Ch.1 Distance (距離) & Displacement (位移) Tai Wai Station Kowloon Tong Station displacement 4 km to the north distance 4.4 km

Class Exercise 1.1 Distance travelled = ?

Class Exercise 1.1 Distance travelled = 12 + 98 + 70 = 180 m

Displacement (位移) = change in position The change in position is a straight line joining the start position (起點) and the end position (終點). Magnitude (量值): Length of the straight line joining the start position and the end position. Direction (方向): Direction of the straight line from the start position to the end position.

North (北) West (西) East (東) South (南)

N 40° Direction = ?

N 40° 50 ° E Direction = 050° (or N 50° E)

N 60° Direction = ?

N 90° + 60° = 150° E 60° S Direction = 150° (or S 30° E)

N 40° Direction = ?

N 90° + 90° + 50° = 230 ° W 40° 50° S Direction = 230° (or S 50° W)

N 40° Direction = ?

N 40° W 270° + 50° = 320° Direction = 320° (or N 40° W)

Magnitude of displacement = ? Direction of displacement = ? Class Exercise 1.2 Magnitude of displacement = ? Direction of displacement = ?

Magnitude of displacement = 128 m Class Exercise 1.2 Magnitude of displacement = 128 m Direction of displacement = 180º + 40° = 220° (S 40º W)

Class Exercise 1.3 (a) & (b)

Class Exercise 1.3 (a) & (b) 6 m 6 m East (90°) East

Class Exercise 1.3 (c) & (d)

Class Exercise 1.3 (c) & (d) 6 m 6 m East East

Note: If the starting position and the end position are the same, the displacement is the same.

Class Exercise 1.4 (a) You walk from LPSMC to the Entrance C of the Tsuen Wan MTR station. Find the magnitude and direction of your displacement.

Class Exercise 1.4 (a) 6.0 cm 3.1 cm Magnitude = ? Direction = ?

Direction = 180° + 78° = 258° (S 78° W) Class Exercise 1.4 (a) 6.0 cm 78° 3.1 cm Magnitude = (6.0 ÷ 3.1 ) × 100 = 194 m Direction = 180° + 78° = 258° (S 78° W)

Class Exercise 1.4 (b) You meet your friend, Tommy, at Entrance C. He is going to the YMCA in Luk Yeung Sun Chuen. What is his displacement from Entrance C to the YMCA?

Class Exercise 1.4 (b) 3.5 cm Magnitude = ? Direction = ? 3.1 cm

(N 50° E) Magnitude = 3.5 cm (3.5 ÷ 3.1 ) × 100 = 113 m 50° Class Exercise 1.4 (b) Magnitude = (3.5 ÷ 3.1 ) × 100 = 113 m Direction = 50° (N 50° E) 3.5 cm 50° 3.1 cm

(a) In the space below, draw arrows to represent these displacements. Class Exercise 1.5 Mr. Chan walks 3 m from A to the East. He turns 90 and walks 6 m to the South. Finally he walks a further distance of 9 m to the West. (a) In the space below, draw arrows to represent these displacements. (Use 1 cm to represent 1 m.)

Class Exercise 1.5 (a)

Class Exercise 1.5 (a)

(b) direction and magnitude of the total displacement = ? Class Exercise 1.5 (b) (b) direction and magnitude of the total displacement = ?

Class Exercise 1.5 (b) 45° magnitude of the total displacement = 8.5 m direction of the total displacement = 180° + 45° = 225° (SW)

Class Exercise 1.6 (a)

Class Exercise 1.6 (a) East 12 m 33

Class Exercise 1.6 (a) East 12 m East 6 m 34

Class Exercise 1.6 (a) East 12 m East 6 m 18 m East

Class Exercise 1.6 (b)

Class Exercise 1.6 (b) S 37

Class Exercise 1.6 (b) S -10 m or 10 m (West)

Class Exercise 1.6 (b) S -10 m or 10 m (West) 8 m (East) 39

Class Exercise 1.6 (b) S -10 m or 10 m (West) 8 m (East) 28 m 40

Class Exercise 1.7

Class Exercise 1.7 distance = 3 + 4 = 7 m

Class Exercise 1.7 distance = 3 + 4 = 7 m 43

Class Exercise 1.7

magnitude = m tan a = ¾ a = 36.9° direction = 127° (S 53° E) Class Exercise 1.7 a magnitude = tan a = ¾ a = 36.9° direction = 127° (S 53° E) m

Class Exercise 1.7

Class Exercise 1.7 P

Class Exercise 1.7 P -4.0 m or 4.0 m (West) 3.0 m (South)

Scalar (無向量) Vector (向量/矢量) 1.3 Scalar and Vector Scalar (無向量) Vector (向量/矢量) has magnitude (量值) only. has both magnitude and direction. Example: distance (距離) time (時間) temperature (溫度) speed (速率) volume (體積) displacement (位移) velocity (速度) acceleration (加速度) force (力) …