Linear Motion Intermediate Physics Notes (PDF)

INTERMEDIATE LEVEL PHYSICS 1ST YEARS LINEAR MOTION 0  MATSEC IM Syllabus Mechanics Section 2.1: Rectilinear Motion 2.1.1 Distance, displacement, speed, velocity and acceleration Candidates...

INTERMEDIATE LEVEL PHYSICS 1ST YEARS LINEAR MOTION 0  MATSEC IM Syllabus Mechanics Section 2.1: Rectilinear Motion 2.1.1 Distance, displacement, speed, velocity and acceleration Candidates should be able to: (a) Distinguish between distance and displacement. (b) Define speed, velocity and acceleration. !" !$ (c) Identify !# as the velocity and !# as the acceleration where s is the displacement, v is the velocity and t is the time. 2.1.2 Distance (displacement)-time, speed (velocity)-time and acceleration- time graphs Candidates should be able to: (a) Draw distance (displacement)-time, speed (velocity)-time and acceleration-time graphs. (b) Identify which section(s) in a distance (displacement)-time graph indicate(s) a state of rest, or constant speed (velocity) or acceleration. (c) Identify which section(s) in a velocity-time graph indicate(s) an object moving with constant velocity, an object moving with a constant acceleration and an object moving with a variable acceleration. (d) Identify which section(s) in a acceleration-time graph indicate(s) an object moving at constant velocity, at constant acceleration and varying acceleration. (e) Determine the speed (velocity) and the average speed (velocity) from a distance (displacement)-time graph. 1 (f) Determine the distance travelled, the displacement and the acceleration from a velocity-time graph. (g) Determine the change in velocity from an acceleration-time graph. (h) Relate the sign of the velocity, displacement and acceleration obtained from displacement-time and velocity-time graphs to directions specified by a sign convention. 2.1.3 Equations for uniformly accelerated motion Candidates should be able to: (a) Derive the equations of uniformly accelerated motion along a straight line. (b) Use the equations of rectilinear motion for uniformly accelerated motion to solve problems. (c) Use the equations of uniformly accelerated motion to solve problems involving projectiles, projected either vertically or horizontally. 2.1.4 Experiment Candidates should be able to: (a) Describe in detail an experiment to measure the acceleration of free fall. This includes providing a diagram, the procedure, adequate precautions, measurements made and how to determine the acceleration of free fall from the graph. 2  Kinematics Mechanics is the area of Physics which deals with the study of the motion of physical objects. Kinematics is the branch of mechanics which describes the motion of objects using words, diagrams, numbers, graphs and equations. In this topic, only linear motion, that is motion along a straight line, is going to be considered. Such motion is said to be translational. Linear Motion Distance, displacement, speed, velocity, acceleration and time are words used to describe the motion of objects. Ø Time Time, t, is one of the seven fundamental quantities in physics. It is a scalar quantity. Its SI unit is the second (s). Unit Symbol Fact second s minute min 1 min = 60 s hour hr 1 hr = 60 min 1 hr = (60 x 60) s = 3600 seconds day d 1 d = 24 hrs 1 d = (24 x 60 x 60) s = 86400 s week wk 1 week = 7 days month mo year yr 1 year = 52 weeks 1 year = 12 months 1 year = 365.25 days 3 Note: To convert from seconds to minutes divide by 60 To convert from minutes to hours divide by 60 To convert from hours to days divide by 24 To convert from days to years divide by 365.25 To convert from years to days multiply by 365.25 To convert from days to hours multiply by 24 To convert from hours to minutes multiply by 60 To convert from minutes to seconds multiply by 60 Ø Distance and Displacement Distance and displacement are both ways of measuring how far an object has moved. The distance is the total length covered when moving between two points (the red path). § The distance, s, is a scalar quantity: only the magnitude is important; the direction is not considered. § Distance measured is always positive. 4 § The total distance covered is equal to the sum of all the distances travelled in different directions. § There is always a distance covered whenever there is motion. § The SI unit of distance is the metre (m). Unit Symbol Fact millimetre mm centimetre cm 1 cm = 10 mm metre m 1 m = 100 cm 1 m = (10 x 100) mm = 1000 mm kilometre km 1 km = 1000 m Note: To convert from millimetres to centimetres divide by 10 To convert from centimetres to metres divide by 100 To convert from millimetres to metres divide by 1000 To convert from metres to kilometres divide by 1000 To convert from kilometres to metres multiply by 1000 To convert from metres to millimetres multiply by 1000 To convert from metres to centimetres multiply by 100 To convert from centimetres to millimetres multiply by 10 The displacement is the length and direction of an imaginary line joining the object to some fixed initial point (the blue arrow). The displacement is the direction and how far a body is from some initial fixed point. Displacement is distance moved in a particular direction 5 § The displacement, s, is a vector quantity: the magnitude of the displacement is the length of the blue arrow; the direction of the displacement is the direction of the blue arrow. § Displacement can be positive or negative depending on the reference point. Example: Two cars start exactly from the same location. The blue car moves 500 m west; whilst the red car moves 500 m east starting point The displacement of the blue car is – 500m or 500 m west. The displacement of the red car is 500 m or 500 m east. § Displacement will be zero if the body comes back to its original position. Example: The person moves from A to B to C to D and then back to A, along the indicated path. Distance travelled = 3 + 5 + 3 + 5 = 16 m Displacement = 0 m. § Its SI unit is the metre. 6 Example: A delivery man starts at the post office, drives 20 km west, then 40 km north to stop for lunch. Find (a) the total distance covered by the delivery man (b) his resultant displacement. 7 Ø Speed and Velocity The speed of an object refers to the distance that an object moves per second. It indicates how fast or slow the object is moving. Definition: Speed is the rate at which distance is covered. Speed is the distance travelled in unit time. 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑆𝑝𝑒𝑒𝑑 = 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 This formula can only be used if the object travels at a constant speed. In fact, it is hardly possible for an object to cover a journey at a constant speed. So, it is more useful to find the average speed. The average speed can be found from the following formula: 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 = 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑠 ∆𝑠 𝑣= = 𝑡 ∆𝑡 A speedometer shows the actual or instantaneous speed of an object. This varies throughout the journey as the object decelerates or accelerates. The instantaneous speed is the speed at a particular moment in a journey. It is found by dividing the distance moved by a very small interval of time. The smaller the time interval, the closer would the answer be to the instantaneous speed. 𝑑𝑠 𝑣= 𝑑𝑡 The average speed and the instantaneous speed would be the same only if the object would be moving with a constant speed. 8 § Speed is a scalar quantity. § It is measured in metres per second (written as: m/s or m s-1) Unit Symbol Fact metres per second m s-1 kilometres per hour km h-1 1 km h-1 = 0.278 m s-1 Note: To convert from km h-1 to m s-1 divide by 3.6 To convert from m s-1 to km h-1 multiply by 3.6 The velocity is the distance travelled per unit time in a stated direction. Velocity is the speed in a specified direction. Definition: Velocity is the rate of change of displacement. For uniform velocity: 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 If velocity changes: 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 ∆𝑠 𝑣= ∆𝑡 The velocity at a particular moment in a journey is known as the instantaneous velocity. The instantaneous velocity can be calculated using the following formula (in calculus notation): 𝑑𝑠 𝑣= 𝑑𝑡 9 This means that dividing the smallest imaginable change in displacement at a particular instant by the corresponding change in time would give the instantaneous velocity. § Velocity is a vector quantity: It has a magnitude and a direction. Remember: Velocities in opposite directions should be given opposite signs. Velocities to the right are usually taken as positive. Velocities to the left are usually taken as negative. The two cars are moving with the same speed, but with different velocities as they are moving in opposite directions. Note that velocity changes if: Ü the object’s speed changes but its direction does not change Ü the object moves with a constant speed, but it changes direction Ü the object’s speed and direction change. § Velocity is measured in metres per second (written as: m/s or m s-1) 10 Example: A boat sails 300 m due west and then 300 m due north. This takes a total time of 1.5 minutes. Find (i) the boat’s average speed, and (ii) its average velocity. Maths Note: D means ‘change in …’ d means ‘small change in …’ ∆" ∆# is the average speed or velocity during a significant time interval. &" &# is the average speed or velocity during a very small time interval. !" !# is the limiting value this ratio approaches as dt gets closer to zero. It represents instantaneous speed or velocity. 𝑑 7 pronounced as 𝑑𝑒𝑒 − 𝑏𝑦 − 𝑑𝑒𝑒 − 𝑡𝑒𝑒 𝑐an be read as 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 … H 𝑑𝑡 11 Ø Acceleration Definition: Acceleration is the rate of change of velocity. Acceleration is the change in velocity per unit time. Note: Acceleration does not necessarily mean ‘an increase in speed’ For uniform acceleration: 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 ∆𝑣 𝑎= ∆𝑡 12 If acceleration changes, the instantaneous acceleration (the acceleration at a particular moment in a journey) can be calculated using the following formula (in calculus notation): 𝑑𝑣 𝑎= 𝑑𝑡 This means that the smallest imaginable change in velocity divided by the corresponding time in which it takes places gives the acceleration at that instant. § Acceleration is a vector quantity: it has a magnitude and a direction. The direction of the acceleration is the same as the direction of the velocity change. Acceleration is a vector. Its sign shows its direction. Therefore, a negative acceleration does not necessarily mean a deceleration. Note: § The velocity of an object increases, if the acceleration is in the direction of motion of the object. § The velocity of an object decreases, if the acceleration is in the opposite direction of motion of the object. Ø Indicating direction Displacement, velocity and acceleration are vectors. It is important that their direction is stated. When the motion is in a straight line (linear motion), the + and – signs can be used to indicate direction. For example: With horizontal motion: § motion to the right can be taken as positive § motion to the left can be taken as negative With vertical motion: § upward motion can be taken as positive § downward motion can be taken as negative It is important to keep the same convention for the entire calculation. 13 Ø Motion Graphs DISTANCE – TIME GRAPHS s/m s/m s/m s/m t/s t/s t/s t/s The object is at The object is The object is The object is rest away from a moving with a moving with moving with reference point. constant speed increasing speed. decreasing speed. away from a The object is The object is reference point. accelerating. decelerating. For a distance – time graph, the distance never decreases. On a distance – time graph Ø A horizontal line implies that the object is stationary. Ø A straight line which is not horizontal implies constant speed. Ø A curved line implies acceleration – the object would be travelling at a non-uniform speed. The y-intercept of a distance – time graph gives the initial distance away from a reference point. The gradient of a distance – time graph gives the speed of the object. 14 The graph line is The gradient of The gradient of the The gradient of the horizontal. the graph is graph is graph is constant increasing. decreasing. The gradient is 0 m s-1. Thus, the speed Thus, the speed of Thus, the speed of of the object is the object is the object is Thus, the speed is constant - the increasing – it is decreasing – it is 0 m s-1. object is moving moving with a moving with a with uniform non-uniform non-uniform The object is at speed. speed. The object speed. The object rest. is accelerating. is decelerating. The gradient ∆" gives the value of The gradient at a The gradient at a Gradient = ∆# = 0 m s-1 the constant point gives the point gives the speed. instantaneous instantaneous speed. speed. ∆" Gradient = ∆# = v The gradient of The average speed The average speed the line is always is found by is found by positive as for a dividing the total dividing the total distance-time distance s by the distance s by the graph, distance time taken t. time taken t. never decreases. 𝑠 𝑠 𝑣= 𝑣= 𝑡 𝑡 The steeper the gradient, the greater the speed. 15  Maths Note: To find the gradient: Choose two points lying on the graph. The points must be as far y away from each other as possible. Read their coordinates. The gradient is found by dividing the change in y by the change in x. ∆" "! $ "" Gradient = ∆# = #! $ #" x The unit of the gradient is found by dividing the units of the y- axis by the units of the x-axis. To find the gradient at a point: Draw a tangent to the curve at the point A tangent is a straight line which touches the curve at one point. The gradient of this tangent is found by choosing two points lying on the tangent. The points must be as far away from each other as possible. Read their coordinates. The gradient is found by dividing the change in y by the change in x. ∆" "! $ "" Gradient = ∆# = #! $ #" DISPLACEMENT – TIME GRAPHS s/m s/m s/m s/m t/s t/s t/s t/s The object is at The object is The object is The object is rest away from a moving with a moving with moving with reference point. constant velocity increasing decreasing velocity. away from a velocity. The The object is reference point. object is decelerating. accelerating. 16 On a displacement – time graph Ø A horizontal line implies that a state of rest. Ø A straight line which is not horizontal implies uniform velocity. Ø A curved line implies acceleration – the object would be travelling at a non-uniform velocity. The y-intercept of a displacement – time graph gives the initial displacement from a reference point. The gradient of a displacement – time graph gives the velocity (both magnitude and direction) of the object. The graph line is The gradient of the The gradient of the The gradient of the horizontal. graph is constant graph is increasing. The graph is decreasing. steeper the gradient, The less steep the The gradient is 0 m s-1. Thus, the velocity of the the greater the velocity. gradient is, the smaller object is constant - the is the velocity. Thus, the velocity is 0 object is moving with Thus, the velocity of the m s-1. uniform velocity. object is increasing – it is Thus, the velocity of the moving with a non- object is decreasing – it is The object is at rest. The gradient gives the uniform velocity. The moving with a non- value of the constant object is accelerating. uniform velocity. The ∆" velocity. object is decelerating. Gradient = ∆# The gradient at a point = 0 m s-1 ∆" gives the instantaneous The gradient at a point Gradient = ∆# = v velocity. gives the instantaneous Positive gradients velocity. (sloping upwards) The average velocity is indicate velocity in one found by dividing the total The average velocity is direction (the positive displacement s by the found by dividing the total direction). time taken t. displacement s by the 𝑠 time taken t. Negative gradients 𝑣= 𝑡 𝑠 (sloping downwards) 𝑣= 𝑡 indicate velocity in the opposite direction (the negative direction) 17 Example: The graph shows the displacement along a horizontal line of a moving object as it changes with time. Describe the motion of the object. 18 SPEED – TIME GRAPHS v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The object is at The object is The object is The object is rest. moving with a moving with moving with constant speed. increasing speed. decreasing speed. The object is The object is accelerating decelerating uniformly. uniformly. v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The object is The object is The object is The object is moving with moving with moving with moving with increasing speed. increasing speed. decreasing speed. decreasing speed. The acceleration The acceleration The deceleration The deceleration is is increasing – is decreasing – is not constant. It non uniform. It is acceleration is acceleration is is increasing. decreasing. not uniform. not constant. On a speed – time graph Ø A horizontal line on the x – axis implies that the object is stationary. Ø A horizontal line (not on the x – axis) implies constant speed Ø A straight line which is not horizontal implies constant acceleration. Ø A curved line implies non - uniform acceleration. 19 The y-intercept of a speed – time graph gives the initial speed. The gradient of a speed – time graph gives the magnitude of the acceleration of the object. The area under a speed – time graph gives the distance travelled. v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The graph line is The gradient is 0 m s-2. The gradient of the The gradient of the graph horizontal. graph is constant. is constant. This shows that the The gradient is 0 m s-2. magnitude of the Thus, the magnitude of Thus, the magnitude of acceleration is 0 m s-2. the acceleration of the the acceleration of the Thus, the magnitude of object is constant. object is constant. the acceleration is ∆$ Gradient = ∆# 0 m s-2. ∆$ ∆$ = 0 m s-2 Gradient = ∆# = a Gradient = ∆# = a The object is at rest, The gradient of the The gradient of the graph since its speed is graph is positive is negative indicating 0 m s-1. indicating that the that the object has a -ve object is accelerating - acceleration ∆$ Gradient = the speed of the object (deceleration) - the speed ∆# = 0 m s-2 is increasing. is decreasing. The area under the The area under the The area under the The area under the graph is 0 m indicating graph between the (v-t graph between the (v-t graph between the (v-t that the object has not graph line and the time graph line and the time graph line and the time moved. The object is at axis) gives the distance axis) gives the distance axis) gives the distance rest. travelled. travelled. travelled. Area of rectangle Area of triangle Area of triangle + Area % = length x breadth = base x height of rectangle & % = base x height & + length x breadth 20 v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The gradient is The gradient is The gradient is The gradient is positive and positive and negative and negative and increasing. This decreasing. This increasing. This decreasing. This shows that the shows that the shows that the shows that the magnitude of the magnitude of the magnitude of the magnitude of the acceleration is acceleration is deceleration is deceleration is increasing – the decreasing – the increasing – the decreasing – the acceleration is not acceleration is not acceleration is not acceleration is not uniform. uniform. uniform. uniform. The gradient at a The gradient at a The gradient at a The gradient at a point gives the point gives the point gives the point gives the magnitude of the magnitude of the magnitude of the magnitude of the instantaneous instantaneous instantaneous instantaneous acceleration. acceleration. acceleration. acceleration. &' &' &' &' Gradient = a = &( Gradient = a = &( Gradient = a = &( Gradient = a = &( The area under the The area under the The area under the The area under the graph gives the graph gives the graph gives the graph gives the distance distance travelled. distance travelled. distance travelled. travelled. Maths Note: 21  Maths Note: Finding the area under the graph: Count the number of squares under the graph. Either combine fragments of squares or else use the following simple rule. Rule: if more than half the square is under the graph, count it as a whole square; if less than half a square, don’t count the square. Multiply the number of squares by the scale of 1 box. VELOCITY – TIME GRAPHS v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The object is at The object is The object is The object is rest. moving with a moving with moving with constant velocity. increasing decreasing velocity. velocity. The The object is object is decelerating accelerating uniformly. uniformly. 22 v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The object is The object is The object is The object is moving with moving with moving with moving with increasing increasing decreasing decreasing velocity. velocity. The velocity. The velocity. The The deceleration is acceleration is acceleration is deceleration is not non uniform. It is increasing – the decreasing – the constant. It is decreasing. acceleration is acceleration is increasing. not uniform. not constant. On a velocity – time graph Ø A horizontal line on the x – axis implies that the object is stationary. Ø A horizontal line (not on the x – axis) implies constant velocity Ø A straight line which is not horizontal implies constant acceleration. Ø A curved line implies non - uniform acceleration. The y-intercept of a velocity – time graph gives the initial velocity. The gradient of a velocity – time graph gives the acceleration of the object. The sign of the gradient indicates the direction of the acceleration. The area under a velocity – time graph gives the displacement. The area under a velocity – time graph can also give the distance travelled. 23 v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The graph line is The gradient is 0 m s-2. The gradient of the The gradient of the graph horizontal. graph is constant. is constant. The acceleration is The gradient is 0 m s-2. 0ms. -2 This shows that the Thus, the magnitude of acceleration is constant the acceleration of the The acceleration is 0 ∆$ or uniform. object is constant. Gradient = ∆# m s-2. a = 0 m s-2 ∆$ ∆$ Gradient = ∆# = a Gradient = ∆# = a The object is at rest, The object is not since its velocity is accelerating but it is The gradient of the The gradient of the graph 0 m s-1. moving with a uniform graph is positive This is negative. This velocity. indicates that the indicates that the ∆$ Gradient = ∆# acceleration is in the acceleration is in the a = 0 m s-2 same direction as the opposite direction to the motion (velocity is motion (velocity is increasing). decreasing). The area under the The area under the The area under the The area under the graph is 0 m indicating graph between the (v-t graph between the (v-t graph between the (v-t that the object has not graph line and the time graph line and the time graph line and the time moved. The object is at axis) gives the axis) gives the axis) gives the rest. displacement. displacement. displacement. Area of rectangle Area of triangle Area of triangle + Area % = length x breadth = & base x height of rectangle % = & base x height + length x breadth 24 v/m s-1 v/m s-1 v/m s-1 v/m s-1 t/s t/s t/s t/s The gradient is The gradient is The gradient is The gradient is positive and positive and negative and negative and increasing. This decreasing. This increasing. This decreasing. This shows that the shows that the shows that the shows that the acceleration is acceleration is acceleration is acceleration is increasing – the decreasing – the increasing – the decreasing – the acceleration is not acceleration is not acceleration is not acceleration is not uniform. The uniform. The uniform. It is in the uniform. It is in the acceleration is in the acceleration is in the opposite direction to opposite direction to same direction as the same direction as the the motion. the motion. motion. motion. The gradient at a The gradient at a The gradient at a The gradient at a point gives the point gives the point gives the point gives the instantaneous instantaneous instantaneous instantaneous acceleration. acceleration. acceleration. acceleration. &' &' Gradient = a = &( Gradient = a = &( &' &' Gradient = a = &( Gradient = a = &( The area under the The area under the The area under the The area under the graph gives the graph gives the graph gives the graph gives the displacement. displacement. displacement. displacement. 25 Example: The velocity – time graph represents the motion of an object. a) Describe the motion of the object. b) Sketch the corresponding acceleration – time graph. c) What is the displacement of the object after 30 s? d) What is the total distance travelled in 30 s? e) What is the average velocity? f) What is the average speed? Note: From the velocity – time graph the following information can be obtained: § The direction in which the object is moving. § How fast it is moving. § Whether the object is accelerating or not. 26 ACCELERATION – TIME GRAPHS a/m s-2 a/m s-2 a/m s-2 t/s t/s t/s The object is not The object is moving with The acceleration is accelerating. uniform acceleration. increasing at a uniform The velocity of the object rate – non uniform a = 0 m s-2 increases uniformly. acceleration. The velocity of the object increases non-uniformly. It is either at rest if its velocity is 0 m s-1 or moving at uniform velocity. a/m s-2 a/m s-2 t/s t/s The object is increasing velocity – the The object is increasing velocity – the acceleration is non uniform. acceleration is non uniform. The acceleration is increasing. The rate of The acceleration is increasing. The rate of increase in acceleration is increasing. increase in acceleration is decreasing. 27 On an acceleration – time graph Ø A horizontal line on the x – axis implies that the object is not accelerating. The object would either be at rest if its velocity is 0 m s-1 or moving with constant velocity. Ø A horizontal line (not on the x – axis) implies constant acceleration. Zero slope implies constant acceleration. Ø A straight line which is not horizontal implies non - uniform acceleration – the acceleration would be changing at a constant rate. Ø A curved line implies non - uniform acceleration – the acceleration would be changing at an increasing or decreasing rate. If the slope is not zero, then the acceleration is varying. The y-intercept of an acceleration – time graph gives the initial acceleration. The area under an acceleration – time graph gives the change in velocity. Example 1: a/m s-2 The acceleration – time graph represents the motion of a car. Given that the initial velocity of the car is 0 m s-1 and that the car is accelerating at 5 m s-2 for 6 s, find the 5 velocity of the car after 6 s of motion. 6 t/s 28 Example 2: A sailboat is sailing in a straight line with a velocity of 10 m s-1. Then at time t = 0 s, a wind blows causing the sailboat to accelerate as seen in the following acceleration – time graph. Find the velocity after 9 s. a/m s-2 4 3 7 9 t/s -2 29 Motion Graphs for a ball which is thrown vertically upwards and returns to the point of projection (Ignore air resistance): Upward directed vectors are taken as positive. Downward directed vectors are taken as negative. Note: AB is the path taken on going up B is the point where the ball momentarily stops. BC is the return path. Distance / m As the ball moves upwards, the force of gravity acts against its motion. The ball’s speed decreases, until the speed becomes zero. As the ball moves downwards, the force of gravity acts in the same direction of motion. The ball’s speed increases. From A to B, the slope decreases gradually – the speed of the ball decreases. (The slope of a distance-time graph gives the speed.) At B, the slope is zero. The ball stops momentarily. From B to C, the slope increases gradually – the speed of the ball Time / s increases. Displacement / m The ball is above the point of projection at all times. Therefore, the displacement is taken as positive, as it is upward directed. From A to B, the slope is positive, and it decreases gradually. Thus, the velocity of the ball is upward directed (ball is moving upwards) and is decreasing. (The slope of a displacement - time graph gives the velocity.) At B, the slope is zero. The ball stops momentarily. From B to C, the slope is negative and increases gradually. The velocity of the ball is downward directed (ball is moving downwards) and is increasing. Time / s 30 speed / m s-1 From A to B, the ball is moving upwards and decelerating as its speed is decreasing. The deceleration is constant; thus the graph is a straight line with a negative gradient. At B, the speed of the ball is 0 m s-1. The ball momentarily stops. From B to C, the ball is moving downwards and accelerating as its speed is increasing. The acceleration is constant – the graph is a straight line with a positive gradient. The area under the graph gives the total distance travelled. Time / s Velocity / m s-1 From A to B, the ball is moving upwards. Its velocity is decreasing. Since the velocity is upward directed, it is given a positive sign. At B, the velocity of the ball is 0 m s-1. The ball momentarily stops. From B to C, the ball is moving downwards. Its velocity is + increasing. Since the velocity is downward directed, it is given a negative sign. The force of gravity is the force acting on the ball whilst it is in motion. So, the ball moves with a constant acceleration – the time /s acceleration due to gravity. This acceleration is downward directed. The graph is therefore a straight line with a negative gradient. - The area under the graph gives the displacement of the ball. Acceleration / m s-2 Since air resistance is assumed to be negligible, the only force acting on the ball whilst it is in motion in air is the force of gravity. time /s Thus, the acceleration of the ball is constant and is equal in magnitude to 9.81 m s-2. The acceleration due to gravity is a downward directed vector. - Therefore, it is negative. 31 Motion Graphs for one bounce (Ignore air resistance and assume that on collision no kinetic energy is lost): Upward directed vectors are taken as positive. Downward directed vectors are taken as negative. Note: AB is the downward path B is the point where the ball hits ground. BC is the return upward path. Distance / m As the ball moves downwards, the force of gravity acts in the same direction as the motion. The ball’s speed increases. As the ball moves upwards, the force of gravity acts against motion. The ball’s speed decreases. From A to B, the slope increases gradually – the speed of the ball increases. (The slope of a distance-time graph gives the speed.) At B, the ball hits the ground. From B to C, the slope decreases gradually – the speed of the ball decreases. Time / s The ball is above the ground at all times. Therefore, the Displacement displacement is taken as positive, as it is upward directed. from ground / m From A to B, the slope is negative and increases gradually. The velocity of the ball is downward directed (ball is moving downwards) and is increasing. (The slope of a displacement - time graph gives the velocity.) At B, the ball hits the ground. From B to C, the slope is positive, and it decreases gradually. Thus, the velocity of the ball is upward directed (ball is moving upwards) and is decreasing. Time / s 32 speed / m s-1 From A to B, the ball is moving downwards and accelerating as its speed is increasing. The acceleration is constant – the graph is a straight line with a positive gradient. At B, the ball hits the ground. From B to C, the ball is moving upwards and decelerating as its speed is decreasing. The deceleration is constant; thus the graph is a straight line with a negative gradient. The area under the graph gives the total distance travelled. Time / s Velocity / m s-1 From A to B, the ball is moving downwards. Its velocity is increasing. Since the velocity is downward directed, it is given a negative sign. At B, the ball hits the ground and it changes direction. The sign of the velocity has to change. + From B to C, the ball is moving upwards. Its velocity is decreasing. Since the velocity is upward directed, it is given a positive sign. The force of gravity is the force acting on the ball whilst it is in motion. So, the ball moves with a constant acceleration – the time /s acceleration due to gravity. This acceleration is downward directed. The graph is therefore a straight line with a negative gradient. - The area under the graph gives the displacement of the ball. Acceleration / m s-2 Since air resistance is assumed to be negligible, the only force acting on the ball whilst it is in motion in air is the force of gravity. time /s Thus, the acceleration of the ball is constant and is equal in magnitude to 9.81 m s-2. The acceleration due to gravity is a downward directed vector. - Therefore, it is negative. 33 Ø Equations of Motion There are 4 equations that can be used whenever an object moves with constant uniform acceleration in a straight line. The equations are written in terms of: s = displacement (m) u = initial velocity (m s-1) v = final velocity (m s-1) a = constant acceleration (m s-2) t = time (s) DERIVATION OF THE EQUATIONS OF MOTION ALTERNATIVE DERIVATION (i) The following is a v – t graph for constant acceleration. By definition: v/m s-1 acceleration is the rate of change of velocity. 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ∆𝑣 v 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = = 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 ∆𝑡 u 𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 Writing this in symbols: t t/s 𝑣−𝑢 The gradient of a v – t graph gives the 𝑎= constant acceleration. 𝑡 ∆" ∆' '$) So, at = v – u Gradient = ∆# = ∆( = ( 𝑣−𝑢 ∴𝑎= 𝑡 v = u + at … … … … … equation 1 So, at = v – u v = u + at 34 (ii) ALTERNATIVE DERIVATIONS By definition: velocity is the rate of change of displacement, s. The following is a v – t graph for constant acceleration. !'"()*+,-,.# " 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = #'-, #*0,. = # v/m s-1 If in time t, an object accelerates uniformly from velocity v u to velocity v, then 12$ 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 3 u " 12$ \ # = 3 t t/s 4 s = 3 (𝑢 + 𝑣) 𝑡 … … … … … equation 2 The area under a v – t graph gives the displacement, s Area = Area of rectangle + Area of triangle * \ s = 𝑢𝑡 + + (𝑣 − 𝑢) 𝑡 '( )( = 𝑢𝑡 + − + + )( '( = + + + * s = + (𝑢 + v)t or Area = Area of trapezium * s = + (𝑢 + v)t (iii) Equation 1: v = u +at The area under a v – t graph gives the 4 Equation 2: s = 3 (𝑢 + 𝑣) 𝑡 displacement, s Area = Area of rectangle Using equation 1 to replace v in equation 2: + Area of triangle * 4 \ s = 𝑢𝑡 + + (𝑣 − 𝑢) 𝑡 s = 3 (𝑢 + 𝑢 + 𝑎𝑡) 𝑡 4 But from equation 1 s = 3 (2𝑢 + 𝑎𝑡) 𝑡 v – u = at 4 * s = 𝑢𝑡 + 𝑎𝑡 3 … … … … … equation 3 \ s = 𝑢𝑡 + + (𝑎𝑡) 𝑡 3 ! s = 𝑢𝑡 + " 𝑎𝑡 " 35 (iv) Equation 1: v = u +at 4 Equation 2: s = 3 (𝑢 + 𝑣) 𝑡 $51 From equation 1: 𝑡 = * Using this to replace t in equation 2: (12$) $51 s= 3 N * O 2as = (u + v)(v – u) 2as = v2 – u2 v2 = u2 + 2as … … … … … equation 4 Note: The SUVAT equations can only be used if the acceleration is constant. Each equation contains only 4 out of 5 variables. So, if any 3 of the variables are known, the equations can be used to find the missing ones. Stopping Distances: The total stopping distance of vehicles which are in motion is made up of two parts: § The thinking distance – the distance travelled before the driver reacts and starts to brake. § The braking distance – the distance travelled while decelerating to a stop once the brakes are pressed. In practice, the thinking and braking distance depend on a number of factors such as the attention of the driver, the road surface, the weather and the vehicle’s condition. 36 Ø Motion under the force of gravity If air resistance is negligible, all bodies falling freely under the force of gravity, do so with a uniform acceleration. This acceleration is called the acceleration due to gravity, g. Its value varies over the Earth. However, generally the acceleration due to gravity on Earth is treated as a constant. On Earth, the acceleration due to gravity is considered to have a value of 9.81 m s-2. It is a downward directed vector. Ø Experiment: To measure the acceleration due to gravity, g A value for the acceleration due to gravity, g can be calculated by timing the fall of a small iron ball through a known distance. Note: A small iron ball is used as very little air resistance acts on it. An electronic timer is used to eliminate reaction time and get accurate measurements. Method: § The apparatus is set up as shown in the diagram with the trapdoor placed exactly underneath the electromagnet. § The electromagnet is switched on by closing the circuit at A. The electronic timer is reset. § The iron ball is placed close to the electromagnet so as to be attracted to it. § The distance s between the bottom of the iron ball and the trapdoor is measured by means of a metre ruler. 37  § The switch is quickly moved to B; simultaneously the electromagnet is switched off, the iron ball is released, and the electronic timer is switched on. § When the iron ball hits the trapdoor, the circuit is broken at C and the electronic timer stops. § The time of fall t is noted from the electronic timer. § The experiment is repeated three times in order to minimize random errors. § All the steps are repeated for a range of different heights. § Results are tabulated. Table: s/m t1 / s t2 / s t3 / s t̅ / s t2 / s2 ± 0.001 m Note: The average value of the time of fall, 𝑡̅ can be found using the following equation: 𝑡4 + 𝑡3 + 𝑡8 𝑡̅ = 3 The value of t2 is obtained by squaring the value of 𝑡.̅ Calculations: s is the distance the iron ball falls = s in metres t is the average time taken for the iron ball to fall through distance s = t in seconds a is the acceleration due to gravity = g in m s-2 u is the initial velocity of the iron ball = 0 m s-1 Using the equation of motion: s = ut + ½ at2 s = 0 + ½ gt2 s = ½ gt2 Compare: s = ½ gt2 with the equation for a straight line: y = mx + c s = ½g t2 + 0 y = m x + c 38 If s is plotted on the y – axis and t2 on the x – axis, a straight line graph which passes through the origin is obtained. The gradient of the graph would be equal to ½ g. The acceleration due to gravity g can thus be found: ∆𝑦 𝑦4 − 𝑦3 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = = ∆𝑥 𝑥4 − 𝑥3 9 gradient = 3 acceleration due to gravity, g = 2 x gradient s/m 9 gradient = 3 t2 / s2 Precautions ¨ A plumbline is used to make sure that the trapdoor is placed directly below the electromagnet and that the metre ruler is held in a vertical direction. ¨ Readings from the ruler are taken at eye level to avoid parallax errors. ¨ Repeated readings are taken to minimize random errors. ¨ The electromagnet is switched on for a short time in order to avoid residual magnetism in the iron core. 39 Ø Projectiles A projectile is an object which moves under the force of gravity only. Examples of projectiles: § An object dropped from rest (air resistance is negligible) § An object that is thrown vertically upward (air resistance is negligible) § An object which is thrown horizontally from the top of a cliff (air resistance is negligible) § An object which is thrown upward at an angle to the horizontal (air resistance is negligible). A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity. Free- Body Diagram of a Projectile: Force of gravity The path the projectile follows is called the trajectory. 40 Types of projection: (A) Vertical Projection The object is dropped or thrown in the vertical direction. The angle of projection is 90o. The range is 0 m. The angle of projection is the angle between the initial direction of travel and the horizontal. The range is the horizontal displacement of the object from the point of projection. (i) Ball dropped from rest: The ball is dropped, so its initial Acceleration due to gravity velocity is 0 m s-1. − 9.81 𝑚 𝑠 $+ g is a downward directed vector, so it is given a negative sign. If air resistance is negligible, the only force acting on the ball is the force of gravity. The force of gravity is in the same direction as the motion of the ball. The ball’s velocity increases. The ball accelerates downwards at a rate of 9.81 m s-1 every second. The ball reaches its maximum velocity v just before it hits the ground. Taking the downward direction as negative: u 0 m s-1 a - 9.81 m s-2. It is a downward directed vector. It is considered negative. s a downward directed vector. It is considered negative. t v a downward directed vector. It is considered negative. Any unknown values can be found by the SUVAT equations. 41 (ii) Ball is thrown vertically up: At maximum height the ball’s Maximum height is reached. velocity is 0 m s-1 but it is still The velocity of the ball is 0 m s- accelerating downwards. Its 1. The ball stops momentarily. velocity starts to increase straightaway. If air resistance is negligible, the If air resistance is negligible, the only force acting on the ball is only force acting on the ball is the force of gravity. The force of the force of gravity. The force of gravity acts opposite to the gravity is in the same direction Acceleration direction of the motion of the as the motion of the ball. The due to gravity ball. The ball’s velocity ball’s velocity increases. The ball decreases. The ball decelerates accelerates downwards at a rate − 9.81 𝑚 𝑠 $+ upwards at a rate of 9.81 m s-1 of 9.81 m s-1 every second. g is a every second. downward directed vector, so it is given a negative sign. The ball is thrown upwards with The ball reaches its maximum an initial velocity u. velocity v just before it hits the ground. Upward motion Downward motion Considering upward direction as positive and Considering upward direction as positive and downward direction as negative: downward direction as negative: u An upward directed vector. u 0 m s-1 Considered positive. a - 9.81 m s-2. a - 9.81 m s-2. A downward directed vector. A downward directed vector. s A downward directed vector. s Height. Considered negative. An upward directed vector. t Considered positive. v A downward directed vector. t Considered negative. v 0 m s-1 Any unknown values can be found by the Any unknown values can be found by the SUVAT equations. SUVAT equations. 42 (B) Horizontal Projection The projectile is launched with an initial horizontal velocity uH from an elevated position. Provided that air resistance is negligible, the projectile follows a parabolic path to the ground. Free Body Diagram The angle of projection is 0o. The range is R m. uH Force of gravity Parabolic Path Height h Range R The horizontal velocity of the projectile does not affect its vertical motion. This is shown in the multiflash photograph. The photograph shows two balls which were simultaneously launched. One ball was dropped from rest while the other was projected horizontally. Their vertical accelerations (due to gravity) are equal. This indicates that a projectile falls like an object which is dropped from rest. The horizontal and vertical motions of a projectile are independent. They do not influence each other. Thus, they can be treated separately in calculations. 43  uH Acceleration uV = 0ms -1 due to gravity − 9.81 𝑚 𝑠 $+ uH vV uH Range R q v vV Note: Upward directed vectors are considered to be positive. Downward directed vectors are considered to be negative. Vertical Horizontal The velocity of projection is in a horizontal The velocity of projection, uH is in the direction. There is no component in the horizontal direction. Therefore, the initial vertical direction. So, the initial velocity in velocity in the horizontal direction is uH the vertical direction is 0 m s-1. uV = 0 m s-1 uH Assuming that air resistance is negligible, the There is no force in the horizontal direction. only force acting on the object is the force of So, the horizontal velocity does not change. gravity. This force acts in the vertical Horizontally, the projectile moves with a direction and is in the same direction of constant velocity. motion. Therefore, the vertical velocity vv 44 increases. The maximum vertical velocity vv is reached just before the object reaches the ground. vv is a downward directed vector, so it is considered negative. vv Considered negative vH = u H The force of gravity acts in the same direction Since in the horizontal direction there is no of motion, so vertically the object force acting on the object, the acceleration of accelerates. This acceleration is the the object in the horizontal direction is 0 m s- 2 acceleration due to gravity, g.. Horizontally the object moves with g is a downward directed vector; it is constant velocity. considered negative. av = - 9.81 m s-2 aH = 0 m s-2 The projectile falls through a height, h. The The projectile covers a horizontal range, R. displacement is downwards, so it is considered negative sH = R sv = h Considered negative The time taken to fall through height h is The time taken to cross the horizontal range known as the time of flight. R. This is the time of flight. t t To find any unknown values in the vertical To find any unknown values in the horizontal direction use the SUVAT equations. direction use the equation: 𝒔 𝒗= 𝒕 The velocity v of the object at any point on the trajectory is made up of the horizontal component uH and the vertical component vv. The components, uH and vV are to be found 45 separately. They can then be added using the parallelogram law or the triangle law of vector addition. The magnitude of the velocity v can be found using Pythagoras Theorem: 𝑣 = U𝑢< 3 + 𝑣= 3 The direction can be found using trigonometry: 𝑣= 𝑇𝑎𝑛 𝜃 = 𝑢< IMPORTANT: When solving questions, all vectors (displacement, velocity and acceleration) acting along the horizontal are to be kept apart from those acting along the vertical. References: Ordinary Level Physics - Abbott, Heinemann Educational Books, London The World of Physics - John Avison, Thomas Nelson and Sons, Surrey GCSE Physics (4th Edition)- Tom Duncan, Heather Kenneth, John Murray. Advanced Physics - Muncaster, R., Stanley Thomas (Publishers) Ltd. Understanding Physics for Advanced Level- Breithaupt, J., Nelson Thornes. Advanced Physics - Adams, S. & Allday, J., Oxford Press. Advanced Physics for You – Keith Johnson, Simmone Hewett, Sue Holt, John Miller, Oxford University Press. Intermediate Physics 16 – 18 (5th Edition) - Martin Peter Farrell, Agenda Publishers. http://www.s-cool.co.uk/alevel/physics.html https://stepbystepscience.com https://www.physicshigh.com https://www.fizzics.org 46