Kinematics Vector and Scalar Definitions Scalar: a physical quantity that can be defined by magnitude (size) only. Vector: a physical quantity that can be defined by magnitude and direction. Scalar examples: distance, time, mass, speed, energy. Vector examples: displacement, acceleration, velocity, force, momentum. NOTES p.1

Distance and Displacement A walker takes the following path (green line). N S EW The total distance walked is shown in green. The total displacement is shown in red as a straight line from start to finish with magnitude, x, in direction,  East of North. Start Finish N x 

Calculating Speed and Velocity These are two different quantities. Remember, speed is a scalar and velocity is a vector. speed = distance time velocity = displacement time NOTE – to find the direction of the velocity use the direction of the displacement. NOTES p.2

TRY THIS IN PAIRS Find the distance travelled, the overall displacement, the average speed and the average velocity for a person who walks 5km, North followed by 3km, South in a time of 2 hours. 3km 5km

TRY THIS IN PAIRS Find the distance travelled, the overall displacement, the average speed and the average velocity for a person who walks 5km, North followed by 5km, East followed by 7km, South in a time of 6 hours. 7km 5km

A car moves 30m North, then 50m West, then 30m South in 1 minute. a)What’s the speed of the car? b)What’s the velocity of the car? Example 1 NOTES p. 2

a) Speed = dist / t = 110 / 60 = 1.83 ms -1 3 cm = 30m 5cm = 50m 3cm = 30m 5cm = 50m Distance = = 110m Displacement = 50m, West b) Velocity = disp / t = (50, West) / 60 = 0.83 ms -1, West 1cm = 10m N

Example 2 Use a scale diagram to calculate the displacement of a man who walks 40m, North then 100m at 120 o. 4 cm = 40m 10cm = 100m 120 o x  x = cm = m.  = o. Displacement = m at o

Example 3 Use Pythagoras and trigonometry to calculate the displacement of a man who walks 40m, North then 50m, East then 10m, South. 10m 50m 40m x 

50m 40 – 10 = 30m x  resultant displacement magnitude (x): x 2 = = 3400 x = 58.3m direction : tan  = opp = 50 adj 30  = 59 o (= 059 o from North) SO…displacement = 58 m at a bearing of 059 o

Homework for Wednesday 18 June 2014 Physics: Past Paper 1995 Q.1