 3, 4, 5 * Displacement/m: 20, 20, 20, 20, 20, 20 * The graph has zero gradient. * The displacement of the car is a constant for every second. #### Car travelling at a uniform velocity of 10 m/s * Time/s: 0, 1, 2, 3, 4, 5 * Displacement/m: 0, 10, 20, 30, 40, 50 * The graph has a constant gradient. * The displacement of the car increases 10 m every second. #### Car travelling with increasing velocity (non-uniform velocity) * Time/s: 0, 1, 2, 3, 4, 5 * Displacement/m: 0, 5, 20, 45, 80, 125 * The graph has an increasing gradient. * The velocity of the car increases. * The instantaneous velocity of the car at any time t is given by the gradient of the tangent at the point. #### Car travelling with decreasing velocity (non-uniform velocity) * Time/s: 0, 1, 2, 3, 4, 5 * Displacement/m: o, 45, 80, 105, 120, 125. * The graph has a decreasing gradient. * The velocity of the car decreases. #### Velocity-Time Graphs * Velocity-time graphs can be used to show uniform and non-uniform acceleration of a car that is travelling along a straight line in one direction. Figures 2.17 and 2.18 show the velocity-time graphs of the car at rest and travelling at a uniform velocity. * The gradient of a velocity-time graph of an object gives the acceleration of the object. #### Car at rest * Time/s: 0, 1, 2, 3, 4, 5 * Velocity/m/s: 0, 0, 0, 0, 0, 0 * The velocity of the car remains at 0 m/s so the car has zero acceleration. #### Car travelling at a uniform velocity of 10 m/s * Time/s: 0, 1, 2, 3, 4, 5 * Velocity/m/s: 10, 10, 10, 10, 10, 10 * The velocity of the car remains at 10 m/s so the car has zero acceleration. #### Car travelling with uniform acceleration * Time/s: 0, 1, 2, 3, 4, 5 * Velocity/m/s: 0, 10, 20, 30, 40, 50 * The velocity of the car increases by 10 m/s every second. * The graph has a positive and constant gradient, and the acceleration is constant. #### Car travelling with uniform deceleration * Time/s: 0, 1, 2, 3, 4, 5 * Velocity/m/s: 50, 40, 30, 20, 10, 0. * The velocity of the car decreases by 10 m/s every second. * The graph has a negative and constant gradient, and the deceleration is constant. #### Car travelling with increasing acceleration (non-uniform acceleration) * Time/s: 0, 1, 2, 3, 4, 5 * Velocity/m/s: 0, 2, 8, 18, 32, 50 * The increase in velocity of the car increasing with time. * The graph has a positive and increasing gradient, and the acceleration increases. #### Car travelling with decreasing acceleration (non-uniform acceleration) * Time/s: 0, 1, 2, 3, 4, 5 * Velocity/m/s: 0, 18, 32, 42, 48, 50 * The increase in velocity of the car decreasing with time. * The graph has a positive and decreasing gradient, and the acceleration decreases. #### Comparisons Between Displacement-Time and Velocity-Time Graphs * Displacement-time graphs and velocity-time graphs look very similar but they give different information. We can differentiate the graphs by looking at the labels on the y-axes. Assume that a car starts from rest and accelerates uniformly in one direction to a constant velocity. The car then slows down and comes to a stop at a red light. Figure 2.23 shows the displacement-time and velocity-time graphs of the car and how they are related. ##### From A to B * Car accelerates uniformly from rest. ##### From B to C * Car moves at constant velocity. ##### From C to D * Car decelerates uniformly to a stop. #### Displacement-time Graph * The y-axis indicates displacement. Thus, this is a displacement-time graph. * Displacement increases at an increasing rate. Hence, the gradient increases (represented by a concave curve). * Displacement increases uniformly over time. Hence, the gradient is a positive constant (represented by a straight line). * Displacement increases at a decreasing rate. Hence, the gradient decreases (represented by a convex curve). #### Velocity-time Graph * The y-axis indicates velocity. Thus, this is a velocity-time graph. * Velocity increases uniformly over time. Hence, the gradient is a positive constant (represented by a straight line). * Velocity is constant. Hence, the graph here is a horizontal line. * Velocity decreases uniformly over time. Hence, the gradient is a negative constant (represented by a straight line). #### Worked Example 2F * Figure 2.24 shows the displacement-time graph of a car. Assume that the direction of the car moving away from origin O is positive. Describe the motion of the car at each stage: * (a) A * (b) B * (c) C * (d) D to E * (e) E to F #### Answer * **Motion of the Car** * (a) A: Displacement is 40 m from O. * Velocity is zero. * (b) B: Displacement is 66 m from O. * Velocity is uniform. * (c) C: Displacement is 90 m from O. * Velocity is zero. * (d) D to E: Car travels in the opposite direction back towards O. * Displacement decreases at a decreasing rate. * Velocity is non-uniform and decreasing. * (e) E to F: Displacement remains at 28 m from O. * Velocity is zero. #### Worked Example 2G * The velocity-time graph of a car is shown in Figure 2.25. Describe the motion of the car. #### Answer * **Motion of the Car** * 0-5 s: * Velocity increases uniformly from 0 m/s to 15 m/s. * Acceleration is uniform at a = (15-0) m/s / (5-0) s = 3.0 m/s². * 5-10 s: * Velocity increases from 15 m/s to 20 m/s at a decreasing rate. * Acceleration is non-uniform and decreasing. * 10-15 s: * Velocity is uniform and is at a maximum. * Acceleration is zero. * 15-20 s: * Velocity decreases uniformly from 20 m/s to 0 m/s. * Acceleration is uniform at a = (0-20) m/s / (20-15) s = -4.0 m/s². * 20-25 s: * Both velocity and acceleration are zero. * The car is stationary. * 25-30 s: * Velocity increases uniformly from 0 m/s to 10 m/s. * Acceleration is uniform at a = (10-0) m/s / (30-25) s = 2.0 m/s². * 30-35 s: * Velocity decreases from 10 m/s to 0 m/s at a decreasing rate. * Acceleration is non-uniform and negative. #### Area Under Velocity-Time Graph * Figure 2.26 shows the velocity-time graph for an object moving with a uniform velocity. * The velocity is 6 m/s from time t = 0 s to t = 16 s. What is the total displacement from t = 0 s to t = 10 s? * We can calculate it as follows: * v = 6 m/s, t = 10 s * v = d / t * d = v x t = 6 m/s x 10 s = 60 m * The product of velocity and time gives the displacement. From Figure 2.27 you can see that the product of 6 m/s and 10 s is the area of the shaded rectangle. #### Helpful Note * Only the area under a velocity-time graph gives the displacement. The area under a displacement-time graph does not give displacement. We can derive the total displacement from a displacement-time graph by reading the values directly off the y-axis. * Now consider the following velocity-time graph for an object that accelerates, moves with a uniform velocity, and then decelerates (Figure 2.28). #### Calculation of Total Displacement * Total displacement of the object = total area under the velocity-time graph = area of the trapezium = (1/2) x (sum of parallel sides) x height = (1/2) x (9 s + 24 s) x 36 m/s = 594 m #### Calculation of Average Speed * Average speed of object = total distance travelled /total time taken = 594 m / 24 s = 24.8 m/s #### Worked Example 2H * An MRT train moves off from Aljunied station and travels along a straight track towards Paya Lebar station. Figure 2.29 shows how the velocity of the train varies with time over the whole journey. * (a) Determine the average speed of the train between t = 0 s and t = 70 s. * (b) Describe the motion of the train between: * (i) t = 0 s and t = 10 s * (ii) t = 10 s and t = 70 s * (iii) t = 70 s and t = 90 s #### Answer * (a) Displacement between t = 0 s and t = 70 s = area under velocity-time graph between t = 0 s and t = 70 s = area of shaded trapezom (Figure 2.29) = (1/2) x (60 s + 70 s) x 15 m/s = 975 m * Average speed = total distance travelled / total time taken = 975 m / 70 s = 13.9 m/s. #### Helpful Note * Instead of using the formula for finding the area of a trapezium we can also find the same shaded area under the velocity-time graph by adding the area of the small shaded triangle and area of the shaded rectangle. #### Worked Example 2I * A motorist approaches a traffic light junction at 54 km/h. The traffic light turns red when he is 30 m from the junction. If he takes 0.4 s before applying the brakes, and his car blows down at a rate of 3.75 m/s², determine whether the motorist can stop his car in time. #### Thought Process * For the car to stop in time, it needs to stop within 30 m. Thus, we need to find the displacement of the car from the junction. #### Answer * Step 1: The displacement of the car can be determined by finding the area under its velocity-time graph. * Note: First covert 54 km/h into a speed in m/s. * 54 km/h = 54 km x 1000 m / 1h x 3600 s = 15 m/s. * Step 2: To find the area under the velocity-time graph, we need to find Δt. * Find: time interval Δt between the point at which the motorist applies the brakes and the point at which the car stops. * Given: uniform deceleration = 3.75 m/s² (i.e. acceleration a = -3.75 m/s²) change in velocity Δv = final velocity - initial velocity = -15 m/s. * a = Δv / Δt = 3.75 m/s² = - 15 m/s / Δt = Δt = 4 s. * Displacement = area under velocity-time graph = area of trapezium = (1/2) x (0.4 s + 4.4 s) x 15 m/s = 36 m * Since the displacement of his car is more than 30 m from the junction, the motorist is unable to stop his car in time. ### 2.3 What is Acceleration of Free Fall? #### Learning Outcome * State that the acceleration of free fall for a body near to Earth is constant and is approximately 10 m/s². * If we drop a large stone and a small pebble from the same height at the same time, which object will hit the ground first? * In the past, Aristotle claimed that a heavier object fell faster than a lighter object. His claim was widely accepted. However, in the, 17th century, Galileo Galilei had a different finding after conducting a series of experiments. He discovered that all objects, regardless of mass or size, fell at the same acceleration due to the Earth's gravity. * Figure 2.35 shows the difference between Aristotle's claim and Galileo's finding. #### Acceleration Due to Gravity * Acceleration due to gravity g is a constant. For objects close to Earth's surface, the value of g is generally taken to be 9.8 m/s². For simplicity in calculations, we take this value to be 10 m/s² throughout this book, unless otherwise stated. #### Let's Map It * Kinematics * can be described in terms of: * Speed (m/s) (scalar) * Velocity (m/s) (vector) * acceleration (m/s²) (vector) * Involves: * Distance (m) (scalar) * Displacement (m) (vector) * **Speed** * Speed = distance/time taken * Average speed = total distance/total time taken * Gradient of distance-time graph * **Velocity** * Velocity = displacement/time taken * Acceleration = change in velocity/time taken * Gradient of velocity-time graph * **Example:** * Acceleration of free fall for a body near Earth is constant and is approximately 10 m/s² * Area under velocity-time graph. #### Worked Example 2J * A sandal fell off a bamboo pole from the third floor while it was being put out to dry. The time taken for the sandal to reach the ground was 1.34 s (Figure 2.37). If air resistance was negligible: * (a) Find the velocity of the sandal just before it hit the ground; * (b) Find the height of the third floor from the ground; and * (c) Do you expect any change in the velocity-time graph if a sock fell off instead? #### Thought Process * Since the air resistance is negligible, the sandal is in free fall (i.e. accelerating at 10 m/s²). #### Answer * (a) Gradient of v-t graph = constant acceleration due to gravity = (v1 - 0) m/s / (1.34 - 0) s = 10 m/s² = v1 = 13.4 m/s. The velocity of the sandal just before it hit the ground was 13.4 m/s. * (b) Displacement = area under v-t graph = height of third floor from the ground = (1/2) x 13.4 m/s x 1.34 s = 9.0 m. * (c) No. Both the sandal and sock would give the same velocity-time graph, since acceleration is constant at 10 m/s² and air resistance is negligible. #### Let's Practise 2.3 * 1. Two objects of different masses were released from rest near to Earth’s surface. Assuming negligible air resistance, are the gradients of the velocity-time graphs for both of the objects the same or different? Explain your answer. * 2. An object is released from an unknown height and falls freely for 5 s before it hits the ground. * (a) Sketch the velocity-time graph for a time interval of 5 s, assuming there is negligible air resistance. * (b) What is the velocity of the object just before it hits the ground? * (c) What is the unknown height? #### Problem-based Learning Activity * You are an engineer in the Land Transport Authority (LTA) committee in charge of planning the transport system in Singapore. Your team receives a news report on the rise in the number of traffic accidents especially those involving elderly pedestrians * From the report, your team realises that many of the accidents happened as the pedestrians were not observing traffic rules. Some elderly pedestrians got into accidents when they jaywalked or crossed the road when the green man was blinking as they overestimated their walking speed. * To reduce the number of traffic accidents your team is tasked to propose how pedestrian crossings can be improved. * Use the following questions to guide you in coming up with your proposal. You may do a search on the Internet to find the answers. * Besides elderly pedestrians, what are some other groups that take a longer time to cross roads? * What are some technologies currently used at pedestrian crossings? * How can we improve these technologies? #### Cool Career: Sports Engineer * If you are someone who loves sports and technology a career in sports engineering may worthwhile exploring. A sports engineer designs apparel for athletes that improves fluid (air or liquid) flow across the athletes' body to increase the speed of the athletes (Figure 2.39). * As speed is a major factor in sports such as running, skating and swimming, a sports engineer will need a good understanding of kinematics, fluid dynamics and material science. This will help them develop sports apparel that will minimise resistive forces during movement. * Beside apparel, a sports engineer also does research on other equipment such as helmets and goggles. They also research on the postures that athletes need to be mindful of to achieve minimal drag (resistance) and therefore complete the race in the best possible times. #### Look at Figure 2.36. Based on what you have learnt, which observer(s) is/are correct? * **Both the man and girl will fall with a constant speed.** * **No, both the man and girl will fall with a constant acceleration.** * **I think the acceleration of the man and girl will increase as they fall.** #### Answer * Observer X and observer Z are incorrect as both the man and girl are undergoing constant acceleration of approximately 10 m/s² due to their weights. * Observer Y is correct as from the time they jump to the time just before the bungee cord becomes stretched and pulls on them, they are undergoing constant acceleration. #### Worked Example 2J * A sandal fell off a bamboo pole from the third floor while it was being put out to dry. The time taken for the sandal to reach the ground was 1.34 s (Figure 2.37). If air resistance was negligible: * (a) Find the velocity of the sandal just before it hit the ground; * (b) Find the height of the third floor from the ground; and * (c) Do you expect any change in the velocity-time graph if a sock fell off instead? #### Thought Process: * Since the air resistance is negligible, the sandal is in free fall (i.e. accelerating at 10 m/s²). #### Answer * (a) Gradient of v-t graph =constant acceleration due to gravity = (v1-0) m/s / (1.34-0) s = 10 m/s = v1 = 13.4 m/s. The velocity of the sandal just before it hit the ground was 13.4 m/s. * (b) Displacement = area under v-t graph = height of third floor from the ground = (1/2) x 13.4 m/s x 1.34 s = 9.0 m. * (c) No. Both the sandal and sock would give the same velocity-time graph, since acceleration is constant at 10 m/s² and air resistance is negligible. #### Let's Practise 2.3 * 1. Two objects of different masses were released from rest near to Earth’s surface. Assuming negligible air resistance, are the gradients of the velocity-time graphs for both of the objects the same or different? Explain your answer. * 2. An object is released from an unknown height and falls freely for 5 s before it hits the ground. * (a) Sketch the velocity-time graph for a time interval of 5 s, assuming there is negligible air resistance. * (b) What is the velocity of the object just before it hits the ground? * (c) What is the unknown height?