1. L.O. Work out acceleration in a straight line using v i, v f, d and t. L.O.Explain the following terms (and use appropriate units): Distance, displacement, speed, velocity, average speed, average velocity & acceleration. 1.What is the symbol for distance? 2.What is the symbol for speed? 3.What is the symbol for time? 4.What is the unit for distance? 5.What is the unit for time? Do now Quietly, answer the following questions in your book d v t meters seconds L.O.Interconvert between a, v i and v f using a, d and t.

2. Physical Quantities QuantitySymbolUnit average velocityvms -1 initial velocityvivi ms -1 final velocityvfvf ms -1 distancedm timets accelerationams -2

3. Determining the velocity of an object. The average velocity of an object can be determined by measuring the distance traveled and timing how long it takes to travel the distance. Distance = 10 m STARTFINISH time = 2 s

4. Work out the average velocity of the following. 1.A car traveling 100 m in 4.0 s. 2.A cycle moving 20 meters in 8.0 seconds. 3.A mouse steadily accelerates from 5.0ms -1 to 10 ms -1. 4.The space shuttle moving 480 km in one minute. 5.A car traveling 300 km in 4.0 hours.

6. d = 480 km = 480 000 m t = 1 min = 60 s d = 300 km = 300 000 m t = 4 hr = 240 min = 14400 s

7. L.O. Use kinematic equations to solve problems involving constant acceleration in a straight line. (vertical motion) Acceleration Acceleration is the rate of change of velocity. In other words it is a measure of how quickly an object changes speed. Acceleration is always caused by an unbalanced force. The symbol for acceleration is a and the unit for it is ms -2 (Meters per second per second). An object that speeds up, slows down or changes direction has accelerated. When a object slows down we say it decelerates. Deceleration is negative acceleration. We use the following equation to work out the acceleration of an object.

8. Final velocity (v f ). The final velocity of an object is an instantaneous velocity and we can work it out by adding the change in the velocity to the initial velocity (v i ). Change in velocity as ∆v = at. We use this equation to work out the final velocity of an object that has undergone constant acceleration for a known time. e.g. What is the final velocity of a car that was traveling at 20 ms -1 and accelerated constantly at 4 ms -2 for 6 s?

9. Distance (d) or height. The equation on the right is used to work out the distance an object is travelling when it is accelerating at a constant rate. Often the equation is used to work out the height an object has fallen using the time it took to fall and the acceleration due to gravity (g = 9.8 ms -2 ) d = ? A ball is dropped from the top of the Eiffel tower it takes 8.13 s to hit the ground. What if the height of the tower?

10. As the ball is not moving before it is dropped its initial velocity is zero ms -1.

11. The equation on the right is used to work out the final velocity of an object if only the distance and acceleration is known. Often the equation is used to work the velocity of falling object just before they hit the ground, remember is the case of falling acceleration is always 9.8 ms -2. d = 324 m A ball is dropped from the top of the Eiffel tower. What is the balls velocity just before it hits the ground?

13. The distance an object travels can be worked out by multiplying it’s average velocity by the time Average velocity.

14. A curve indicates the object is accelerating Horizontal line indicates the object is stationary A straight line indicates constant speed L.O. Explain the motion taking place and obtain information from displacement/time and velocity/time graphs.

15. Change in time Change in distance Working out the Speed from a Distance/time graph

16. Object is moving at a constant velocity. Object is decelerating. Object is accelerating

17. Area of A = ½ base x height = ½ 6 x 6 = 18 m Area of B = base x height = 3 x 6 = 18 m Area of C = ½ base x height = ½ 2 x 6 = 6 m Total Area = 18m + 18 m + 6m = 42 m Distance travelled = 42 m The total area under a velocity/time graph is equal to the distance and object has travelled B C A

18. Do Now 1.Work out the acceleration for parts A and B on the graph. 2.Work out the total distance the skateboard travelled. Answers Distance = ½ 4x2+ +½4x4+ 6x2 =4+4+8+12 =24

20. Motion due to Gravity The gravity of the earth can have a huge effect on the motion of objects. All objects at sea level on the surface of the earth have a uniform gravitational field of 9.8 NKg -1 acting on them. That is a force, of 9.8 Newtons for every Kg of their mass, pulls them to the centre of the Earth. When objects are unsupported this force causes objects to accelerate at 9.8 ms -2 towards the centre of the Earth.

21. Projectile motion An object undergoes projectile motion when the only force acting on it is gravity. The force of gravity (g) causes all objects to always accelerate vertically towards the ground at 9.8 ms -2. There are no force acing horizontally on an object, so there is no acceleration and the object moves at a constant speed. 1.Describe the horizontal motion of this ball. 2.Describe the vertical motion of this ball. v i = 20 ms -1 θ = 50°

22. v h = 13 ms -1 v i = 20 ms -1 The motion of an object moving through the air we can simplified by separating it into two components: Vertical motion and horizontal motion. v v = 15 ms -1 v i = 20 ms -1

23. Manipulating formula d v t Distance is the subject of the formula Time is the subject of the formula Speed is the subject of the formula

24. Work out the distance travelled by the following. 1.A car travelling for 100 seconds at 20 meters per seconds. 2.A motor cycle moving for 200 seconds at 6 meters per seconds. 3.A cat running 50 seconds in 10 metres per. Work out the time taken to travel the following. 1.A car travelling 50 metres at 15 meters per seconds. 2.A boat moving 500 metres at 6 meters per seconds. 3.A dog running 20 metres in 4 metres per.

26. Speed Average speed is calculated for the complete journey. Instantaneous speed is the actual speed of any part of the journey. What does the speedometer in the car measure? Actual (instantaneous) speed.

27. Speed Average speed (V av ) is calculated by the formula: V average = distance travelled total time taken V av = d t

28. Units involved DistanceTimeSpeed Centimetre (cm)Second (s)cms -1 Metre (m)Second (s)ms -1 Kilometre (km)Hour (hr)Kmh -1

29. Example John travels from Dunedin to Christchurch. It is 325km and it takes him 4.5 hours. What is his average speed?

30. Answer the following questions. 1.What is the average speed of a car travelling 250 metres in 10 seconds? 2.How long does it take a boat to sale 500 metres at 12 meters per seconds? 3.How for will a snail moving at 0.05 ms -1 travel in 2 min? Draw distance/time or speed/time graphs from data Use the gradient (slope) of a distance/time graph to describe the speed of an object 250/10 = 25 ms 500/12 = 42.5 s 0.05 x 120 = 6 m

31. L.O. Define acceleration and state its unit and symbol Acceleration Acceleration is the rate of change of velocity. In other words it is a measure of how quickly an object changes speed. Acceleration is always caused by an unbalanced force. The symbol for acceleration is a and the unit for it is ms -2 (Meters per second per second). An object that speeds up slows down or changes direction has accelerated. When a object slows down we say it decelerates. Deceleration is negative acceleration. We use the following equation to work out the acceleration of an object.

32. Use the acceleration equation to answer the following questions. What is the acceleration of a car that takes 7.4 seconds, from standing still, to reach 45 ms -1 ? Change in speed = 45 – 0 = 45 ms -1 Change in time = 7.4 s What is the acceleration of a girl that takes 4.3 seconds to go from 0 ms -1 to 5.6 ms -1 ? Change in speed = 5.6 – 0 = 5.6 ms -1 Change in time = 4.3 s Exercise Complete exercises 1-7 on page 163 of the New Directions in Science text book..

33. Calculate the time taken to reach a velocity of 13.5 ms -1 is a car is acceleration at 3.2 ms -2.

34. Can draw free body force diagrams to show the forces acting on an object. Can work out the net force acting on an object by combining the force vectors from a force diagram. Can break force vectors into their horizontal and vertical (or North and West ) components using soh cah toa. Can use the equation F= m x a to work out one component when the others are known. Can interconvert between mass and weight using 9.8 ms -2 as the acceleration due to gravity. Can describe two examples to illustrate each of Sir Isaac Newton’ three Laws of Motion.

35. Force, mass and acceleration. Force A force is a push, pull or a twist to an object that causes it to accelerate. The symbol for force is F and the unit is the Newton, N. Force is a vector quantity and this is often depicted by an arrow. The resultant force is the product of the vectors of all the forces acting on an object. 15 N 9.8 N Resultant force Vector triangle Diagram of ball being thrown

36. Law 3 For every action force there is an equal and opposite reaction force. (the action force acts on an object and the reaction force acts on a different object) Law 2 When the resultant force acting on an object is non-zero the object will accelerate in the direction of the force. Law 1 If the resultant force acting on an object is zero then the object will have a constant velocity. Newton’s laws of motion.

37. 4N Support 6N 2N 6N Gravity Friction Thrust The net force is 2N to the left This means that the toy car will accelerate to the left. Net force 2N Unballanced forces

38. The size of the force needed to accelerate an object at a certain rate is dependent on two factors: The rate of acceleration acting on the object. A faster acceleration will need a larger force. The mass of the object the force is acting on. The larger the mass, the large the force need to accelerate it. Force, mass and acceleration. Example What force is needed to accelerate and 10 kg ball 3 ms -2 ? F=m x a = 10 x 3 = 30 N

39. All objects, at sea level, on the earth are accelerated by gravity (g) at a rate of 9.8 ms -2 : The force that is produced by this acceleration is called weight and it is measured in Newtons (N). This force of gravity (weight) can be calculated by using the following equation. g = 9.8 ms -2 Force of gravity. Example What is weight acting on a 60 kg girl at sea level? F g =m x g = 60 x 9.8 = 588 N

40. Work is done when an object is moved a distance (d) by and force (f) that is parallel to the movement. Only the component of the force that is parallel to the direction of the movement. d = 12 m f = 20 N θ = 30˚

41. Power Power is how much energy(work) is produced per second. When using this equation the energy could come from work, kinetic energy or E p. Example What is the power of a motor that does 100 J of work in 4 seconds? Energy = work = 100J P = E ⁄ t = 100J / 4 s = 25 W Power measured in Watts(W) Energy measured in Joules(J) Time measured in seconds(s)

42. Example How much energy is used when a 100 W light bulb is on for 30 s? E = P x t = 100 W x 30 s E = 3000 J Example How long does it take a 500 W motor to do 10 000 J work? t = E / P = 500 W / 10 000 j t = 20 s Rearranging the Equation To find the energy To find the time

43. Kinetic energy Kinetic energy is the energy that moving objects have. We use the following equation to work out the kinetic energy an object has. Exercise What is the kinetic energy of a 8.0 kg ball moving at 5.0 ms -1 ? E k = ½ mv 2 = ½ 8.0 x 5 2 = 4.0 x 25 = 100 J mass(Kg) speed(ms -1 ) kinetic energy (J)

44. In this equation speed is squared. This has the effect of increasing the energy by four times when the speed is doubled. Example Ball traveling at 4 ms -1 E k = ½ mv 2 = ½ 8 x 4 2 = 4 x 16 = 64 J Ball traveling at 8 ms -1 (Double the speed) E k = ½ mv 2 = ½ 8 x 8 2 = 4 x 64 = 256 J (The kinetic energy is quadrupled)

45. Gravitational potential energy Gravitational potential energy (GPE) is the energy an object above the ground has. We use the following equation to work out the gravitational potential energy an object has. Exercise What is the GPE of a 2.0 kg ball 8.0 m above the ground? Ep = 2.0 kg x 9.8 ms -2 x 8 m = 160 J mass(Kg) Force of gravity(9.8 NKg -1 ) Gravitational potential energy (J) Height of object(m)

46. Converting energy Any time that an object is lifted the work done to lift the object is equal to the gravitational potential energy (E p ) gained by the object. 2 m E p = mgh = 0.5 x 10 x 2 = 10 J Mass of toy car 0.5 Kg W = f x d = 5 x 2 = 10 J What is the gravitational potential energy gained when a box is lifted with a constant force of 50 N for 1.5 m? W = f x d = 50 x 1.5 = 75 J E p = work done = 75J

47. When work is done accelerating an object by a force over a distance, this energy is transformed into kinetic energy(E k ). However, because of friction some of this work is turned into heat energy and only some of the work is transformed into kinetic energy. Any time that energy is transformed some energy is always lost as heat energy. The amount of heat energy lost can be reduced by decreasing the amount of friction acting on an object. We can work out the efficiency of an energy transformation using the following equation.

48. When on object falls, the gravitational potential energy it had at the top of the fall is equal to the kinetic energy it has just before it hits the ground (if we ignore air friction). 4 m At the top E p = mgh = 2 x 10 x 4 = 80 J Just before hitting the ground E k = E p E k = 80 J

50. Describe Hooke’s law Interconvert between the force, spring constant or extension of a spring using Hooke’s law states that “the distance a spring extends is proportional to the tension force acting on the spring.” 10 N 0.12 m The diagram on the right shows what happens when a 10 N force is applied to an un-stretched spring. The tension force (F) is always in the opposite direction to the force extending the spring. For this spring the tension force, F = - 10 N and the extension, x = 0.12 m. Hooke’s law can be used to work out the spring constant for this spring. Hooke’s Law

51. The potential energy gained when a spring is compressed or extended can be calculated by using the equation on the right. The potential energy of a spring is dependent only on the extension of the spring and the spring constant.

52. Exercises Complete the following exercises from page 150 of NCEA Level 1 Science book. Questions1, 2, 3 and 4 Answers