ENGINEERING SCIENCE

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Video description: Uniform & Non –Uniform motion

Distance, Displacement, Speed and Velocity

1. When an object is in motion, it changes position. The total length of the path travelled
by the object from one location to another location is called distance.

2. Distance does not take into account the direction travelled by the object. Hence, we can
say that the distance is scalar quantity.

3. Displacement is the distance travelled by the object in a specific direction.
Displacement is a vector quantity.

4. Distance and Displacement are measured in metres (SI units).

Chapter 2 - Linear Motion/ Page 44

Smart Tips

Displacement is the shortest distance that links the initial position to final position
of an object that has moved.
Distance travelled = displacement, if the motion is linear.
However, for displacement, the direction of motion needs to be stated.

P
QB

R
S

AT

ARB = displacement
APB, AQB, ASB and ATB is known as distance

5. Speed and velocity describe how fast an object is moving, but there has difference between
these quantities.

6. Speed is the distance travelled per unit time or the rate of change of distance. Speed is
a scalar quantity.

= ( )

( )

7. Velocity is the speed of an object in a specified direction or the rate of change of
displacement. Velocity is a vector quantity.

( )
= ( )

Chapter 2 - Linear Motion/ Page 45

8. Both speed and velocity have the same SI unit, that is meter per second (m/s or ms-1 ).
9. Average velocity is used when the object is not moving at a constant velocity for the

whole journey.
Example:

t=0s t=1s t=2s

15 m 25 m

The car moves 15 meters for 1 second and 25 meters for 2 seconds.
On average the car moves 13.33 meter per second

10. The instantaneous velocity shows the velocity of an object at one point. For example,
when you driving a car and its speedometer swings to 90 km/h, then the instantaneous
velocity of the car is 90 km/h.

Chapter 2 - Linear Motion/ Page 46

QUICK CHECK

1. A car moved 25 km to the North. What is its displacement?

Answer: s = 25 km North

2. A car moved 25 km East and 70 km west. What is the distance?

Answer: d = 95 km

3. A car moved 60 km east in 2 hours and 20 km west in 2 hours. What is its
average speed?

Answer: Average speed = 20 km/hr

4. How far will a car travelled in 10 min at 20 m/s?

Answer: s = 12 000 m or 12 km

Chapter 2 - Linear Motion/ Page 47

Acceleration and Deceleration

1. When the velocity of an object changes, the object is said to be accelerating.

2. Acceleration is defined, as the rate of change in velocity with time.

= ( ) − ( )
= ( )

−
=

3. Acceleration is a vector quantity. The SI unit for acceleration is metre per second per
second or m/s2.

4. If the value of initial velocity, u is less than the value of final velocity, v then the object’s
acceleration is positive.

5. If the value of initial velocity, u is higher that the value of final velocity, v then the object’s
acceleration is negative. Negative acceleration is called deceleration.

6. If velocity is constant or uniform (the value of u = v), the change in velocity is zero.
Therefore, the object’s acceleration is zero.

Example 1

A rocket, moving with a constant velocity of 2500 m/s, increases the rate of combustion in the
combustion chamber to increase its velocity to 3500 m/s in 5s. What is the acceleration of the
rocket?

Solution: , ? = 0 = 5

−
= = 2500 = 3500

3500 − 2500
= 5
= /

Chapter 2 - Linear Motion/ Page 48

Smart Tips

The acceleration of the rocket is 200 m/s2. This means its velocity increases by 200 m/s for
every 1 second as illustrated in the diagram below.

Example :

= 0 = 1 = 2 = 3 = 4 = 5
= 3500 /
= 2500 / = 2700 / = 2900 / = 3100 / = 3300 /

Example 2

Azman cycles at a uniform speed of 30 m/s. He then stops pedaling and finally he and his
bicycle stop after 8s. What is his average deceleration?

Solution:

Given: = 30 = 0 = 8 = ?



−
=

0 − 30
= 8


= −3.75 2

= − . − ( ℎ )

Chapter 2 - Linear Motion/ Page 49

The Equation of Linear Motion

1. There are 4 equation of linear motion. They are :

a) = + or = − Where :

= ( )
= ( )
b) = +

c) = +

= ( )
= ( / 2)
d) = ( + )
= ( )


Example 3

A sprinter reaches his top velocity in 4 seconds after starting from rest. He covers a
displacement of 32 meters in 4 seconds. What is his acceleration?

Solution:

Given: = 0 = 32 = 4 = ?



= + 1 2
2

32 = (0 × 4) + 1 × × 42 )
(2

32 = 0 + 8

32
= 8

or −
=

Chapter 2 - Linear Motion/ Page 50

Example 4

By applying the brakes, a driver reduces his car’s velocity from 80 m/s to 10 m/s after
travelling a distance 50 m. Find the deceleration of the car.

Solution: Given: = 80 = 10 = 50 = ?



2 = 2 + 2
102 = 802 + (25 × × 50 )

100 = 6400 + 100

100 = 100 − 6400

100 = −6300

−6300
= 100


= −
= − − ( ℎ )

Example 5

A light plane needs a velocity of 35 m/s for take-off. If the plane can accelerate at 3 m/s2,

calculate the minimum length of the runway for the plane to reach the proper speed to

take-off. = 3 m/ 2
Solution:

2 = 2 + 2
352 = 02 + (2 × 3 × ) = 0 = 35
1225 = 6 s?

1225
= 6
= .

Chapter 2 - Linear Motion/ Page 51

Example 6

A ball is thrown vertically upwards with an initial velocity of 20 m/s. by neglecting an air

resistance, find : Take g = 9.81 m/s2

a) The maximum height that the ball can reach.

b) The time taken before the ball reaches the ground.

Solution:

a) Given: = 20 , = 0 ( ℎ ℎ ℎ ) , , ?




= − = −9.81 2 ( , ℎ )

2 = 2 + 2

02 = 202 + (2 × −9.81 × )

400 = 19.62

= .

b) time for upward, t ?
= +
0 = 20 + (−9.81 × ) *For upward flight, the value of acceleration is -9.81 m/s2
= .

time for downward, t ?

= + 1
2
The direction of an acceleration of the
20.39 = (0 × ) + 1 × 9.81 × 2) ball is the same with acceleration of
(2 gravity (downwards) and the ball is
free fall with acceleration of 9.81 m/s2

20.39 = 0 + 4.905

= √20.39 = .

4.905

Total time taken to reach the ground, t = time for upward + time for downwards
t = 2.04 +2.04 = 4.08 s

Chapter 2 - Linear Motion/ Page 52

QUICK CHECK

1. A sport car accelerates from rest and covers a distance of 95 m in 5 s. What is
its acceleration?

Answer: a = 7.6 m/s2

2. A rocket travelling at 2000 m/s increases its velocity to 6000 m/s after moving
a distance 80 km. Calculate the time for the rocket to reach this velocity.

Answer: t = 20 s

3. A coconut falls down higher to ground with velocity 15 m/s by neglecting an
air resistance, find:
i) The height of coconut from ground
ii) The time taken to reach the ground

Answer: i) s = 11.47 m ii) t = 1.53
s

Chapter 2 - Linear Motion/ Page 53

QUICK CHECK

4. A body is thrown vertically from ground level with velocity 55 m/s.
i) Find the body displacement at 3 s?
ii) How many maximum heights achieved?
iii) Calculate the time taken to reach the maximum heights

Answer: i) s = 82.5m ii) s = 154.18m iii) t = 5.61s

Chapter 2 - Linear Motion/ Page 54

2.2 VELOCITY-TIME GRAPH

1. A velocity – time graph is a graph showing how the velocity of an object varies with time.
2. The graph in Figure 2.4 shows how the velocity of the car changes over time.

u= 0 m/s v= 50m/s v= 50 m/s v= 0m/s

t = 0s t = 20s t = 40s t = 60s
Start Stop
a=0
v (m/s)
Constant
50 velocity

AB C

t (s)

0 20 40 60
Figure 2.4: Velocity of the car changes over time

3. A car starts from rest and accelerates for 20 s until it reaches a velocity of 50 m/s. The

driver maintains this velocity for 20 s. The velocity of the car is then gradually reduced

until it stops at t = 60 s.

4. In Section A of the graph , the acceleration of the car ,

− 50 − 0 = . −
= = 20 = 2.5 2

5. In section B of the graph, the car travels at a constant velocity of 50 m/s from t = 20 s until

t = 40 s. On the velocity – time graph, a horizontal line (gradient = 0) represent a constant

velocity (where a = 0 m/s).

6. In section C of the graph, the deceleration of the car ,

− 0 − 50 = − . −
= = 20 = −2.5 2

7. On velocity – time graph, the gradient of the graph represents the acceleration of the

object.

Chapter 2 - Linear Motion/ Page 55

Example 7

Figure below shows the velocity –time graph of a car travelling along a straight road between
two traffic lights.

Velocity (m/s)

12 A B

O5 C
Time (s)

15 25

a) Explain the motion of the car at :
i) OA
ii) BC

b) What is the time interval during the car is moving at a constant velocity?
c) What is the distance between the two traffic lights?

Solution:

a) Explain the motion of the car at :
i) OA shows a car is accelerating from 0 m/s to 12 m/s
ii) BC shows a car is decelerating from 12 m/s to 0 m/s (stop)

b) ℎ , = 15 − 5 =

c) , = ℎ = 1 ( 25 + 10)(12) = or

2

, 1 + (12 × 10) + 1 × 12) = 30 + 120 + 60
= ( 2 × 5 × 12) ( 2 × 10

=

Chapter 2 - Linear Motion/ Page 56

Example 8
The velocity time graph shows the movement of a particle.

Velocity (m/s)

12

Time (s)

O3 6 10

a) What is the acceleration of the particle in the first 3 seconds?
b) For how long was the particle moving at a constant velocity?
c) Calculate the average velocity of the particle.

Solution:

a) = 12−0

3

= /

b) = 6 − 3 =

c) =



, = ℎ
11

= ( 2 × 3 × 12) + (3 × 12) + (2 × 4 × 12)
= 18 + 36 + 24
= 78

78
= 10 = .

Chapter 2 - Linear Motion/ Page 57

Example 9
The following is velocity-time graph of a car.

Velocity (mB/s) C
12

15 20 25 28
Time (s)
AD G
5 10

- 12
EF

Calculate:

a) The acceleration of the car during the first 5 s.
b) The acceleration of the car at t = 15 s.
c) The total displacement of the car.
d) The total distance travelled by the car.

Solution:

a) , =
−

=
12 − 0

= 5
= . or . −



b) , =

−12 − 12
= 20 − 10

or − . −
= − .

Chapter 2 - Linear Motion/ Page 58

c) , = −

= 1 × (15 + 5) × 12 ] − 1 × (13 + 5) × 12] Displacement
[2 [2 considered direction.
Reverse direction
= 120 − 108 means negative

=

d) The distance travelled, d = +

= 120 + 108

= Distance does not take into account the
direction. That why the calculation is

adding.

QUICK CHECK

1. The velocity – time graph below shows the motion of the cart on the straight
railway.

Velocity (m/s)

B
6

E
A C Time (s)
0 5 10

-6
D

a) Calculate the total distance travelled by the cart.
b) Calculate the displacement of the cart.

Answer: a) d = 30 m b) s = 0 m

Chapter 2 - Linear Motion/ Page 59

Smart Tips Type of motion
Motion with a constant velocity
Velocity time – graph
Motion with uniform acceleration
Velocity Motion with uniform deceleration

Time Object at rest

Velocity

Time

Velocity

Time

Velocity

Time

Chapter 2 - Linear Motion/ Page 60

REFRESHER
Velocity

No horizontal straight line Time
Gradient
Horizontal line Object moves at a constant
Intersection on the time axis Represent the acceleration of the object
Sign of gradient Object moves at a constant velocity
Object stops
Area under graph Positive = acceleration
Negative = deceleration

Represent the distance travelled by the
object

Chapter 2 - Linear Motion/ Page 61

2.3 EXPERIMENT RELATED TO LINEAR MOTION
Ticker- timer

1. A ticker –timer is a device used in the lab to study the motion of a moving object, usually
a trolley.( Figure 2.5)

Figure 2.5: Ticker-timer

2. A vibrating metal strip with a pin is set to vibrate up and down 50 times per second ( i.e
50 hertz)

3. Each time the metal pin moves down, it makes a dot on the pre-carbonated ticker tape
which passes underneath.

4. The ticker tape is attached to the trolley which moves on the runaway.
5. The ticker-timer can be used to determine the following:

i) Time interval of the motion
ii) Displacement of the object
iii) Velocity of the object
iv) Acceleration of the object
v) Type of the motion of the object

Chapter 2 - Linear Motion/ Page 62

Analysis Ticker Tape
Uniform Velocity : Figure 2.6 shows ticker tape with uniform velocity:

The distance of the dots is equally
distributed. All length of tapes in the chart
are in equal length. Then, the object is
moving in uniform velocity.

Figure 2.6: Ticker tape with uniform velocity

Uniform Acceleration : Figure 2.7 shows ticker tape with uniform acceleration:

The distance between dots increase
uniformly. The length of the strips of tape
in the chart increase uniformly. Then, the
object is moving in a constant acceleration.

Figure 2.7: Ticker tape with uniform acceleration

Uniform Deceleration: Figure 2.7 shows ticker tape with uniform deceleration:

The distance between dots decrease
uniformly. The length of the strips of tape
in the chart decrease uniformly. Then, the
object is moving in a constant
deceleration.

Figure 2.8: Ticker tape with uniform deceleration Chapter 2 - Linear Motion/ Page 63

To determine displacement, velocity and acceleration of trolley

Figure 2.9: Set up linear motion experiment
1. The apparatus is set up as shown in Figure 2.9.
2. The inclination of the runway is adjusted so that the trolley, which is at rest, will roll

down freely without any applied force.
3. A length of ticker tape is passed through the ticker timer and attached to the trolley.
4. The ticker tape is switch on and the trolley is released from the top end of runaway.
5. The ticker tape is then cut into 6 pieces (or more pieces depends on the length of ticker

tape) of 10-tick strips.
6. The strips are pasted side by side on graph paper to form a tape chart.
7. The displacement, velocity and acceleration of the trolley are determined from the graph.

Chapter 2 - Linear Motion/ Page 64

Displacement X is a length of strip

X6
X5
X4
X3
X2
X1

Time (s)
Figure 2.10: Tape chart obtained as the result from experiment

From tape chart (Figure 2.10), some calculation can be done to determine displacement of
the trolley, time taken, velocity and acceleration of the trolley:

a) Displacement = X1 + X2 + X3 + X4 + X5 + X6

b) Time taken = 6 strips x 10-tick = 6 x 0.2 s = 1.2 s

c) Average velocity = = X1 + X2 + X3 + X4 + X5 + X6
1.2

d) Initial velocity = X1
0.2

e) Final velocity = X6

0.2

f) Time taken for the change of velocity , t = (6-1) x 0.2 s = 1.0 s

g) Acceleration, a = −


Chapter 2 - Linear Motion/ Page 65

ACTIVITY CHAPTER 2

1. A student carried out an experiment using a trolley and a ticker – timer that vibrates
at a frequency of 50Hz.
Velocity (cm per 10 ticks)

7
6
5
4

3

Time (s)

0.2 0.4 0.6 0.8 1.0

Figure above shows a tape chart consisting of 10-tick strips he obtained. Calculate :
a) The total displacement travelled
b) The average velocity
c) The initial velocity
d) The final velocity
e) The time for acceleration
f) The acceleration of the trolley.

Chapter 2 - Linear Motion/ Page 66

Answer: a) s = 25 cm b) velocity = 25 cm/s c) u = 15 cm/s d) v = 35 cm/s e) t = 0.8 s f) a = 25 cm/s2

2. A boy takes 10 s to finish the path AB. Calculate : 8m
a) The total distance travelled 6m
b) The displacement
c) The average speedof the path.
4m

2m B
A

Answer : a) d = 28 m b) s = 12 m c) average speed = 2.8 m/s

Chapter 2 - Linear Motion/ Page 67

3. A girl is running in a race covers 60 meters in 12 seconds.
a) What is her speed in
i) m/s
ii) km/hr
b) if she takes 40 seconds to complete the race, what is her distance covered?

Answer : i) v = 5 m/s ii) v = 18 km/hr iii) d = 200 m

4. Alina is driving her car at a velocity of 10 m/s. On seeing an obstacle in front, she
applies the brakes to stop her car. If the deceleration of the car is 2 m/s2, what is the
distance her car travels before it comes to a halt?

Answer : s = 25 m

Chapter 2 - Linear Motion/ Page 68

5. A bicycle is moving at a velocity of 25 m/s, and it brakes with a deceleration of
2 m/s2 before it stop. Calculate :
a) The time taken for the bicycle to stop
b) The braking displacement of the bicycle

Answer: a) t =12.5 s b) s = 156.25 m

6. A train is accelerates from rest to 100 km/hr in 4 seconds. Find the average
acceleration.

Answer: a = 6.945 m/s2

Chapter 2 - Linear Motion/ Page 69

7. An object accelerates uniformly along a straight line from a velocity of 5 m/s to 30
m/s in 5 s. Calculate :
a) The acceleration
b) The total displacement travelled by the object

Answer: a) a = 5 m/s2 b) s = 87.5 m

8. A girl swims in a 100 m length swimming pool. She takes 25 s from A to B and
another 30 s from B back to A. Calculate :
a) The average speed
b) The average velocity
AB

Answer: a) 3.64 m/s b) 0 m/s
Chapter 2 - Linear Motion/ Page 70

9. A runner runs 4 km towards south in 0.5 hour and 5 km towards east in 1.5 hours.
What is his average speed?

Answer: average speed = 1.25 m/s

10. Zahid is going to his friend’s home but he loses his way. He phones his friend, and
his friend told him “When you arrived at the supermarket, just turn right and walk
for 400 m. That is my home”. In this conversation, which of the following quantity
is NOT used?
A) Direction
B) Displacement
C) Acceleration

Answer : C

11. Find the total distance travelled by the girl as shown in figure.

10 m

10 m

20 m

Answer : d = 71.42 m

Chapter 2 - Linear Motion/ Page 71

12. Figure below shows a velocity- time graph. Determine:
a) the acceleration in time interval between
i) t = 0 s and t = 5 s
ii) t = 10 s and t =12 s
b) the distance travelled in the first 12 s

Velocity (ms-1 )

20 time (s)
05
10 12

Answer : a) i) a = 4.0 m/s2 ii) a = -10 m/s2 b) s = 170 m

Chapter 2 - Linear Motion/ Page 72

13. A train at rest starts to move from one station and stops at another station in 18
minutes. In first 2 minutes, it moves with a constant acceleration of 0.20 m/s2 and
after that its speed is constant until it is stopped by a constant force from the brakes,
which applied for 1 minute.
a) Sketch a velocity –time graph for the motion.
b) Determine the distance between the two station

Answer: d = 23760 m

14. An object moves along a straight line with constant acceleration. At a particular
moment, the velocity of the object is 30 m/s. After travelling through 50 m, the
object has velocity 20 m/s. How far does the object have to travel before it comes
to a stop?

Answer: d = 40 m

Chapter 2 - Linear Motion/ Page 73

15. A car which is initially at rest starts to move along a straight line with constant
acceleration. It reaches a velocity of 60 m/s after travelling through a distance of
100 m. Determine:
a) The acceleration
b) The time taken to reach the velocity of 60m/s
c) The velocity at the third second

Answer: a) a = 18 m/s2 b) t = 3.3 s c) v = 54 m/s

Chapter 2 - Linear Motion/ Page 74

16. Explain TWO (2) differences between distance and the displacement.

[Final Exam Collection – December 2013]

Answer: Refer notes

17. A bus starting from rest moves up a hill with a constant acceleration of 5 m/s2.
Calculate the time taken for the bus to move 170 m up the hill.

[Final Exam Collection – December 2013]

Answer: t = 8.25 s

Chapter 2 - Linear Motion/ Page 75

18. The figure below shows the velocity–time graph of a railway coach from A to D.
Initially, the railway coach decelerates from 40 m/s to 10 m/s in 10 seconds. Then,
the coach moves with uniform velocity and finally accelerates for 7 seconds to reach
a velocity of 30 m/s. The whole journey from A to D is 1000 m. Calculate :

Velocity (m/s)

A D
40
30 B C
7s
10

Time (s)

0 10

a) The acceleration of the coach
b) The distance travelled when the coach is accelerating.
c) The distance travelled when the coach is decelerating.
d) The time taken when the coach is moving at a constant velocity.

[Final Exam Collection – December 2013]

Answer: a) a = 2.86 m/s2 b) d =140 m c) d = 250 m d) t = 61 s

Chapter 2 - Linear Motion/ Page 76

19. Define the following terms with its SI units:
a) Displacement
b) Acceleration

[Final Exam Collection – June 2013]

Answer: Refer notes

20. Give TWO (2) differences between speed and velocity.

[Final Exam Collection – June 2013]

Answer: Refer notes

Chapter 2 - Linear Motion/ Page 77

21. A car moving at a velocity of 2.7 km/hr is stopped after its brakes are applied. The

distance when the brakes are applied till it stopped is 200 m. Determine:

i) the acceleration of the car

ii) time taken to stop the car

[Final Exam Collection – June 2013]

Answer: i) a =−1.41 × 10−3 ii) = 531.91
−2

Chapter 2 - Linear Motion/ Page 78

22. A train moves from station A to D in 40 seconds. During the journey it passed at

station B and C. The time taken for moving from station A to B is 10 seconds with

acceleration of 10 m/s2 and from station B to C in 15 seconds with an acceleration

of 8 m/s2. For the last 15 seconds the train moves with uniform deceleration until it

stops at station D. Determine:

i) Draw a velocity time- graph for the journey

ii) Maximum velocity during the journey

iii) Deceleration for the last 15 seconds

iv) Total distance travelled by the train

[Final Exam Collection – June 2013]

Answer: ii) vmax = 220 m/s iii) a = -14.7m/s2 iv) d = 4550 m

Chapter 2 - Linear Motion/ Page 79

23. Give TWO (2) examples for each of the following terms:
i) Uniform Motion
ii) Non Uniform Motion

[Final Exam Collection – June 2012]

Answer : any answers related in the topics

24. Define and state the SI units for the following terms:

i) Displacement

ii) Velocity

iii) Acceleration

[Final Exam Collection – June 2012]

Answer: Refer notes

Chapter 2 - Linear Motion/ Page 80

25. A car starts from rest and accelerates at a constant acceleration of 5 m/s2 for 10 s.

The car then travels at constant velocity for 15s. The brakes are then applied and the

car stops in 5 seconds.

i) Sketch a velocity-time for the whole journey

ii) Calculate the maximum velocity attained by the car

iii) Calculate the total distance travelled.

[Final Exam Collection – June 2012]

Answer: ii) vmax= 50 m/s iii) d = 1125 m

Chapter 2 - Linear Motion/ Page 81

26. Calculate acceleration of a car that moves from rest and achieves a velocity of 100

km/hr in 15 s.

[Final Exam Collection – June 2012]

Answer : a =1.85 m/s2

27. A car is accelerated at 8 m/s2 from initial velocity of 4 m/s for 10 seconds.

Calculate:

i) The final velocity

ii) The distance travelled by the car

[Final Exam Collection – June 2011]

Answer: i) v = 84 m/s ii) s = 440 m

Chapter 2 - Linear Motion/ Page 82

28. A runner runs from the starting line and achieves a velocity of 20m/s in 2.6
seconds.
i) Calculate his acceleration
ii) If the acceleration is constant, calculate the displacement travelled in 6.5
seconds.

Answer : i) a = 7.69 m/s2 ii) s = 162.45 m

29. Determine the average speed of the following situations :
i) A motorcycle moves 0.5 km to the north within 50 s. Then, it move 1200 m
to the east within 1 minute.
ii) An athlete runs about 1500 m, swims about 1200 m and cycles about 8300 m
within 40 minutes and 10 seconds.

Answer: i) Average speed = 15.45 m/s ii) Average speed = 4.56
m/s

Chapter 2 - Linear Motion/ Page 83

30. A car is moving with the velocity of 80 km/hr. It reduce it’s velocity until it stop
within 15 s. Determine:
i) Acceleration of the car
ii) Velocity of the car after 5s
iii) Displacement of the car within 15s.

Answer: i) = −1.48 / 2 ii) = 14.82 / iii) = 166.8
Chapter 2 - Linear Motion/ Page 84

31. An object started moving with a velocity of 10 m/s. After travelling a distance of
5 m, it gets a velocity of 20 m/s. Find its:
i) Acceleration
ii) Time taken for the journey

Answer : i) a = 30 m/s2 ii) 0.333 s

Chapter 2 - Linear Motion/ Page 85

32. A car starting from rest accelerates uniformly to reach velocity of 72 km/hr after
10 seconds. It continues at this velocity for 10 seconds. It then accelerates again
to reach 35 m/s after a further 15 seconds. Then it starts to decelerate for 3.5 m/s2
until it rest.
i) Calculate the time taken for the car to totally rest
ii) Plot a velocity –time graph for the car.
iii) Calculate the total distance travelled by the car.

Answer : i) t = 45 s iii) d = 887.5 m

Chapter 2 - Linear Motion/ Page 86

Concept Map

Is the rate of
change of

Displacement Velocity

Is the Is the rate of
change of
Distance Acceleration

Travelled Is the rate of
in a change of
specific
Speed
Direction

Is the total
length of the
path travelled
from one

Location To another

And

Scalar quantity Is a quantity that
has

Magnitude

Is a quantity
that has

Vector quantity

Chapter 2 - Linear Motion/ Page 87

3 Force

3.1 Concept of force
3.2 Concept of moment of force

An airplane in flight is always in the middle of
a tug-of-war with the four forces. For an
airplane to takeoff, thrust must be greater than
drag and lift must be greater than weight. To
maintain level flight, lift must equal weight
and thrust must equal drag. For landing, thrust
must be less than drag, and lift must be less
than weight.

Chapter 3 – Force/ Page 88

CHAPTER Force

3

Learning Outcomes
OutOutcomes
Outcomes

A student should be able to:

❖ Define force and its units
❖ Identify the type of force
❖ State the effect of net force
❖ Define the Newton’s First Law (∑ = 0) and Newton’s Second Law (∑ = )
❖ Describe between weight and mass
❖ Define forces in equilibrium
❖ Calculate resultant force by using resolution method
❖ Apply the concept of force in solving problems related to equilibrium of forces
❖ Define moment of force and its unit
❖ Describe principle of moment of force
❖ Apply the concept and formula of moment of force in solving the related problems

Chapter 3 – Force/ Page 89

MIND MAP

FORCE

Newton’s Law of Concept of
Motion Moment

Balanced force Unbalanced force

Body remains Body moved with Body
stationary constant velocity accelerates/decelerates

KEYWORDS Resultant force – Daya paduan
Resolution of force – Leraian daya
Force – Daya Moment – Momen
Newton’s Law – Hukum Newton Principle of Moment – Prinsip
Weight – Berat Momen
Inclined plane – satah condong Centre of gravity – Titik graviti
Balanced force – Daya seimbang
Unbalanced force – Daya tidak
seimbang
Forces in equilibrium – Daya-daya
dalam keseimbangan

Chapter 3 – Force/ Page 90

3.1 CONCEPT OF FORCE

Definition A force is a push or pull from one object to another object

Type of Vector quantity which is have magnitude and direction
Quantity
Unit Newton (N) or −2
Formula
F = ma W = mg

F= Force (N) W= Weight (N)

m = mass (kg) m= mass (kg)

a= acceleration ( −2) g= acceleration of gravitational (9.81 −2)

Effect of force The action of a force can cause:

a) a stationary object to move
b) a moving object to change its speed
c) a moving object to change its direction of motion
d) an object to change in size and shape
e) stop a moving object

Video description: Force and Chapter 3 – Force/ Page 91
weight

An Example Effects of Force

1) A man uses a force to cause a 2) A footballer uses a force to change
stationary trolley to move. the direction of motion of a ball.

3) A cyclist uses a bigger force to pedal 4. A baker uses a force on his dough to
his bicycle and increase the speed. shape it into a curry puff.

5. A driver stops a car at traffic light 6. A plastic ruler can be bent when a
when light turned to red. force is exerted on it.

Chapter 3 – Force/ Page 92

The Differences Between Weight and Mass

The Table 3.1 show the differences between weight and mass.

Table 3.1: Differences between weight and mass

Mass Weight

It is the amount of matter contained in It is the gravitational force acting on an

substance object

It is the same everywhere It changes from place to place

It is measured by equal arm balance It is measured by spring

Its unit is kg Its unit is N

Scalar quantity Vector quantity

Example 1

The Hubble telescope has a mass of 11 600 kg
a) Determine its weight when it is resting on Earth which has gravitational field strength of

9.81 −2
b) What is the value of the gravitational field strength if its weight at a particular orbit above

the Earth is 95 000 N

Solution:
a) Given: m = 11600kg , g = 9.81ms−2 , , ?

= mg
= 11600 × 9.81
=

b) Given: m = 11600kg , w = 95 000 N , , ?
W

= m
95000

= 11600
= . −

Chapter 3 – Force/ Page 93