Physics Module Form 4 Chapter 2 : Force and Motion

2. 2.1 1.

FORCE AND MOTION ANALYSING LINEAR MOTION Types of physical quantity: has only a magnitude (i) Scalar quantity: …………………………………………………………………. has both magnitude and direction (ii) Vector quantity: ………………………………………………………………… The difference between distance and displacement: length of the path taken (i) Distance: ………………………………………………………………………… distance of an object from a point in a certain direction (ii) Displacement: …………………………………………………………………… Distance always longer than displacement. Example: The following diagram shows the location of Johor Bahru and Desaru. You can travel by car using existing road via Kota Tinggi, or travel by a small plane along straight path. Calculate how far it is from Johor Bahru to Desaru if you traveled by: a. The car b. The plane Kota Tinggi 41 km 53 km

Distance and displacement

2.

3. 4.

Solution: a. b. by car = 41 + 53 = 94 km

Johor Bahru

by plane = 60 km

Desaru The path traveled by the plane is shorter than travelled by the car. So, Distance = 94 km Displacement = 60 km

60 km

Hands-on Activity 2.2 pg 10 of the practical book. Idea of distance and displacement, speed and velocity. Speed and velocity 1. 2. 3. 4. the distance traveled per unit time or rate of change of distance Speed is ..………………………………………………………………………………… the speed in a given direction or rate of change of displacement Velocity is: ..……………………………………………………………………………... total distance traveled, s (m) , v = s m s-1 Average of speed: ……………………………………………………………………… time taken, t (s) t -1 Average of velocity:displacement, s (m) , v = s ms ……………………………………………………………………... Time taken, t (s) t

1

JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

5.

Example: An aero plane flies from A to B, which is located 300 km east of A. Upon reaching B, the aero plane then flies to C, which is located 400 km north. The total time of flight is 4 hours. Calculate i. The speed of the aero plane ii. The velocity of the aero plane Solution: i. Speed = Distance Time = 300 + 400 4 = 175 km h-1 ii. velocity = displacement time (Determine the displacement denoted by AC and its direction) = . 500 . 4

C 400 km

A

300 km

B C 400 km

A 300 km Acceleration and deceleration 1.

B

= 125 km h-1 (in the direction of 0530)

Study the phenomenon below;

0 m s-1

20 m s-1

40 m s-1

2. 3.

The velocity of the car increases. Observation: ……………………………………………………………………………… the rate of change of velocity Acceleration is, ………………………………………………………………………. Final velocity – initial velocity Or, a = v – u Then, a = Time of change t Example of acceleration; t=2s t=2s A B C

0 m s-1

20 m s-1 2

40 m s-1

JPN Pahang

20 – 0 2 = 10 m s-2

Physics Module Form 4 Chapter 2 : Force and Motion

Calculate the acceleration of car; i) from A to B aAB = 20 – 0 2 ii) 4. 5. From B to C aBC = 40 – 20 2

= 10 m s-2 = 10 m s-2

when the velocity of an object decreases, In calculations, a will Deceleration happens ...………………………………………………………………… be negative ……………………………………………………………………………………………… Example of deceleration; A lorry is moving at 30 m s-1, when suddenly the driver steps on the brakes and it stop 5 seconds later. Calculate the deceleration of lorry. Answer : v = 0 m s-1, u = 30 m s-1, t = 5 s Then , a = 0 – 30 5 = -6 m s-2

Analyzing of motion 1. Linear motion can be studied in the laboratory using a ticker timer and a ticker tape. Refer text book photo picture 2.4 page 26. (i) Determination of time:

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the frequency of the ticker timer = 50 Hz ( 50 ticks in 1 second) so, 1 tick = 1 second = 0.02 seconds 50 (ii) Determination of displacement as the length of ticker tape over a period of time. x y (iii)

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xy = displacement over time t measure by ruler s

. . . . . . . . Uniform velocity ……………………………………………………………………………………….. . . . . . . . . Acceleration...