 # Kinematics – studying how objects move. What do we know from Nat 5? In your group write down as many different things as you can to do with movement. Note.

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Kinematics – studying how objects move. What do we know from Nat 5? In your group write down as many different things as you can to do with movement. Note we are not looking at forces yet. Your group has 5mins.

Scalars and Vectors With forces direction is important so force is a vector quantity. Time has no direction so it is a scalar quantity.

Scalar – magnitude (size) only eg distance, d = 34 km Vector – magnitude - direction - reference point eg. Displacement, s = 34 km south of Edinburgh Vectors always added nose to tail.

quantitysymbolunitsymbolScalar vector dmetres Displacementsmetremv Speedv Metres per second ms -1 s Velocityv Metres per second ms -1 v Acceleration Metres per second per sec Time Fnewtons EnergyJ mkilogramkg Back of notes jotter

quantitysymbolunitsymbolScalar vector Distancedmetrems Displacementsmetremv Speedv Metres per second ms -1 s Velocityv Metres per second ms -1 v Accelerationa Metres per second per sec ms -2 v Timetsecondss ForceFnewtonsNv EnergyEjoulesJs Massmkilogramkgs

Tutorial Questions Nat 5 Review and Vectors pages 2 and 3 In rough work jotter Now some more challenging vectors question.

Directions Left/right, up/down30 o above horizontalNorth, south etc 3 figure bearings 315 o north west

Example 1 (You might want to copy this example in your notes jotter) (a) Draw a scale diagram to find the distance travelled and the final displacement of someone following this route. 4 km at 045 o 2 km due south 6 km at 60 o south of east N resultant Distance = 12 km Displacement = 7.4 km at 128 o Scale 1cm: 1km Now do the extra example sheet in rough jotter.

(b) If a boy took 4 hours to complete this course, what was his average speed in km/hour? (c) What was his average velocity in kmh -1 ? (d) What was his velocity in ms -1 ? Distance = 12 km Displacement = 7.4 km at 128 o (b) Average speed = d ÷ t = 12 ÷ 4 = 3 kmh -1 ( c) average velocity = s ÷ t = 7.4 ÷ 4 = 1.85 kmh -1 At 128 o (d) 1.85 km/h = 1850 m/h = 30.83 m/min = 0.51 ms -1 at 128 o

2010 Higher paper Qu 21 A helicopter is flying at a constant height above the ground. (a)The helicopter flies 20km on a bearing of 180 o (due south). It then turns on to a bearing of 140 o and travels a further 30km. The helicopter takes 15 minutes to travel the 50km. (i) By scale diagram or otherwise find the resultant displacement of the helicopter.(3) (ii) Calculate the average velocity of the helicopter during the 15 minutes. (3) 47km ± 1 at 156 o ± 2 or 24 o east of south or 66 o south of east V= 52.3 ms -1 at 156 o ( or 52.2ms -1 or 188 kmh -1 )

Example 2 Find the resultant vector (add vectors nose to tail.) 65 N140 N (a) For forces the resultant vector is called the unbalanced force. (b)The wind blows with a force of 3N to the North. The current is 4N to the East. What is the resultant force? 30 o 10 N (c)

40 N east 5 N south a 500 N horizontal 100 N down b c40 N east 20 N at 120 o d 30 o 10 N 20 N Now you try some examples resolving vectors (rough jotter) Now do Purple book Ex 1.1 Qu 3 to 10 Past paper Questions 2007 Qu 21 2005 Qu 1 Homework number 3

Some more examples resolving vectors (rough jotter) The plane’s engines are pushing the plane due north with a force of 250 kN. The wind is pushing it west with a force of 30 kN. What is the resultant force on the plane? 3000 N 1000 N 140 o e f Find the resultant force on Santa’s sledge.

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