topik 6 maska

CHAPTER 2

2.0Linear Motion

2.1 APPLY THE CONCEPT OF LINEAR MOTION
2.1.1 Define linear motion

i. Motion means movement.
ii. Linear motion is when an object travels in a straight line.
iii. Non-linear motion is that is not in a straight line.
iv. An object is in motion when it is continuously changing its position relative to a reference point and as

observed by a person

End
En
d

Table 2.1 : Example Linear And Non-Linear Motion

Linear motion example Non-linear motion example

Rolling the bowling ball Snake crawling over branch
The ants are walking in straight line Roller coaster

Bullet was shot Cycling on a winding road

A car is moving in straight line A train moving uphill and downhill

2.1.2 Define uniform and non-uniform motion
i. Uniform motion is a movement in a straight line at a constant speed.
ii. Non-uniform motion is when the direction and the speed of an object are in motion changes.

CHAPTER 2 | fatma azura

Types of View of movement
movement

uniform velocity

Constant velocity =Zero acceleration

uniform
acceleration

Constant acceleration in the same direction as v =Increasing velocity

uniform
deceleration

Constant acceleration in the same direction as v = Decreasing velocity

iii. A vehicle slows down short distance become closer
iv. A vehicle speeds up long distance away each other

2.1.3 Define distance, displacement, speed, velocity, average velocity, acceleration and deceleration

Physical Quantities

Scalar Vector

Scalar quantities are physical Vector quantities are physical
quantities that have magnitude only. quantities that possess both
magnitude and direction.

Table 2.2 : Example Linear and Non-linear motion

Scalar Quantities Vector Quantities
Distance Displacement

CHAPTER 2 | fatma azura

Power Force
Work Momentum
Speed
Velocity

Table 2.3 : Distance and displacement

Quantity Distance (d) Displacement(s)
Definition
Distance is the total length Displacement is the distance
SI Unit travelled by a moving object. travelled in a specified direction
Type of Quantity
meter (m) meter (m)

Scalar quantity Vector quantity

House

Example 1 :
State the situation travels from A to B of bunny and fox.

CHAPTER 2 | fatma azura

Solution :

From A to B :
Distance travelled = 30 m (fox)
Displacement travelled = 16 m, in the direction of East (bunny)

Example 2 :

What was the displacement of the football team if they begin at the school, and run 2 km and finish back at the
school?

Solution :

0 km
Example 3 :
Figure shows a path travelled by Jordon from A to D. What is the displacement of Jordon from point A?

B A
7m
15 m

C D
1m

Solution :

Displacement = √( − ) +
=√ +
= .

Example 4 :

Determine the position and distance of two people pictured below

Solution 2 :

The position of the boy = -3 m or 3m to the left. The position of the man = +4m or 4m to the right.The distance
between men is 7 m.

REMARK!!!
Position is shown with respect to the reference point using either
positive and negative signs or words describing the direction (left,

right, up, down etc)

CHAPTER 2 | fatma azura

2.1.4 Solve the related problems by using kinematic formula: (a. = + , b. 2 = 2 + 2 ,c. = + 1 2
2,d. = 1 2 ( + ) )

i. Speed and Velocity

Table 2.3: Differences between speed and velocity

Quantity Speed Velocity

Definition Speed is the rate of change of Velocity is the rate of change of
distance or distance moved per unit displacement or the displacement of an
time. object per unit time.

Formula =
=



= =

SI Unit
( / ) ( / )

Type of Quantity Scalar quantity Vector quantity

Example The car is travelling at speed 70 km/h The car is travelling at speed 70 km/h
(north)

EXAMPLE 5 :

A long distance runner ran 8 km south and then 10 km east. If the run took him 3 hours, what was his average
speed?

SOLUTION

d1 = 8km x 103
= 8000m

d2 = 10km x 103
= 10000m

3h = 3 x 3600s
= 10800 s

Average speed, v = 8000+10000

10800

= 1.67 ms−1

EXAMPLE 6 :
An object moves from rest with an uniform acceleration of 2 ms −2. What is the velocity of the object after 30 s?
SOLUTION

v uat

v = 0 + (2x30)
= 60 ms −1

ii. Acceleration and Deceleration

CHAPTER 2 | fatma azura

Table 2.4: Differences between acceleration and deceleration

Quantity Acceleration(positive) Deceleration(negative)

Definition Acceleration is defined as the rate of change of velocity.

Formula , = , − ,
SI Unit
,
Type of Quantity
= ( / 2)
( )2

Vector quantity

Speed Object moves with increase in speed. Object moves with decrease in speed.

EXAMPLE 7:

A car is travelling along a straight road 13 ms-1. It accelerates uniformly for 10 seconds until it is moving at 20 ms-
1.

SOLUTION:
−

=
20 − 13

= 10
= 0.7 ms-2

EXAMPLE 8 :

Zahra throws a stone vertically down a well at 12 ms-1. If she hears the stone hit the water 3 s later.
a. How deep is the well?
b. What is the speed of the stone when it hits the water? ( assume g = 9.81 ms-2 )

SOLUTION

a. How deep is the well?

= + 1 2
2
1
= (−12 3) + 2 (−9.81 32)

s = - 80.1m

b. What is the speed of the stone when it hits the water? ( assume g = 9.81 ms-2 )

= +
v  ( 12 )  ( 9.81x3 )
v  ( 12 )  ( 29.4 )
v  41.4
Stone strikes water at v = 41.4 ms -

CHAPTER 2 | fatma azura

EXAMPLE 9 :
A car starts from rest and accelerates in a straight line at 1.6 ms-2 for 10s.

a =1.6 ms-2

a. What is its final speed?
b. How far has it travelled in this time?
c. If the brakes are then applied and it travels a further 20m before stopping. What is the deceleration

(retardation) ?

SOLUTION :
a. What is its final speed?

Initial speed, u = 0 ms-1 Acceleration, a = 1.6 ms-2 Time, t = 10s

V = u +at
= 0 + (1.6 x 10)
= 16 ms-1

b. How far has it travelled in this time?

to give distance gone
= + 1 2

2

= 0 + 1 (1.6 10)2

2

s = 80 m

c. If the brakes are then applied and it travels a further 20m before stopping. What is the deceleration
(retardation) ?

initial speed, u = 16 ms-1 final speed ,v = 0 ms-1 distance, s = 20 m to find acceleration

v2 = u 2 + 2as
0 = 162 + 2a(20)
a = -6.4ms-2

Deceleration is negative acceleration so: Deceleration = 6.4 ms-2

CHAPTER 2 | fatma azura

EXAMPLE 10 :

A cyclist accelerates uniformly at 1.2 ms-2 in 10 s from rest. What is his displacement at this time?Tips: Rest mean
zero velocity !!

SOLUTION : a=1.2 ms-2 v=?
u = 0 ms-1

= + 1 2
2
1
= (10 10) + 2 (1.2 102)

= 60 m (very often, student forget to mention the unit. )

2.2 Solve problems of linear motion from velocity-time graph

i. Motion graphs can tell you how far a body has travelled, how fast it is moving and all the speed changes there
have been.

ii. Graphs Of Linear Motion
- Constant Velocity

CHAPTER 2 | fatma azura

- Positive Constant Acceleration
- Negative Constant Acceleration
2.2.1 Calculate the distance, displacement, speed, velocity, average velocity, acceleration and
deceleration from the velocity-time graph

CHAPTER 2 | fatma azura

EXAMPLE 11 :

The speed versus time graph below represents the motion of a car. Approximately how far did the
car travel during the first 5 seconds?

SOLUTION :

−
=

0 − 40
= 10
a = - 4 ms-2

For motion with constant acceleration we have

= +


After 5 s we have

= (40 5) + 1 (4 52)
2
s = 150 m

During the first 5 seconds the car traveled 150 m.

CHAPTER 2 | fatma azura

EXAMPLE 12 :

The velocity – time graph of particle starting from rest and travelling towards as shown in figure
above. Calculate :
a. The acceleration of the particle in the initial period of 7 s
b. The total travelled by the moving particle

SOLUTION

a. The acceleration of the particle in the initial b. The total travelled by the moving particle
period of 7 s
1
− = 2 ( + )ℎ
= 1
= 2 (12 + 23)5
5−0
=7 = 37.5
= 1.4 ²

2.2.2 Sketch velocity-time graph
EXAMPLE 13 :
Adam is driving his car at 15 m/s. He accelerates at 2 m/s2 for 5 seconds. Adam then maintains a
constant velocity for 15 more seconds.

a) Sketch a velocity-time graph which represents the 20 seconds of Adam’s motion.
b) Calculate the deceleration of the car during the first 5s.

SOLUTION :

a) Sketch a velocity-time graph which represents the 20 seconds of Adam’s motion.

CHAPTER 2 | fatma azura

b) Calculate the deceleration of the car during the first 5s.

= +
= 2(5) + 15
= 10 + 15
= 25 /

= 1 ( + ) = 1 (15 + 25)(5) = 100
2 2

= ( ) = 15(25)= 375 m

Total distance = +
= 100 m + 375
= 475 m

EXAMPLE 14 :

A car starts from rest and accelerates with constant acceleration of 2.4 m/s2 until it reaches a
velocity of 12 m/s after 5s. The car maintains its velocity for 5s and then reduces its velocity
uniformly to 10 m/s in 5s. The car then accelerates uniformly to a velocity of 15 m/s in 5s and
reduces its velocity uniformly for 5s until it stops.

a) Draw a velocity-time graph for the whole journey
b) Calculate the deceleration of the car during the first 5s.
c) Calculate the total distance travelled by the car.

SOLUTION :

a) Draw a velocity-time graph for the whole journey

CHAPTER 2 | fatma azura

b) Calculate the deceleration of the car during the first 5s.

= −



10 − 12
=5
= − . /

c) Calculate the total distance travelled by the car.

11
= 2 ( ) = 2 (12)(5) = 30
= ( ) = 12(5) = 60

11
= 2 ( + ) = 2 (12 + 10)(5) = 55

11
= 2 ( + ) = 2 (10 + 15)(5) = 62.5

11
= 2 ( ) = 2 (15)(5) = 37.5
= + + + +

= 30 + 60 + 55 + 62.5 + 37.5
= 245

EXERCISE :

1. A bus starting from rest moves up a hill with constant acceleration of 5 ms-2. Calculate the
time taken for the bus to move 170 m up the hill.( 8.25s)

2. A car increases its velocity steadily from 72 kmh −1 to 108 kmh −1 in 5 seconds. What is its
acceleration in ms−2? (2ms-2)

3. A school bus accelerates with an acceleration of 4 ms-2 after picking up some students at a bus
stop. Calculate the

a. Velocity of the bus after 5 s (20 ms-1)
b. Distance travelled by the bus after 5s.( 50 m)
4. A car is accelerated at 6 ms-2 from an initial velocity of 2 ms-1 for 10 seconds. Calculate
a. The final velocity, (62ms-1
b. The distance moved(320 m)
5. A remote control toy’s car starts from rest and moves with uniform acceleration of 2 ms-2 for

20 s. What is thedistance moved by the toy’s car? (400 m)

6. A long jumper was running at a velocity of 5 ms-1 towards the long jump pit. He needed to
achieve a velocity of 10 ms-1 after covering a distance of 4.5 m before lifting himself off the
ground from the jumping board.

a. Calculate the required acceleration for him to do so. (8.3 ms-2)
b. Calculate the time taken for him to cover the horizontal distance of 4.5 m. (0.6 s)

CHAPTER 2 | fatma azura

7. Adlina traveling at a velocity of 108 kmh-1 notice a sheep in the middle of the road 80 m in
front of her. Onseeing the sheep, she instantly applies the brakes and is able to bring the car
to a stop after 6 s.

a. What is the deceleration of the car (-5 ms-2)
b. Calculate the distance traveled by the car from the time the driver applies the brakes until it

comesto stop.
(90 m)
c. Is she able to avoid knocking the sheep (can’t avoid)

8. If a van moving at 50 kmh-1 uniformly accelerates up to 70 kmh-1 in 20 s. How far along the
straight road willit travel in that process? (333.3 m)

9. A car can be brake from 100 kmh-1 along 45 m. Find
a. Time for the car stop (3.24 s)
b. The deceleration of the car (-8.57 ms-2)

10. A car starts from rest and reaches at destination at 8 minutes with velocity of 72 kmh-1,
appraise

a. The acceleration of the car in ms-2 (-0.042 ms-2)
b. The distance travelled in km (4.8 km)

11. A car starts from rest and accelerates at a constant acceleration of 2ms-2 for 5 second. The car
then travels ata constant velocity for 9 second. The brake is then applied and the car stops in 6
second.

a. What is the maximum velocity attained by the car (10 ms-1)
b. Plot a velocity-time graph for the whole journey
c. From the graph plotted, determine the total distance travelled (145 m)

12. A train moves from station Sunflower to an Orchid in 40 s. During the journey it stops at
station Rose and Hibiscus. The time taken for moving from station Sunflower to Rose is 10 s
with acceleration of 10 ms-2 andfrom Rose to Hibiscus in 15 s with an acceleration of 8 ms-
2. For the last 15 s the train moves with uniformdeceleration until it stops at station an
Orchid. Determine
a. Maximum velocity during the journey(220 ms-1)
b. Deceleration for the last 15 s(-14.67 ms-2)
c. Total distance travelled by the train(4550 m)
d. Draw the velocity time graph for this journey

13. Four objects move with the velocity versus time graphs shown. Which object has the largest
displacement between t = 0 s and t = 2 s? (2 m2, 4 m2, 2 m2, 2.25m2, B)

CHAPTER 2 | fatma azura

14. Calculate the total distance of the car from the given graph. (100 m)

15. Following the graph shown, calculate acceleration over OP, QR and RS at the given graph
shown. (10 ms-2, -10 ms-2, -10 ms-2)

16. How far did the object (in the graph shows) travel
a. Between 0 and 2 seconds (10 ms-2)
b. Between 2 and 5 seconds (-10 ms-2)
c. Total travelled from 0 to 5seconds(-10 ms-2)

17. A car starts from rest and accelerates at a constant acceleration of 2ms-2 for 5 second. The car
then travels at aconstant velocity for 9 second. The brake is then applied and the car stops in
6 second.
a. What is the maximum velocity attained by the car (10 ms-1)
b. Plot a velocity-time graph for the whole journey
c. From the graph plotted, determine the total distance travelled (145 m)

CHAPTER 2 | fatma azura

18. A lorry moves from a stationary state, undergoes uniform acceleration for 200 m in 5 s. After
5 s, the lorrymoves at a constant velocity for half a minute. The lorry then stops in 10 s. Based
on the given situation, ,
a. Sketch the velocity time graph
b. Calculate the acceleration of the lorry for the first 5 s (16 ms-2)
c. Calculate the deceleration of the lorry (-8 ms-2)
d. Calculate the total travelled of the lorry (3000m)

19. A train travelling at 60 kmh-1 decelerates uniformly to rest at a rate of 2 ms-2. Calculate,
a. The time(8.335 s)
b. Distance taken to stop(694.3 m)

20. vehicle travelling at 1.5 ms-1 suddenly accelerates uniformly to 5ms-1 in 30 s. Calculate

a. The acceleration ( 0.117 ms-2)
b. The average velocity (3.25 ms-1)
c. Distance travelled (97.5 m)

21. A speed of a car travelling along a straight road decreases uniformly from 50 ms-1 to 20 ms-1
over 100m.
Calculate the,

a. The retardation of the car (retardation is equivalent to deceleration) (-10.5 ms-2)
b. Time taken for the speed to decrease from 50 ms-1 to 20 ms-1 (2.86 s)

22. A car starts from 30 ms-1 and accelerates at a constant acceleration of 4 ms-2 for 10 s. Then it
travels at aconstant velocity for 15 s. Then, the brake is applied and the car stops in 9 s.
a. Calculate the maximum velocity attained by the car (7 ms-1)
b. Sketch the velocity time graph for whole journey
c. From the velocity time graph, determine the total distance travelled (1865 m)

23. Figure shows a car moves from point U to point O which is located 50 meter, to the north in
60 second. Then the car moves to point A, 120 meter to the east within 40 second Calculate:
a. Displacement of the car (130m)
b. Average speed of the car (1.70ms-1)

O 120m A
50 m
U

CHAPTER 2 | fatma azura

24. An aircraft accelerates uniformly from rest at 3.1 m/s2 to reach its take-off velocity of 100
m/s.
a. How long does it take for the aircraft to leave the ground? (32.258s)
b. How far does it travel during the take -off? (1612.9m)

25. The graph represents the movement of a car. Answer the questions according to the graph.
a. Determine the initial velocity (10m/s)
b. Calculate the initial acceleration. (3.5m/s2)
c. Determine the value of V if the acceleration is -8 m/s2. (40m/s)
d. Calculate the total distance travelled by the c ar. (1500 m)

Velocity (m/s)

Time (s)

CHAPTER 2 | fatma azura