Edexcel IGCSE Physics Movement & Position PDF

Head to www.savemyexams.com for more awesome resources Edexcel IGCSE Physics Your notes Movement & Position Contents Distance-Time Graphs Speed Core Practical: Investigating Motion Acceleration Velocity-Ti...

Head to www.savemyexams.com for more awesome resources Edexcel IGCSE Physics Your notes Movement & Position Contents Distance-Time Graphs Speed Core Practical: Investigating Motion Acceleration Velocity-Time Graphs Area under a Velocity-Time Graph Calculating Uniform Acceleration Page 1 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Distance-Time Graphs Your notes Distance-time graphs A distance-time graph shows how the distance of an object moving in a straight line (from a starting position) varies over time: This graph shows a moving object moving further away from its origin Constant speed on a distance-time graph Distance-time graphs also show the following information: Whether the object is moving at a constant speed How large or small the speed is A straight line represents constant speed The slope of the straight line represents the magnitude of the speed: A very steep slope means the object is moving at a large speed Page 2 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources A shallow slope means the object is moving at a small speed A flat, horizontal line means the object is stationary (not moving) Your notes This graph shows how the slope of a line is used to interpret the speed of moving objects. Both of these objects are moving with a constant speed, because the lines are straight. Examiner Tips and Tricks The Edexcel IGCSE Physics specification states that you need to be able to explain distance-time graphs and to be able to plot them. Plotting graphs is an important skill in physics and usually comes up in at least one exam paper. If you are not confident in plotting graphs, you should definitely practice before your exam! Changing speed on a distance-time graph Objects might be moving at a changing speed This is represented by a curve In this case, the slope of the line will be changing Page 3 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources If the slope is increasing, the speed is increasing (accelerating) If the slope is decreasing, the speed is decreasing (decelerating) Your notes The image below shows two different objects moving with changing speeds Changing speeds are represented by changing slopes. The red line represents an object slowing down and the green line represents an object speeding up. Calculating speed from a distance-time graph The speed of a moving object can be calculated from the gradient of the line on a distance-time graph: ∆y speed = gradient = ∆x Page 4 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The speed of an object can be found by calculating the gradient of a distance-time graph ∆ y is the change in y (distance) values ∆ x is the change in x (time) values Worked Example A distance-time graph is drawn below for part of a train journey. The train is travelling at a constant speed. Page 5 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Calculate the speed of the train. Answer: Step 1: Draw a large gradient triangle on the graph The image below shows a large gradient triangle drawn with dashed lines ∆ y and ∆ x are labelled, using the units as stated on each axes Page 6 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Step 2: Convert units for distance and time into standard units The distance travelled = 8 km = 8000 m The time taken = 6 mins = 360 s Step 3: State that speed is equal to the gradient of a distance-time graph The gradient of a distance-time graph is equal to the speed of a moving object: ∆y speed = gradient = ∆x Step 4: Substitute values in to calculate the speed 8000 speed = 360 speed = 22. 2 m/s Page 7 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked Example A student decides to take a stroll to the park. They find a bench in a quiet spot and take a seat, Your notes picking up where they left off reading a book on Black Holes. After some time reading, the student realises they lost track of time and runs home. A distance-time graph for for the student's trip is drawn below: a) How long does the student spend reading the book? b) Which section of the graph represents the student running home? c) What is the total distance travelled by the student? Answer: Part (a) The student spends 40 minutes reading the book The flat section of the line (section B) represents an object which is stationary - so section B represents the student sitting on the bench reading This section lasts for 40 minutes - as shown in the graph below Page 8 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Part (b) Section C represents the student running home The slope of the line in section C is steeper than the slope in section A This means the student was moving with a larger speed (running) in section C Part (c) The total distance travelled by the student is 0.6 km The total distance travelled by an object is given by the final point on the line - in this case, the line ends at 0.6 km on the distance axis. This is shown in the image below: Page 9 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Examiner Tips and Tricks Use the entire line, where possible, to calculate the gradient. Examiners tend to award credit if they see a large gradient triangle used - so remember to draw these directly on the graph itself! Remember to check the units of variables measured on each axis. These may not always be in standard units - in our example, the unit of distance was km and the unit of time was minutes. Double-check which units to use in your answer. Page 10 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Speed Your notes Calculating average speed The speed of an object is the distance it travels every second Speed is a scalar quantity because it has a magnitude but not a direction Average speed The speed of an object can vary throughout its journey Therefore, it is often more useful to know an object's average speed A hiker might have an average speed of 2.0 m/s, whereas a particularly excited bumble bee can have average speeds of up to 4.5 m/s Average speed formula The equation for calculating the average speed of a moving object is: Page 11 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources distance moved average speed = time taken Your notes Average speed considers the total distance moved and the total time taken Average speed formula triangle Formula triangle for average speed, distance moved and time taken How to use formula triangles Formula triangles are really useful for knowing how to rearrange physics equations To use them: 1. Cover up the quantity to be calculated, this is known as the 'subject' of the equation 2. Look at the position of the other two quantities If they are on the same line, this means they are multiplied If one quantity is above the other, this means they are divided - make sure to keep the order of which is on the top and bottom of the fraction! In the example below, to calculate average speed, cover-up the variable speed so that only distance and time are left The equation is revealed as: distance speed = time Page 12 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes To use a formula triangle, simply cover up the quantity you wish calculate and the structure of the equation is revealed Worked Example Planes fly at typical average speeds of around 250 m/s. Calculate the distance travelled by a plane moving at this average speed for 2 hours. Answer: Step 1: List the known quantities Average speed = 250 m/s Time taken = 2 hours Step 2: Write the relevant equation distance moved average speed = time taken Step 3: Rearrange to make distance moved the subject distance moved = average speed × time taken Step 4: Convert any units The time given in the question is not in standard units Convert 2 hours into seconds: Page 13 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 2 hours = 2 × 60 × 60 2 hours = 7200 s Your notes Step 5: Substitute the values for average speed and time taken distance moved = 250 × 7200 distance moved = 1 800 000 m Examiner Tips and Tricks Rearranging equations is an important skill in Physics. You can use the equation triangles to help you practice, but it is better not to rely on them because they do not work for all equations you may need to rearrange in the exam. Page 14 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Core Practical: Investigating Motion Your notes Core practical 1: investigating motion Aim of the experiment The aim of this experiment is to investigate the motion of some everyday objects, by measuring their speed Examples of objects that could be used are: a paper cone a tennis ball Measuring speed directly is difficult to do; therefore, by measuring distance moved and time taken, the average speed of the object can be calculated This is just one method of measuring the speed of different objects - some methods involve the use of light gates to measure speed and acceleration, e.g. for a toy car moving down a slope Variables Independent variable = Distance, d Dependent variable = Time, t Control variables: Use the same object (paper cone, tennis ball etc.) for each measurement Equipment Equipment list Equipment Purpose Paper cone / tennis ball To measure the speed of Stop watch To measure time taken Tape measure / metre rule To measure distance moved Resolution of measuring equipment: Page 15 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Ruler = 1 mm Stop clock = 0.01 s Your notes Method Investigating the motion of a falling paper cone 1. Measure out a height of 1.0 m using the tape measure or metre ruler 2. Drop the object (paper cone or tennis ball) from this height, which is the distance travelled by the object 3. Use the stop clock to measure how long the object takes to travel this distance 4. Record the distance travelled and time taken 5. Repeat steps 2-3 three times, calculating an average time taken for the object to fall a certain distance Page 16 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 6. Repeat steps 1-4 for heights of 1.2 m, 1.4 m, 1.6 m, and 1.8 m Results Your notes Example results table A results table should include spaces for all the measurements taken and space to perform any necessary calculations, such as averages Analysis of results The average speed of the falling object can be calculated using the equation: distance moved average speed = time taken Where: Average speed is measured in metres per second (m/s) Distance moved is measured in metres (m) Time taken is measured in seconds (s) Therefore, calculate the average speed at each distance by dividing the distance by the average time taken Evaluating the experiment Systematic errors Make sure the measurements on the tape measure or metre rule are taken at eye level to avoid parallax error Page 17 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources The average human reaction time is 0.25 s, which is equivalent to half a second per when starting and stopping the timer Your notes This is likely to be significant when small intervals of time are measured To reduce this systematic error, larger distances could be used resulting in larger time intervals Using a ball bearing and an electronic data logger, like a trap door, is a good way to remove the error due to human reaction time for this experiment Consider using an electronic sensor, such as light gates, to obtain highly accurate measurements of time The timer on a light gate starts and stops automatically as it passes the sensors positioned at the start and stop points Random errors Ensure the experiment is done in a space with no draft or breeze, as this could affect the motion of the falling object Safety considerations Place a mat or a soft material below any falling object to cushion its fall Page 18 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Acceleration Your notes Acceleration Rate of change in velocity Acceleration is defined as the rate of change in velocity In other words, it describes how much an object's velocity changes every second The equation below is used to calculate the average acceleration of an object: change in velocity acceleration = time taken ∆v a= t Where: a = acceleration in metres per second squared (m/s2) ∆ v = change in velocity in metres per second (m/s) t = time taken in seconds (s) Formula triangle for acceleration, change in velocity and time Page 19 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes To use an equation triangle, simply cover up the value you wish to calculate and the structure of the equation will be revealed The change in velocity is found by the difference between the initial and final velocity: change in velocity = final velocity − initial velocity ∆v = v − u Where: v = final velocity in metres per second (m/s) u = initial velocity in metres per second (m/s) Therefore, the acceleration, or the rate of change in velocity, equation can be written as: (v − u ) a= t Speeding up and slowing down Page 20 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources An object that speeds up is accelerating An object that slows down is decelerating Your notes The acceleration of an object can be positive or negative, depending on whether the object is speeding up or slowing down If an object is speeding up, its acceleration is positive If an object is slowing down, its acceleration is negative (also known as deceleration) Examples of acceleration and deceleration A rocket speeding up (accelerating) and a car slowing down (decelerating) Worked Example A Japanese bullet train decelerates at a constant rate in a straight line. The velocity of the train decreases from 50 m/s to 42 m/s in 30 seconds. Page 21 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources (a) Calculate the change in velocity of the train. (b) Calculate the deceleration of the train, and explain how your answer shows the train is slowing Your notes down. Answer: Part (a) Step 1: List the known quantities Initial velocity, u = 50 m/s Final velocity, v = 42 m/s Step 2: Write the relevant equation ∆v = v − u Step 3: Substitute values for final and initial velocity ∆ v = 42 − 50 ∆ v = − 8 m/s Part (b) Step 1: List the known quantities Change in velocity, ∆ v = − 8 m/s Time taken, t = 30 s Step 2: Write the relevant equation (v − u ) ∆v a= = t t Step 3: Substitute the values for change in velocity and time −8 a= 30 a = − 0. 27 m/s2 Step 4: Interpret the value for deceleration Page 22 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources The answer is negative, which indicates the train is slowing down Your notes Examiner Tips and Tricks Remember, the units for acceleration are metres per second squared, m/s2 In other words, acceleration measures how much the velocity (in m/s) changes every second, m/s/s. Page 23 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Velocity-Time Graphs Your notes Velocity-time graphs A velocity time graph, or velocity-time graph, shows how the velocity of a moving object varies with time Velocity-time refers to the fact that velocity is plotted against time on the graph The red line represents an object with increasing velocity The green line represents an object with decreasing velocity Velocity-time graph Increasing and decreasing velocity represented on a velocity-time graph Acceleration on a velocity-time graph Velocity-time graphs also show the following information: Whether the object is moving with a constant acceleration The magnitude of the acceleration A straight line represents constant acceleration (or deceleration) The slope of the line represents the magnitude of acceleration Page 24 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources A steep slope means large acceleration The object's speed changes very quickly Your notes A gentle slope means small acceleration The object's speed changes very gradually A positive gradient shows increasing velocity The object is accelerating A negative gradient shows decreasing velocity The object is decelerating A flat line means the acceleration is zero The object is moving with a constant velocity Constant acceleration and constant velocity on a velocity-time graph Page 25 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Flat horizontal lines on a velocity-time graph show periods of constant velocity, and sloping straight line show periods of acceleration Your notes Gradient of a velocity-time graph How to find acceleration on a velocity-time graph The acceleration of an object can be calculated from the gradient of a velocity-time graph ∆y acceleration = gradient = ∆x How to find the gradient of a velocity-time graph ∆ y is the change in y (velocity) values ∆ x is the change in x (time) values Worked Example A cyclist is training for a cycling tournament. Page 26 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources The velocity-time graph below shows the cyclist's motion as they cycle along a flat, straight road. Your notes (a) In which section (A, B, C, D, or E) of the velocity-time graph is the cyclist's acceleration the largest? (b) Calculate the cyclist's acceleration between 5 and 10 seconds. Answer: Part (a) Step 1: Recall that the slope of a velocity-time graph represents the magnitude of acceleration The slope of a velocity-time graph indicates the magnitude of acceleration Therefore, the only sections of the graph where the cyclist is accelerating are sections B and D Sections A, C, and E are flat; in other words, the cyclist is moving at a constant velocity (therefore, not accelerating) Step 2: Identify the section with the steepest slope Section D of the graph has the steepest slope Hence, the largest acceleration is shown in section D Page 27 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Part (b) Your notes Step 1: Recall that the gradient of a velocity-time graph gives the acceleration Calculating the gradient of a slope on a velocity-time graph gives the acceleration for that time period Step 2: Draw a large gradient triangle at the appropriate section of the graph A gradient triangle is drawn for the time period between 5 and 10 seconds Step 3: Calculate the size of the gradient and state this as the acceleration The acceleration is given by the gradient, which can be calculated using: ∆y a= ∆x 5 a= 5 a = 1 m/s2 Page 28 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Therefore, the cyclist accelerated at 1 m/s2 between 5 and 10 seconds Your notes Examiner Tips and Tricks Use the entire slope, where possible, to calculate the gradient. Examiners tend to award credit if they see a large gradient triangle used. Remember to actually draw the lines directly on the graph itself, particularly when the question asks you to use the graph to calculate the acceleration. Page 29 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Area under a Velocity-Time Graph Your notes Area under a velocity-time graph How to find the area under a velocity-time graph The area under a velocity-time graph represents the displacement (or distance travelled) by an object The displacement, or distance travelled, is represented by the area beneath the graph If the area beneath the velocity-time graph forms a triangle (i.e. the object is accelerating or decelerating), then the area can be determined by using the following formula: Area = ½ × Base × Height If the area beneath the velocity-time graph forms a rectangle (i.e. the object is moving at a constant velocity), then the area can be determined by using the following formula: Area = Base × Height How to find distance from a velocity-time graph Enclosed areas under velocity-time graphs represent total displacement (or total distance travelled) in a time interval Page 30 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Three enclosed areas (two triangles and one rectangle) under this velocity-time graph represent the total distance travelled in the total time If an object moves with constant acceleration, its velocity-time graph will consist of straight lines In this case, calculate the distance travelled by working out the area of enclosed rectangles and triangles The area of each enclosed section represents the distance travelled in that particular interval of time The total distance travelled is the sum of all the individual enclosed areas Worked Example The velocity-time graph below shows a car journey that lasts for 160 seconds. Page 31 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Calculate the total distance travelled by the car. Answer: Step 1: Recall that the area under a velocity-time graph represents the distance travelled In order to calculate the total distance travelled, the total area underneath the line must be determined Step 2: Identify each enclosed area In this example, there are five enclosed areas under the line These can be labelled as areas 1, 2, 3, 4 and 5, as shown in the image below: Page 32 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Step 3: Calculate the area of each enclosed shape under the line Area 1 = area of a triangle 1 A1 = × base × height 2 1 A1 = × 40 × 17. 5 2 A 1 = 350 m Area 2 = area of a rectangle A 2 = base × height A 2 = 30 × 17. 5 A 2 = 525 m Area 3 = area of a triangle 1 A3 = × base × height 2 Page 33 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 1 A3 = × 20 × 7. 5 2 Your notes A 3 = 75 m Area 4 = area of a rectangle A 4 = base × height A 4 = 20 × 17. 5 A 4 = 350 m Area 5 = area of a triangle 1 A5 = × base × height 2 1 A5 = × 70 × 25 2 A 5 = 875 m Step 4: Calculate the total distance travelled by finding the total area under the line Add up each of the five areas enclosed: total distance = A + 1 A2 + A3 + A4 + A5 total distance = 350 + 525 + 75 + 350 + 875 total distance = 2175 m Examiner Tips and Tricks Some areas will need to be split into a triangle and a rectangle to determine the area for a specific time interval, like areas 3 & 4 in the worked example above. If you are asked to find the distance travelled for a specific time interval, then you just need to find the area of the section above that time interval. For example, the distance travelled between 70 s and 90 s is the sum of Area 3 + Area 4 Page 34 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Calculating Uniform Acceleration Your notes Calculating uniform acceleration Uniform acceleration is constant acceleration The following equation applies to objects moving with uniform acceleration: (final speed)2 = (initial speed)2 + (2 × acceleration × distance moved) v2 = u2 + 2as Where: s = distance moved in metres (m) u = initial speed in metres per second (m/s) v = final speed in metres per second (m/s) a = acceleration in metres per second squared (m/s2) This equation is used to calculate quantities such as initial or final speed, uniform acceleration, or distance moved in cases where the time taken is not known Examiner Tips and Tricks This is an example of an equation that cannot be rearranged with a formula triangle. It is really important that you learn to rearrange equations without the help of a triangle for your exam. To rearrange any equation, follow these simple rules: What ever you do to the equation, you must do to both sides To undo an operation, perform the opposite operation To undo a subtraction, you must add (and vice versa) To undo a multiplication, you must divide (and vice versa) To undo a square, you must square root (and vice versa) Always show your working out, there is usually a mark awarded for rearranging an equation in an exam question. Page 35 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked Example A car accelerates steadily from rest at a rate of 2.5 m/s2 up to a speed of 16 m/s. Your notes Calculate how far the car moves during this period of acceleration. Answer: Step 1: List the known quantities Initial speed, u = 0 m/s Because the car starts from rest Final speed, v = 16 m/s Acceleration, a = 2. 5 m/s Step 2: Identify and write down the equation to use The question says that the car 'accelerates steadily' - so the equation for uniform acceleration can be used: v 2 = u 2 + 2as Step 3: Rearrange the equation to work out the distance moved Subtract u 2 from each side v 2 − u 2 = 2as Divide both sides by 2 a v2 − u2 s= 2a Step 4: Substitute known quantities into the equation and simplify where possible 162 − 02 s= 2 × 2.5 256 s= 5 s = 51 m Examiner Tips and Tricks Page 36 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Writing out your list of known quantities and labelling the quantity you need to calculate is really good exam technique. It helps you determine the correct equation to use, and sometimes examiners award credit for showing this working. Your notes Page 37 of 37 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers