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- 1. Y12 Physics Course 2010 Monday, 24 May 2010
- 2. Monday, 24 May 2010
- 3. Monday, 24 May 2010
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- 7. REVISION 1. Define a force as a push or a pull and be able to describe loosely the effect on motion that a force has. 2. Know that there are two different types of forces and recall examples of these. 3. Use v=d/t to calculate speed, distance and time. (Rearranging simple equations) 4. Use a = ∆v/∆t to calculate acceleration, speed and time. 5. Describe the units of both speed and acceleration (ms-1) and (ms-2) 6. Make labelled diagrams showing all the forces acting on an object in motion or an object at rest. 7. Recognise that the change in the velocity of an object is a result of the action of an unbalanced force and be able to calculate the size of the unbalanced force in simple 1D situations. 8. Use F=ma to calculate force, mass and acceleration. (Rearranging simple equations) 9. Perform basic unit conversions including the conversion from kmh-1 to ms-1 (Use of prefixes) 10. Rearrange equations that form part of the course Monday, 24 May 2010
- 8. WHAT IS A FORCE? • A force is a push or a pull • You cannot see a force but you can sometimes see the effects of a force (what a force does). • Forces can be either contact or non-contact in nature. 1. Examples of contact forces: Tension (lifting a bucket of water using a rope) Thrust (pushing on a desk) Lift (Air pressure differences holding a plane up in the air) Support (standing on the ground instead of falling to the earth’s centre) Friction (a force that opposes motion) 2. Examples on non-contact forces: Gravity, Magnetism, Electrostatic forces A force can: 1. Cause movement in an object that is initially stationery 2. Change the speed of an object (speed it up or slow it down) 3. change the direction of an object. 4. change the shape of an object. 5. hold an object up (or lift an object) Monday, 24 May 2010
- 9. BALANCED AND UNBALANCED FORCES BALANCED FORCES Back Forward Support force Friction force Driving force speed stays the same v OR speed equals ZERO (STANDING STILL) Force due to gravity Draw diagrams showing all the forces acting in the following situations: 1.An aeroplane travels through the air in level flight at a constant speed. 2.A rock falls vertically with constant speed Monday, 24 May 2010
- 10. UNBALANCED FORCES Back Forward The object will be speed up or it will slow down or change direction Support force Friction force Driving force speed increases Force due to gravity Support force Friction force Driving force speed decreases Force due to gravity Monday, 24 May 2010
- 11. Changing direction Car B is on a “collision course” car A’s path after the crash with car A Car B Car A Monday, 24 May 2010
- 12. CALCULATING AVERAGE SPEED distance travelled Average speed = time taken This formula allows you to calculate the average speed when the distance and time are known. It can be written using symbols: v= d where v = average speed (in metres per second, ms-1) t d = distance travelled (in metres, m) t = time taken (in seconds, s) Use this formula to calculate time when distance and speed are known t= d v and this formula to calculate time when distance and speed are known d= v x t “Here’s an easy way of remembering the formulae” d Just put your ﬁnger over the quantity you want to calculate and the v t formula appears Monday, 24 May 2010
- 13. CALCULATING ACCELERATION change in velocity Acceleration = change in time This formula allows you to calculate the acceleration when the change in velocity and the change in time are known. It can be written using symbols: a = ∆v where a = acceleration (in metres per second per second, ms-2) ∆t ∆v = change in speed (in metres per second,ms-1) ∆t = change in time (in seconds, s) Use this formula to calculate time when change in speed and acceleration are known ∆t = ∆v a and this formula to calculate time when acceleration and change in speed are known ∆v = a x ∆t ∆v “Here’s an easy way of remembering the formulae” a ∆t Just put your ﬁnger over the quantity you want to calculate and the formula appears Monday, 24 May 2010
- 14. FORCE AND ACCELERATION When an object experiences an unbalanced force it will speed up (acceleration), slow down (negative acceleration) or change direction (acceleration because the velocity is changing). SO AN UNBALANCED FORCE CAUSES ACCELERATION 20 N 1500 N Parachutist Parachutist falling slowly falling rapidly at at first the time that the parachute opens 1000 N 1000 N The unbalanced force on the The unbalanced force on the parachutist is 980 N acting parachutist is 500 N acting downwards upwards so he will speed up so he will slow down Monday, 24 May 2010
- 15. Summary FORCE, MASS & ACCELERATION When the forces acting on an object are unbalanced the object will accelerate: F = ma Where F = Force measured in Newtons, N m = mass measured in kilograms, kg a = acceleration measured in metres per second squared, ms-2 Use this formula to calculate acceleration when Force and mass are known a= F m Use this formula to calculate mass when Force and acceleration are known m= F a F Example m a A skyrocket of mass 0.1kg accelerates upwards at 10 ms-2. What is the unbalanced force acting on the rocket? F = ma = 0.1 x 10 = 1 N Monday, 24 May 2010
- 16. UNITS AND CONVERSIONS SI Units Standard units in Physics originate from Europe. These are called SI units (which stands for Systeme Internationale) In Mechanics these are: metres, m kilograms, kg ‘mks’ system of measurement seconds, s These units can be prefixed to show multiples of 10. A prefix shows what the standard unit is multiplied by. For example: kilo ‘k’ means ‘x 103’ (or x 1000) Other Prefixes 3 so 1 km = 1 x 10 m = 1000 m M mega 106 & 20.5 km = 20.5 x 103 m = 20500 m k kilo 103 m milli 10-3 centi ‘c’ means ‘x 10-2’ (or x 1 ) 100 µ micro 10-6 so 1 cm = 1 x 10-2 m = 0.01 m n nano 10-9 & 15.3 cm = 15.3 x 10-2 m = 0.153 m p pico 10-12 Monday, 24 May 2010
- 17. x 103 x 102 x 10 km m cm mm ÷ 103 ÷ 102 ÷ 10 x 100,000 km cm ÷ 100,000 x 1,000,000 km mm ÷ 1,000,000 “Leave it to me. I can cover the distance dude! x 1000 m mm ÷ 1000 Monday, 24 May 2010
- 18. x 1000 x 1000 x 1000 tonne kg g mg ÷ 1000 ÷ 1000 ÷ 1000 x 1,000,000 tonne g ÷ 1,000,000 x 1,000,000,000 tonne mg ÷ 1,000,000,000 x 1000 L mL ÷ 1000 Monday, 24 May 2010
- 19. Examples REARRANGING EQUATIONS vf = vi + at Make vi the subject of the equation d = (vi + vf)t Make t the subject of the equation 2 d = vit + 1at2 Make t the subject of the equation 2 Monday, 24 May 2010
- 20. ACCELERATED MOTION constant acceleration in a straight line; free fall under gravity, projectile motion; Relative motion, change in velocity, velocity vector components, circular motion (constant speed with one force only providing centripetal force). 1. Describe the difference between distance and displacement 2. Describe the difference between speed and velocity and calculation speed and velocity from distance and displacement 3. Calculate the acceleration of moving objects when information relating to speed and time are known. 4. Construct a velocity - time graph from ticker tape 5. Determine distance and acceleration from a velocity - time graph 6. Find the velocity from a distance - time graph and make general interpretations of the nature of the velocity from distance - time sketches 7. Remember these equations for uniformly accelerated motion, and use them to solve problems: d = (vf + vi) t vf = vi + at , 2 vf vi2 + 2ad d = vi t + 1 at 2 = 2 2 Monday, 24 May 2010
- 21. DISTANCE AND DISPLACEMENT Distance is the total path length from the starting point to the end point of a journey. Displacement means the “shift in an object’s position”. When we state the displacement of an object we are describing its distance from the starting point and the direction of the shift from the starting point. We describe the object’s position. (In other words “displacement is distance with direction”) Distance, d is called a scalar quantity A scalar is a physical quantity which has size only (Examples of scalar quantities: Mass, Temperature, Volume) Displacement, d is a vector quantity. ~ A vector is a physical quantity which has both size and direction (Examples of vector quantities: Velocity, Acceleration and Force) “~” shows a vector quantity Example 1 An athlete runs 400 m around an athletics track. He starts and finishes in the same place. The distance travelled is 400 m (this is the total path length). The displacement is 0m (because the runner has not shifted from the starting point) Monday, 24 May 2010
- 22. Example 2 A boat travels from the Whakatane ramp and out to White Island which is 15 km away. White Island is 5o East of North from Whakatane. The distance travelled = 15 km The displacement from the start = 15 km @ 5o E of N We use arrows to show displacement: White Island A bearing is an angle measured clockwise from North 5o 15 km 5o written as bearing would be 005o Whakatane Example 3 Consider an object which has moved from point A to point B by first travelling 3 km North and then travelling 4 km East: For the journey: distance travelled, d = 3 + 4 = 7 km displacement from the start, d = 5 km @ 530 E of N ~ Monday, 24 May 2010
- 23. Example 2 A boat travels from the Whakatane ramp and out to White Island which is 15 km away. White Island is 5o East of North from Whakatane. The distance travelled = 15 km The displacement from the start = 15 km @ 5o E of N We use arrows to show displacement: White Island A bearing is an angle measured clockwise from North 5o 15 km 5o written as bearing would be 005o Whakatane Example 3 Consider an object which has moved from point A to point B by first travelling 3 km North and then travelling 4 km East: For the journey: distance travelled, d = 3 + 4 = 7 km displacement from the start, d = 5 km @ 530 E of N A ~ Monday, 24 May 2010
- 24. Example 2 A boat travels from the Whakatane ramp and out to White Island which is 15 km away. White Island is 5o East of North from Whakatane. The distance travelled = 15 km The displacement from the start = 15 km @ 5o E of N We use arrows to show displacement: White Island A bearing is an angle measured clockwise from North 5o 15 km 5o written as bearing would be 005o Whakatane Example 3 Consider an object which has moved from point A to point B by first travelling 3 km North and then travelling 4 km East: For the journey: distance travelled, d = 3 + 4 3 km = 7 km displacement from the start, d = 5 km @ 530 E of N A ~ Monday, 24 May 2010
- 25. Example 2 A boat travels from the Whakatane ramp and out to White Island which is 15 km away. White Island is 5o East of North from Whakatane. The distance travelled = 15 km The displacement from the start = 15 km @ 5o E of N We use arrows to show displacement: White Island A bearing is an angle measured clockwise from North 5o 15 km 5o written as bearing would be 005o Whakatane Example 3 Consider an object which has moved from point A to point B by first travelling 3 km North and then travelling 4 km East: 4 km For the journey: distance travelled, d = 3 + 4 3 km = 7 km displacement from the start, d = 5 km @ 530 E of N A ~ Monday, 24 May 2010
- 26. Example 2 A boat travels from the Whakatane ramp and out to White Island which is 15 km away. White Island is 5o East of North from Whakatane. The distance travelled = 15 km The displacement from the start = 15 km @ 5o E of N We use arrows to show displacement: White Island A bearing is an angle measured clockwise from North 5o 15 km 5o written as bearing would be 005o Whakatane Example 3 Consider an object which has moved from point A to point B by first travelling 3 km North and then travelling 4 km East: 4 km B For the journey: distance travelled, d = 3 + 4 3 km = 7 km displacement from the start, d = 5 km @ 530 E of N A ~ Monday, 24 May 2010
- 27. Example 2 A boat travels from the Whakatane ramp and out to White Island which is 15 km away. White Island is 5o East of North from Whakatane. The distance travelled = 15 km The displacement from the start = 15 km @ 5o E of N We use arrows to show displacement: White Island A bearing is an angle measured clockwise from North 5o 15 km 5o written as bearing would be 005o Whakatane Example 3 Consider an object which has moved from point A to point B by first travelling 3 km North and then travelling 4 km East: 4 km B For the journey: distance travelled, d = 3 + 4 3 km = 7 km 53o 5 km displacement from the start, d = 5 km @ 530 E of N A ~ Monday, 24 May 2010
- 28. Example 4 A toy train travels 0.4 m North, 0.6 m East and 1.0 m South. The time for this journey is 5 s. (a) What is the total distance covered by the train? _________________________________________________________________ (b) Use the grid below to determine the displacement of the train over this journey? N W E on grid = 0.1 m displacement = _________________________ Monday, 24 May 2010
- 29. SPEED AND VELOCITY Speed is a scalar quantity. It is how fast the object is travelling. It is the rate of change of distance. Velocity is a vector quantity (it refers to the object’s speed and the direction in which the object moves) Speed = change in distance = ∆d Equation 1 change in time ∆t Velocity = change in displacement = ∆d ~ Equation 2 change in time ∆t ∆ = “change in” v = speed v = velocity ( both have the same unit: ms-1) ~ The equations always calculate the average speed or average velocity Velocity calculations - Process 1. Calculate the size of the change in displacement (using a diagram if you have to) 2. Apply equation 2 3. State your answer, including units and direction Note Speed or velocity can be instantaneous. This is the speed or velocity at an instant in time. Monday, 24 May 2010
- 30. Example 1 A girl walks in a straight line from the hair salon to her home. Her position at 30 s is 20 m North of the salon and her position at 60 s is 50 m North of the salon. Calculate the girl’s speed and her velocity during the section of the journey described. _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ Example 2 A toy train travels 0.3 m North, 0.7 m East and 0.9 m South. The time for this journey is 5 s. (a) Calculate the speed of the train for the entire journey _________________________________________________________________ (b) Calculate the velocity of the train for the entire journey _________________________ N _________________________ _________________________ W E _________________________ velocity = ________________ on grid = 0.1 m Monday, 24 May 2010
- 31. ACCELERATION Acceleration is how fast the velocity changes. It is a vector quantity because velocity is a vector quantity. Acceleration = change in velocity = ∆v ~ change in time ∆t Change in velocity = final velocity - initial velocity ∆v = vf - vi For now we will consider 2 situations: 1. An object travels in the forward direction only, in which case the velocities can both be considered to be positive 2. An object can travel forwards or backwards along a straight line, in which case the forward direction can be considered to be positive and the backward direction can be considered to be negative. Note If the final velocity is smaller than the initial velocity then the object has slowed down. The value of acceleration will be negative. A negative value of acceleration is called deceleration. Monday, 24 May 2010
- 32. Acceleration calculations - Process 1. Show the positive and negative direction as part of your working. 2. Calculate the change in velocity, using positive and negative values in your calculation. 3. Determine the change in time EXAMPLES 4. Use the formula to calculate the acceleration (a) A motorcyclist speeds up from 2 ms-1 to 22 ms-1 in a time of 10 s. Calculate his acceleration. _________________________________________________________________ _________________________________________________________________ _________________________________________________________________ _________________________________________________________________ _________________________________________________________________ (b) The same motorcyclist travels at a constant speed of 22 ms-1 for a few seconds and then slows down for a set of traffic lights which have turned red. He takes 11 s to stop. Calculate his acceleration. _________________________________________________________________ _________________________________________________________________ _________________________________________________________________ _________________________________________________________________ _________________________________________________________________ Ex.7A Q.1 to 8 Monday, 24 May 2010
- 33. DISPLACEMENT - TIME GRAPHS The gradient of a displacement - time graph gives the velocity of the object. This principle can be used to label the following Steady speed Stopped graphs from the list of labels given (right) Acceleration Deceleration d d d d t t t t Example Calculate the velocity from the following distance - time graphs. 1 d Gradient = Rise = 10 = 2 ms-1 10 Run 5 Rise So velocity is 2 ms-1 The gradient of a graph is a measure of how 0 t 5 steep the graph is. Run Monday, 24 May 2010
- 34. 2 d ____________________________________________ 10 ____________________________________________ ____________________________________________ ____________________________________________ 0 2 t The velocity at any given time is given by the 3 A non-linear graph: gradient of the tangent at that point on the d graph ____________________________________________ 10 ____________________________________________ 5 ____________________________________________ 0 2 4 6 t ____________________________________________ Monday, 24 May 2010
- 35. VELOCITY - TIME GRAPHS The area under a velocity - time graph gives the distance travelled Consider a ball rolling on a flat surface (in the absence of friction). It has a constant speed of 4 ms-1. Calculate the distance that the ball has travelled in that 5 s. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ The graph of the ball’s motion for the first 5 s is shown in the speed-time graph below: v (ms-1) 5 A calculation of the area under the graph 4 gives a distance covered of 20 m which 3 is the same as value achieved from 2 A rearranging the equation: v = d 1 t 0 1 2 3 4 5 t (s) When speed is constant the distance travelled can be calculated using: d = v.t This works in this situation because there is a single value for speed which is the same for the entire time, t that the object is moving. Monday, 24 May 2010
- 36. If the ball were to roll down a sloping surface (in the absence of friction) then its speed would increase. The speed would increase at a constant rate. In other words the acceleration would be constant. We would expect this because gravity is causing this increase in speed and because this force is constant then the acceleration would be constant. (Force is linked to acceleration according to the equation F = ma) The graph of the ball’s motion for the first 5 s would be a straight line sloping upwards and might look like this: v (ms-1) A calculation of the area under the graph 5 gives the distance covered: 4 3 d = Area = 1 x 5 x 52 = 12.5 m 2 2 A 1 0 1 2 3 4 5 t (s) When speed is changing the distance travelled can not be calculated using: d = v.t This does not work in this situation because v is changing. Monday, 24 May 2010
- 37. The gradient of a velocity-time graph gives the object’s acceleration Again considering the ball rolling down the slope, we know that the acceleration is constant. Acceleration can be calculated using the following equation: Acceleration = change in velocity a= ∆v => change in time ∆t The graph again will look like this: v (ms-1) vf = 5 5 4 3 ∆ v = vf - vi 2 =5-0 The Rise A 1 vi = 0 =5 0 1 2 3 4 5 t (s) The Run ∆ t = tf - ti = 5 - 0 = 5 a = ∆v = 5 = 1 ms-2 in other words a = rise = gradient of the graph ∆t 5 run Monday, 24 May 2010
- 38. SUMMARY 1. The gradient of a displacement - time graph of an object’s motion gives the velocity. 2. The gradient of a velocity - time graph gives acceleration. 3. The area under an acceleration - time graph gives velocity. 4. The area under a velocity - time graph gives distance. THESE RULES CAN BE USED TO CONSTRUCT ONE GRAPH WHEN THE OTHER IS GIVEN: gradient gradient d-t graph v-t graph a-t graph area area Ex.7B Q.1 to 3 Monday, 24 May 2010
- 39. Examples “Try them yourself first” 1. For the above graph, calculate the distance travelled: (i) between 0s and 4s ________________________________________________________________ (ii) between 5s and 10s ________________________________________________________________ 2. Calculate the acceleration of the object (i) in the first 4s ________________________________________________________________ (ii) between 4s and 7s ________________________________________________________________ (iii) between 7s and 10s ________________________________________________________________ Monday, 24 May 2010
- 40. 3. The velocity - time graph below shows the motion of a cyclist along a straight road. Complete the table (to the right of the graph) to show the acceleration of the cyclist in each time interval: Time interval (s) Acceleration (ms-2) 0 to 10 10 to 25 25 to 35 35 to 40 40 to 45 Calculate the distance travelled by the cyclist during the 45 s ride. ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ Calculate the displacement of the cyclist at 45 s. ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ Monday, 24 May 2010
- 41. CONSTANT SPEED, AVERAGE SPEED Constant speed, v For an object that travels at a constant speed, v: v Distance travelled = Area under the graph d = v.t t Average speed, vave v For an object that travels with constant acceleration, a: Distance travelled = Area under the graph This is the same area as the area under the graph vave of vave vs t (shown) d = vave.t t Monday, 24 May 2010
- 42. RECOGNISING CONSTANT ACCELERATION For an object that has constant acceleration the speed increases evenly. For example An object that is accelerating at 10 ms-2 will have a speed that increases 10 ms-1 every second: If it starts from rest with speed 0 ms-1 then after the first second the speed is 10 ms-1 and after the second the speed is 20 ms-2 and so on as shown in the table (below): time (s) 0 1 2 3 4 5 6 ... speed (ms-1) 0 10 20 30 40 50 60 ... Constant acceleration For an object that travels with constant acceleration, a: v Distance travelled = Area under the graph vf In this instance the area under the graph is the area of a trapezium: vi d = 1(vi + vf).t 2 t Monday, 24 May 2010
- 43. KINEMATIC EQUATIONS These equations apply to situations in which the acceleration is constant: d = displacement (i.e. distance from the start in the positive or negative direction) a = acceleration (positive or negative) Mostly vector quantities t = time vf = final velocity (positive or negative) vi = initial velocity (positive or negative) Monday, 24 May 2010
- 44. Process 1. Read the question carefully and underline the relevant information. 2. List the 5 symbols on your page in specific order and write in the known quantities in SI units: d= vi = vf = a= t= 3. Use a “?” to indicate the quantity that you want to calculate and if a quantity is not given then leave it blank. 4. Only three quantities are required out of the five to perform a calculation. 5. Select the equation that you will use based on which of the five quantities is not given. 6. Rearrange the equation to make the required quantity the subject. 7. Substitute the numerical values into the equation and evaluate. Examples 1. A stone is released from a height of 20 m above the ground. Neglecting air resistance and using the acceleration due to gravity as 9.81 ms-2, find the velocity with which the stone will hit the ground . Monday, 24 May 2010
- 45. “Try them 2. Calculate the uniform acceleration of a sports car which: yourself first. (a) starts from rest and reaches a speed of 15 ms-1 in 10 s. Use pencil” (b) changes its speed from 20 ms-1 to 32 ms-1 in 4.0 s. (c) starts from rest and travels a distance of 98 m in 7.0 s. (d) slows down from a speed of 66 ms-1 and comes to rest in 12 s. Answers: (a) 1.5 ms-2 (b) 3 ms-2 (c) 4 ms-2 (d) -5.5 ms-2 Monday, 24 May 2010
- 46. 3. A ball rolls from rest down an incline with a uniform acceleration of 4.0 ms-2. (a) What is its speed after 8.0 s? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (b) How long will it take to reach a speed of 36 ms-1? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (c) How long does it take to travel a distance of 200 m, and what is its speed after that time? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (d) How far does it travel during the third second of its motion? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________ Answers: (a) 32 ms-1 (b) 9 s (c) 40 ms-1 (d) 10 m Monday, 24 May 2010
- 47. 4. A model train moving at 20 cms-1 increases its speed uniformly to 60 cms-1 in 5.0 s (a) What was the trains acceleration? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (b) What distance was travelled in that time? ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ 5. A car accelerating uniformly, reaches a speed of 30 ms-1 from rest in 3.0 min. Calculate its acceleration and the distance travelled in this time. Later it comes to rest in 20 s when the brakes are applied. Find the deceleration and the distance covered while it is slowing down. ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________ Answers: 4.(a) 8 cms-2 (b) 200 cm 5. 0.17 ms-2 & 2700 m ..LATER.. -1.5 ms-2 & 300m Ex.7C Q.1 to 5 Monday, 24 May 2010
- 48. 12 PHYSICS KINEMATICS MINI ASSIGNMENT Name ___________________ 1. A skateboarder starts a rail slide at 8.0 ms-1 and accelerates at 6.0 ms-2 to reach a speed of 10 ms-1 at the bottom of the rail. How long is the rail? 2. A car slows from 25 ms-1 to 15 ms-1 over a distance of 400 m. (a) What is the deceleration of the car? (b) Calculate the time taken for the car to slow. 3. A car is travelling at 12 ms-1. At t = 0 it accelerates at 1.5 ms-2 to a speed of 18 ms-1. (a) Calculate the time taken for the car to reach the final speed of 18 ms-1. (b) Calculate the distance travelled in this time. (c) What is the car’s speed once it has travelled 100 m? 4. A driver is travelling steadily at 12 ms-1 when he spots a dog 48 m ahead. After a reaction time of 0.75 s, he applies the brakes and stops. The brakes produce a steady deceleration of 2.0 ms-2. (a) Calculate how far he travels during his reaction time (with no braking). (b) Calculate how far he travels while he is braking. (c) What happens to the dog? Monday, 24 May 2010
- 49. 5. A plane drops a Red Cross package from a height of 1200 m. If the package had no parachute (and by this you can assume negligible air friction): (a) How fast will the package be travelling just before it hits the ground? (b) How many seconds will the package take to fall? 6. A 1 kg object is dropped from a tower 120 m high. (a) Calculate the time it will take for the object to fall to the ground. (b) Calculate the objects final speed on reaching the ground. (c) How long does it take to reach a speed of 35 ms-1 ? 7. A shell is fired straight up with an initial speed of 96 ms-1. (a) Calculate the time it will take for the object to fall to the ground. (b) When will the shell have an upwards speed of 48 ms-1 ? (c) Calculate the time for the shell to reach its maximum height. (d) Calculate the maximum height reached by the shell. (e) What is the shells acceleration at the top of its motion? Monday, 24 May 2010
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- 57. VECTORS & RELATIVE MOTION 1. Show vector direction as a bearing, compass direction or angle relative to a fixed axis. 2. Perform simple vector arithmetic (including addition, subtraction) 3. Resolve vectors into components 4. Define the term relative velocity 5. Use vectors to solve relative velocity problems Reading: Chapter 8 (p91 to 103) Monday, 24 May 2010
- 58. VECTOR DIRECTION Definition A vector is a physical quantity that has both size and direction. Examples Displacement, velocity, acceleration and force. A vector has a head and a tail head head tail tail A vector’s direction can be described in a number of different ways: (a)Bearings A bearing is an angle measure clockwise from the North. Eg. N N N 30o 45 o 30o Direction = 045o Direction = Direction = Monday, 24 May 2010
- 59. (b)Compass reference An angle is given from the N, S, E or W direction Eg. N N N 30o o 45 30o Direction = Direction = Direction = (c) Referenced from the vertical or horizontal An angle is given from a vertical or horizontal axis Eg. 20o 60o Direction = Direction = Monday, 24 May 2010
- 60. VECTOR ARITHMETIC Addition Two vectors are added head to tail to produce a resultant vector. The resultant vector is a single vector that has the same effect as the two vectors combined. Eg: Adding two vectors, a and b: head head b ~ a ~ tail tail a b ~+ ~ Ex.8A Q.1 to 8 Monday, 24 May 2010
- 61. Subtraction Vector subtraction is “addition of the opposite”. The opposite of a vector is a vector which has the same magnitude but is opposite in direction. Eg: Subtracting two vectors, a minus b: b ~ a ~ -b ~ a b a - b ~ - ~ = ~ + ~ Monday, 24 May 2010
- 62. VECTOR COMPONENTS Any vector can be drawn as the sum of two other vectors which are drawn at right angles to each other. These two vectors are called components. Examples - Resolving vectors into components: 1 Any vector can be expressed as the sum of two components (a) (b) (c) 2 Horizontal and vertical components are the most useful components 30 N 30 N Vertical component 25o 25o Horizontal component The vector is the sum of its horizontal and vertical components Monday, 24 May 2010
- 63. CALCULATING THE SIZE OF THE COMPONENTS 3 It is possible to calculate the component of a vector along any axis 30 N F ~ 25o f ~ This is the component of the vector, F along this axis ~ f = F.cos25o Example Calculate the size of the component of the following velocity vector along the axis shown: 50 ms-1 Exercises: “Vector arithmetic & components” Monday, 24 May 2010
- 64. CHANGE IN VELOCITY The change in velocity of a moving object is the final velocity minus the initial velocity: ∆v = vf - vi Example 1. A tennis ball falls to the ground, striking at right angles. It bounces off the ground and travels along the same path on the rebound as it travelled as it was falling. The initial velocity is 10 ms-1 vertically downwards. It rebounds with a final velocity of 10 ms-1 upwards. Calculate the change in velocity of the ball. 10 ms-1 10 ms-1 Monday, 24 May 2010
- 65. Example 2. A billiard ball strikes the cushion of a billiard table at an angle of 20o and rebounds at the same angle. Use a vector diagram to calculate the change in velocity. 8 2 ms-1 20o 20o 2 ms-1 Vector subtraction Calculation Ex.8A Q.9 to 14 Monday, 24 May 2010
- 66. Relative velocity is the velocity of an object in relation to another object. This other object can be stationery (like the ground) or moving. RELATIVE VELOCITY along a straight line Consider the example of a train travelling in a straight line along a track. A boy is standing still on the roof of a train which is travelling slowly in a straight line at a speed of 2 kmh-1. 2 kmh-1 A second boy standing on the ground holding a speed gun measures the boy’s velocity at 2 kmh-1. Monday, 24 May 2010
- 67. Relative velocity is the velocity of an object in relation to another object. This other object can be stationery (like the ground) or moving. RELATIVE VELOCITY along a straight line Consider the example of a train travelling in a straight line along a track. A boy is standing still on the roof of a train which is travelling slowly in a straight line at a speed of 2 kmh-1. 2k A second boy standing on the ground holding a speed gun measures the boy’s velocity at 2 kmh-1. Monday, 24 May 2010
- 68. Relative velocity is the velocity of an object in relation to another object. This other object can be stationery (like the ground) or moving. RELATIVE VELOCITY along a straight line Consider the example of a train travelling in a straight line along a track. A boy is standing still on the roof of a train which is travelling slowly in a straight line at a speed of 2 kmh-1. 2k A second boy standing on the ground holding a speed gun measures the boy’s velocity at 2 kmh-1. The boy’s velocity is 2 kmh-1 relative to the ground. Monday, 24 May 2010
- 69. A boy is standing still on the roof of a train which is travelling slowly in a straight line at a speed of 2 kmh-1. 2 kmh-1 A second boy standing on the roof of the train measures the boy’s velocity at 0 kmh-1. Monday, 24 May 2010
- 70. A boy is standing still on the roof of a train which is travelling slowly in a straight line at a speed of 2 kmh-1. 2 A second boy standing on the roof of the train measures the boy’s velocity at 0 kmh-1. Monday, 24 May 2010
- 71. A boy is standing still on the roof of a train which is travelling slowly in a straight line at a speed of 2 kmh-1. 2 A second boy standing on the roof of the train measures the boy’s velocity at 0 kmh-1. The boy’s velocity is 0 kmh-1 relative to the train. Monday, 24 May 2010
- 72. TWO VELOCITIES RELATIVE TO EACH OTHER Consider the example of a car (travelling at 20 kmh-1) overtaking a bus which is travelling considerably slower (at 12 kmh-1). Car 20 kmh-1 Bus 12 kmh-1 Monday, 24 May 2010
- 73. TWO VELOCITIES RELATIVE TO EACH OTHER Consider the example of a car (travelling at 20 kmh-1) overtaking a bus which is travelling considerably slower (at 12 kmh-1). Car 20 kmh-1 Bus 12 kmh-1 The velocity of the car relative to the bus is 8 kmh-1 in the forward direction because an observer in the bus sees the car moving forward (past the bus) at 8 kmh-1. Monday, 24 May 2010
- 74. TWO VELOCITIES RELATIVE TO EACH OTHER Consider the example of a car (travelling at 20 kmh-1) overtaking a bus which is travelling considerably slower (at 12 kmh-1). Car 20 kmh-1 Bus 12 kmh-1 The velocity of the car relative to the bus is 8 kmh-1 in the forward direction because an observer in the bus sees the car moving forward (past the bus) at 8 kmh-1. • The velocity of the car relative to the bus is the velocity of the car minus the velocity of the bus. We can write this as a simple equation: vcb = vc - vb Where c = car ~ ~ ~ b = bus Monday, 24 May 2010
- 75. TWO VELOCITIES RELATIVE TO EACH OTHER Consider the example of a car (travelling at 20 kmh-1) overtaking a bus which is travelling considerably slower (at 12 kmh-1). Car 20 kmh-1 Bus 12 kmh-1 The velocity of the car relative to the bus is 8 kmh-1 in the forward direction because an observer in the bus sees the car moving forward (past the bus) at 8 kmh-1. • The velocity of the car relative to the bus is the velocity of the car minus the velocity of the bus. We can write this as a simple equation: vcb = vc - vb Where c = car ~ ~ ~ b = bus The vector subtraction is shown below: Monday, 24 May 2010
- 76. TWO VELOCITIES RELATIVE TO EACH OTHER Consider the example of a car (travelling at 20 kmh-1) overtaking a bus which is travelling considerably slower (at 12 kmh-1). Car 20 kmh-1 Bus 12 kmh-1 The velocity of the car relative to the bus is 8 kmh-1 in the forward direction because an observer in the bus sees the car moving forward (past the bus) at 8 kmh-1. • The velocity of the car relative to the bus is the velocity of the car minus the velocity of the bus. We can write this as a simple equation: vcb = vc - vb Where c = car ~ ~ ~ b = bus The vector subtraction is shown below: 20 kmh-1 Monday, 24 May 2010
- 77. TWO VELOCITIES RELATIVE TO EACH OTHER Consider the example of a car (travelling at 20 kmh-1) overtaking a bus which is travelling considerably slower (at 12 kmh-1). Car 20 kmh-1 Bus 12 kmh-1 The velocity of the car relative to the bus is 8 kmh-1 in the forward direction because an observer in the bus sees the car moving forward (past the bus) at 8 kmh-1. • The velocity of the car relative to the bus is the velocity of the car minus the velocity of the bus. We can write this as a simple equation: vcb = vc - vb Where c = car ~ ~ ~ b = bus The vector subtraction is shown below: 12 kmh-1 20 kmh-1 Monday, 24 May 2010
- 78. TWO VELOCITIES RELATIVE TO EACH OTHER Consider the example of a car (travelling at 20 kmh-1) overtaking a bus which is travelling considerably slower (at 12 kmh-1). Car 20 kmh-1 Bus 12 kmh-1 The velocity of the car relative to the bus is 8 kmh-1 in the forward direction because an observer in the bus sees the car moving forward (past the bus) at 8 kmh-1. • The velocity of the car relative to the bus is the velocity of the car minus the velocity of the bus. We can write this as a simple equation: vcb = vc - vb Where c = car ~ ~ ~ b = bus The vector subtraction is shown below: 8 kmh-1 12 kmh-1 20 kmh-1 Monday, 24 May 2010
- 79. RELATIVE VELOCITY in 2D Consider the example of an aircraft carrier which is floating down a canal with a speed of 2 ms-1. An officer walks across the flat deck of the carrier at right angles to the sides of the vessel with a speed of 2 ms-1 as shown. N The bank N 2 ms-1 2 ms-1 Monday, 24 May 2010
- 80. RELATIVE VELOCITY in 2D Consider the example of an aircraft carrier which is floating down a canal with a speed of 2 ms-1. An officer walks across the flat deck of the carrier at right angles to the sides of the vessel with a speed of 2 ms-1 as shown. N The bank N 2 ms-1 2 ms-1 The officer’s velocity relative to the moving object (the boat) is 2 ms-1 North The velocity of the moving object is 2 ms-1 East The officer’s velocity is a combination of these two velocities. Monday, 24 May 2010
- 81. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: Monday, 24 May 2010
- 82. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object Monday, 24 May 2010
- 83. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 Monday, 24 May 2010
- 84. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 The velocity of the moving object Monday, 24 May 2010
- 85. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 The velocity of the moving object 2 ms-1 Monday, 24 May 2010
- 86. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 The velocity of the moving object 2 ms-1 2 ms-1 Monday, 24 May 2010
- 87. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 The velocity of the moving object 2 ms-1 2 ms-1 2 ms-1 Monday, 24 May 2010
- 88. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 The velocity of the moving object 2 ms-1 2 ms-1 2 ms-1 The officer’s velocity, v Monday, 24 May 2010
- 89. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 The velocity of the moving object 2 ms-1 2 ms-1 2 ms-1 v = 22 + 22 The officer’s velocity, v Monday, 24 May 2010
- 90. The officer’s direction is North - East. The officer’s velocity can be calculated by adding the two velocity vectors: The officer’s velocity relative to the moving object 2 ms-1 The velocity of the moving object 2 ms-1 2 ms-1 2 ms-1 v = 22 + 22 The officer’s velocity, v = 2.8 ms-1 NE Monday, 24 May 2010
- 91. Intuitively: The officer is moving at 2 ms-1 North at the same time as he is moving at 2 ms-1 East. It follows that the officer’s velocity (relative to the ground) is the sum of these 2 motions. In other words: “The velocity of the officer relative to the ground is equal to the velocity of the officer relative to the boat plus the velocity of the boat relative to the ground” We can write this as an equation: Where o = officer vog = vob + vb g b = boat ~ ~ ~ g = ground PROCESS FOR PROBLEM-SOLVING 1. Read the question carefully ---> Underline relevant information. 2. Assign a variable (a letter) to each object in the system. Eg. let g = the ground 3. Construct the vector symbol equation that relates the relative velocities to each other. 4. Rearrange the equation if required. 5. Draw the vector diagram from the equation. 6. Solve for the unknown quantity. Monday, 24 May 2010
- 92. EXAMPLES 1. A plane flies East with an air speed of 500 kmh-1 (this is the velocity of the plane relative to the air). There is a wind blowing in the opposite direction at a speed of 30 kmh-1. Calculate the velocity of the plane relative to the ground by answering the questions below: (a) Give letters to symbolise the air, plane and ground. (b) Write a vector equation that relates the 3 velocities. Below each vector symbol sketch the vector showing size and direction. (c) Perform the vector calculation to get your answer. Monday, 24 May 2010
- 93. 2. A plane flies at 60 ms-1 due North, relative to the ground. A 20 ms-1 Northwest wind is blowing (i.e. air coming from the Northwest). Find: (a) the airspeed of the plane ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (b) the direction that the pilot has to aim the plane to achieve his 20 ms-1 velocity relative to the ground. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ Monday, 24 May 2010
- 94. 3. A ferry crosses a 500 m wide river in 1.5 minutes. The ferry must travel across the river in a straight line from one ferry landing to the other as shown in the diagram (below). The ferry has a water speed of 10 ms-1. Find the speed of the current. Ferry landing Current 500 m Ferry Ferry landing __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ Ex.8B Q.1 to 6 Monday, 24 May 2010
- 95. 12 PHYSICS RELATIVE VELOCITY ASSIGNMENT Name 1. A man swims in a river which has a current of 2.5 ms-1. He finds that he can swim upstream, against the current, at 1.5 ms-1 relative to the river bank. bank river current bank (a) What is his velocity relative to the water? ________________________________________________________________ (b) He then swims downstream with the same effort. What is his velocity relative to the bank? ________________________________________________________________ ________________________________________________________________ (c) The man then swims (at 4 ms-1 relative to the water) so that he faces directly across the river. As he swims across, the current takes him downstream a little. Draw a vector velocity triangle to show this. Monday, 24 May 2010
- 96. 2. A small plane is flying West at 60 ms-1. A 20 ms-1 South wind springs up. (a) Sketch this situation (b) The pilot has to continue at 60 ms-1 Westwards despite the wind. Sketch a vector triangle which shows what the pilot should do. (c) Calculate the direction of the plane’s air speed (i.e. its speed relative to the air). ________________________________________________________________ ________________________________________________________________ (d) Calculate the magnitude of the plane’s air speed. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ Monday, 24 May 2010
- 97. 3. A man rows his boat North directly across a river at 2.0 ms-1. A current starts to flow East at 1.3 ms-1. bank (a) Sketch a vector triangle of the velocities. Label each vector. river current Boat bank (b) Write a vector equation for the velocities. ________________________________________________________________ (c) Calculate the magnitude of the boat’s velocity relative to the bank. ________________________________________________________________ ________________________________________________________________ (d) Calculate the direction of the boat’s velocity relative to the bank. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ Monday, 24 May 2010
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- 109. Exercises VECTOR ARITHMETIC & COMPONENTS Perform the following vector operations in the spaces provided. a b d ~ ~ c ~ ~ 1 a + b 2 c + d ~ ~ ~ ~ b + d 4 b + c 3 ~ ~ ~ ~ Monday, 24 May 2010
- 110. a b d ~ ~ c ~ ~ 5 a - b 6 c - a ~ ~ ~ ~ b - d 8 b - c 7 ~ ~ ~ ~ Monday, 24 May 2010
- 111. a b ~ ~ 60o c 15 N ~ 18 N 30o 45o 20 N Perform the following calculations of components: 9. Horizontal component of a 10 Horizontal component of b ~ ~ 11 Vertical component of b 12 Component of c along the dotted line ~ ~ Monday, 24 May 2010

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