2.1

ANALYSING

LINEAR MOTION

INTRODUCTION

How fast ?(Speed / velocity)

Does it change its speed ?(Acceleration / deceleration)

How would you describe the motion in word ?

How far does it travel ? (distance/displacement)

Information required:

How fast ?(Speed / velocity)

How far does it travel ? (distance/displacement)

• A straight line motion

LINEAR MOTION

• Not a straight line motion

NON LINEAR MOTION

• Total path travelled in a given time is the same as the shortest path

• Total path travelled in a given time is different from the shortest path

• DISTANCE AND DISTANCE AND DISPLACEMENT DISPLACEMENT

• SPEED AND SPEED AND VELOCITY VELOCITY

• ACCELERATION AND ACCELERATION AND DECELERATIONDECELERATION

Learning areaLearning area

DISTANCE AND DISPLACEMENT

Pontian Kecil Desaru

Johor Bahru

Pontian Kecil

Ayer Hitam

SenaiKota Tinggi

Mawai

Benut

How far is it from Johor Bahru to Desaru ?

Distance = total path length =JB to Desaru via Kota Tinggi

Displacement = shortest path length = JB direct to Desaru

SCALAR

VECTOR

SPEED AND VELOCITY

start

end

path

Distance =

Displacement =

Average Speed =

Average Velocity =

Time taken =

ACCELERATION AND DECELERATION

Velocity increases

Constant velocity

Velocity decreases

Acceleration = Rate of change of velocity

= Change of velocity Time= final velocity – Initial velocity Time

a = v – u t

vector

m s-2

Velocity increases = acceleration

Velocity decreases = deceleration

Carry out Hands-on Activity 2.2

( page 11 of the practical book)

Aim : To differentiate between acceleration and decelerationDiscussion :1. (a) The speed of the trolley increases . (b) The speed of the trolley decreases.

2.Acceleration is the rate of increasing speed in a specified direction.

Deceleration is the rate of decreasing

speed in a specified direction.

Lesson 2Lesson 2

RELATING DISPLACEMENT, RELATING DISPLACEMENT, VELOCITY , ACCELERATION VELOCITY , ACCELERATION

AND TIMEAND TIME

a. Using ticker tape

b. Using Equations of Motion

Learning areaLearning area

ticker timer

ticker tape

A.C. 50 Hz

50 dots made in 1 second

Carbon disc

Time interval between two adjacent dots = 1/50 s

= 0.02 s

1 tick = 0.02 s

dots

1 tick

Slow movementfaster movementfastest movement

PREPARING A TAPE CHART (5 -TICKS STRIP)

0 5 10

First 5-tick strip

2nd 5-tick strip

Velocity, v

(cm /s)

Time / s

INFERENCE FROM TICKER TAPE AND CHART

•Zero acceleration

•constant velocity

• Constant acceleration

• Constant deceleration

Carry out Hands-on Activity 2.3

( page 13 of the practical book)

Aim : To use a ticker timer to identify the types of motion

Discussion 2.3(A):2. Spacing of the dots is further

means a higher speed.

Spacing of the dots is closer means a slower speed.

Discussion Hands-on Activity 2.3(B)

( page 13 of the practical book)

Aim : To determine displacement, average velocity and acceleration

Discussion 2.3(B):1. Prepare a tape chart.2. Determine average velocity using v = Total displacement

time3. Determine acceleration using a = final velocity – initial velocity

time

Lesson 3Lesson 3

TO DETETMINE THE AVERAGE VELOCITY

EXAMPLE The time for each 5-tick strip = 5 x 0.02 s

= 0.1 sLength / cm

Time / s

0

7

10

1415

22

0.10.2

0.30.4

0.50.6

0.7

= (7 +10 +14 +15 +22 +14 +10) cm= 92 cm

= 7 strips = 0.7 s

Total displacement

Total time taken

Average velocity = displacement Time taken

= 92 / 0.7 = 131.4 cm s-1

TO DETERMINE THE ACCELERATION

EXAMPLE The time for each 10-tick strip = 10 x 0.02 s

= 0.2 s

5.8 / 0.2 =29 cm s-1

27.3 / 0.2 = 136.5

Initial velocity, u

Final velocity, v

acceleration = v-u t= (136.5 – 29) cm s-1

1.2 s

Length / cm

Time / s

0 0.20.4

0.6 11.2

Time takenTime taken

=(7-1 )strips

= 6 x 0.2 s= 1.2 s

5.8

27.3

1.40.8

= 89.6 cm s-2

Lesson 4Lesson 4

ss = Displacement = Displacement

uu = Initial velocity = Initial velocity

vv = Final velocity = Final velocity

aa = Constant = Constant acceleration acceleration

tt = Time interval = Time interval

THE EQUATIONS OF MOTION

v u at 21

2s ut at 2 2 2v u as

2

u vs t

EXAMPLE

A car travelling at a velocity 10 m s-1 due north speeds up uniformly to a velocity of 25 m s-1 in 5 s. Calculate the acceleration of the car during these five seconds

u = 10 m s-1 , v = 25 m s-1, t = 5 s, a = ?

v = u + at

s = ut + ½ at2

v2 = u2 + 2as

s = ½ (u + v) t

Using v = u + at

25 = 10 + a(5)

5 a = 15

a = 3 m s-2

Don’t forget the unit

EXAMPLE

A rocket is uniformly accelerated from rest to a speed of 960 m s-1 in 1.5 minutes. Calculate the distance travelled.

u = 0 m s-1 , v = 960 m s-1, t = 1.5 x 60 = 90 s, s = ?

v = u + at

s = ut + ½ at2

v2 = u2 + 2as

s = ½ (u + v) t

Using s = ½ (u + v)t

s = ½ (0 + 960) 90

= 43 200 mWhat is the

unit ?

EXAMPLE

A particle travelling due east at 2 m s-1 is uniformly accelerated at 5 m s-2 for 4 s. Calculate the displacement of the particle.u = 2 m s-1 , a = 5 m s-2, t = 4 s, s = ?

v = u + at

s = ut + ½ at2

v2 = u2 + 2as

s = ½ (u + v) t

Using s = ut + ½ at2

s = 2(4) + ½ (5)(4)2

= 8 + 40

= 48 m What is the unit ?

EXAMPLE

A trolley travelling with a velocity 2 m s-1 slides 10 m down a slope with a uniform acceleration. The final velocity is 8 m s-1. Calculate the acceleration.

u =2 m s-1 , v = 8 m s-1 , s = 10 m , a = ?

v = u + at

s = ut + ½ at2

v2 = u2 + 2as

s = ½ (u + v) t

Using v2 = u2 + 2as

82 = 22 + 2 a (10)

20 a = 64 – 4

= 60

a = 3

What is the unit ?

m s-2

EXAMPLE teks book pg 27

u =0 m s-1, a = 2.5 m s-2 , t = 10 s v = ? , s = ?

Using v = u + at

= 0 + (2.5)(10)

= 25 m s-1

v = u + at

s = ut + ½ at2

v2 = u2 + 2as

s = ½ (u + v) t

Using s = ut + ½ at2

= 0(10) + ½ (2.5)(10)2

= 125 m

EXAMPLE teks book pg 27

u = 25m s-1, v = 0 m s-1 , s = 50 m , a = ?

Using v 2 = u2 + 2as

0 = 252 + 2a (50)

0 = 625 + 100a

a = - 625 100

= - 6.25 m s-2

v = u + at

s = ut + ½ at2

v2 = u2 + 2as

s = ½ (u + v) t

The negative sign shows deceleration.