2.1
2.
Learning Standard
2.1.1 Describe the type of linear motion of an
object in the following states
(i) Stationary (ii) uniform velocity
(iii) non-uniform velocity
2.1.2 Determine:
(i) distance and displacement
(ii) speed and velocity
(iii) acceleration / deceleration
2.1.3 Solve problems involving linear motion
using the following equations:
(i) a = u + at
(ii) s = ½ (u + v)t
(iii) s = ut + ½ at2 (iv) v2 = u2 + 2as
*************************************************************
Introduction
Linear motion is a study of moving object in a
straight line.
How would you describe the motion of the runner
in words? We describe linear motion in terms of
distance, displacement, speed, velocity,
acceleration and deceleration.
Activity 1:
Aim: To study distance and displacement
1. Your task is to put a pencil at point C.
Diagram below shows the difference between
distance and displacement.
Distance = Length of the road
Displacement = Length of the line AB
3. Every day Rahim walks from his house to the
junction which is 1.5 km from his house. Then
he turns back and stops at warung Pak Din
which is 0.5 km from his house.
(a) What is Rahim’s displacement from his
house
(i) when he reaches the junction?
1.5 km to the right / to East
(ii)
(b)
2. With a pencil take a ride from position A to B
and to the final destination, C. Your total path
of length = 1 + 2 = 3 m
this value is known as distance
3.
Repeat step 2 but now from A direct to the
final destination of C. This is your shortest
path to the final destination, C. Value = 1.73.
This value is known displacement
State the definition of distance and displacement
Distance: total path length travel by an object.
Displacement : distance in a specified direction.
When he is at warung Pak Din?
0.5 km to the left / to West
After breakfast, Rahim walks back to his
house. When he reaches home,
(i) what is the total distance traveled by
Rahim?
1.5 + 1.5 + 0.5 + 0.5 = 4.0 km
(ii) what was Rahim’s total displacement
from his house?
1.5 + (-1.5) + (-0.5) + 0.5 = 0 km
4. In a jungle tracking activity, a scout is given a
compass and a map. He starts his journey
from station A and is required to walk to
station B which is located 400 m to the east of
station A. When he reached station B, he is
ordered to go to station C which is 400 m north
from station B.
3.
Give definition of acceleration and its
formula.
Acceleration: Rate of change of velocity
a = change in velocity = final velocity – initial velocity
time
time
a=v–u
t
4.
Observe the motion of the car in Diagram (a),
Diagram (b) and Diagram (c).
(a) When the scout is at station C,
(i) what is the distance he traveled from
station A?
400 + 400 = 800
Rajah (a)
Rajah (b)
(ii) What is the scout’s displacement from
station A? State the magnitude and
direction of the scout from station A.
565.69 to the north east
(b) Why does the answer in (i) and (ii) different?
Displacement has magnitude and direction.
Aktivity 2:
Aim: speed, velocity, uniform velocity, non
uniform velocity, acceleration, deceleration and
zero acceleration
1. Difference between speed and velocity
Speed:
Velocity:
Rate of change of
Rate of change of
distance
displacement
SI unit: ms-1
SI unit: ms-1
Derived quantity and
scalar quantity
Derived quantity and
vector quantity
Formula
v = distance = s
time
t
Formula
v = displacement = s
time
t
Rajah (c)
Describe the changes in displacement and
velocity of a car. Is the car accelerate or
decelerate?
Diagram
Diagram
Diagram
(a)
(b)
(c)
displacement
Equal
Increases decreases
Velocity
Equal
Increases
decreases
Acceleration
Zero
Accelerate
decelerate
5.
A bicyclist starts from rest and increases his
velocity at a constant rate until he reaches a
speed of 4.0 m/s in 5.0 s. What is his average
acceleration?
(a) state the initial velocity = 0
(b) state the final velocity = 4.0
(c) state the time taken = 5.0 s
2.
The speedometer reading for a car traveling
north shows 80 km/hr. Another car traveling
at 80 km/hr towards south. Is the speed of
both cars same? Is the velocity of both cars
same?
Speed equal. Velocity different because
opposite direction.
(d) calculate acceleration = 4.0 – 0 = 0.8 ms-2
5.0
Exercise 1:
1. You walk along a long straight school corridor
for 55 m, then you turn around and walk 30 m
in the opposite. Finally, you turn again and
walk 39 m in the original direction and stop.
What is your displacement from your starting
point?
4. Figure 2.6 shows Radzi’s run from A to B and
then back to C. The total time taken is 20 s.
Determine the
(a) distance
Distance = AB + BC
= 100 + 20 = 120 m
2. A boy walks finish the following path AB.
(c) speed
of Radzi’s motion.
Speed = 120/20
= 6 m/s
(b) displacement
100 + (-20) = 80 to right
(d) velocity
velocity = 80/20 = 4 m/s
5.
Find:
(a) total distance traveled
5 + 7 + 5 + 10 + 10 + 10 + 10 = 57 m
(b) displacement
7 + 10 + 10 = 27 m
3. Fill in the blanks:
(a) Constant speed 10 m/s:
A constant speed of 10 m/s: A distance of 10 m
is traveled every second
(b) Constant velocity 10 m/s:
A steady velocity of 10 m/s: A distance of 10 m
is traveled every second to the right.
(c) Constant velocity – 10 m/s:
A constant velocity of - 10 m/s: A distance of
10 m is traveled every second to the left.
(d) Constant acceleration 4 ms-2 :
Speed increases by 4 m/s every second.
(e) Constant acceleration 4 ms-2 :
A steady deceleration of 4 ms-2 : Speed
decreases by 4 m/s every second.
Muthu moves
from O to B along
the route OAB as
shown in Figure
2.7. The time
taken is 15 s.
Determine the
(a) distance
5 + 12 = 17 m
(b) displacement
13 m
(c) speed
of Muthu’s motion.
v = 17/15 = 1.13 m/s
6.
(d) velocity
v = 13/15 = 0.87 m/s
After landing on the runway, a plane slows
down so that its velocity reduces from 75 m
s–1 to 5 m s–1 in 20 s. What is the acceleration
of the plane?
a = 5 – 75 = - 3.5 ms-2
20
Relating Displacement, Velocity, Acceleration and
Time
Ticker timer:
•
It is connected to an alternating current
power supply of 50 Hz. When it is turned on,
the iron strip will vibrate 50 times per second.
•
The time taken to make 50 dots on the ticker
tape is 1 second. Hence, the time interval
between 2 consecutive dots is 1/50 = 0.02 s.
•
Give definition of one tick.
Time interval between two dots.
Activity 2:
Aim: Method of calculationn
To investigate motion in laboratory mean to
determine distance / displacement, speed/
velocity, time and acceleration/ deceleration
STEP 1: Determination of time taken for 1 tick.
(a) Time taken for 50 ticks = 1.0 s
(b) Time taken for 1 tick = 0.02 s
(c) Time taken from A to B = 10 ticks = 0.2 s
STEP 2 : Determination of displacement
The displacement of the object is determined by
measuring the length of the ticker tape that is
pulled through the ticker time
Displacement from A to B = 8.0 cm
STEP 3 : Determination of velocity
Velocity, v = displacement
Time
= 8.0 = 40.0 cm/s
0.2
STEP 4 : Determination of acceleration
The first strip: Initial velocity, u at AB = 1.5 = 7.5
0.2
The last strip: Final velocity, v at DE = 7.5 = 37.5
0.2
The time interval for, t
= (4 – 1) x 0.2 = 0.6 s
the change in the velocity
Acceleration, a = 37.5 – 7.5 = 50 cms-2
0.6
2.
Calculate the acceleration .
u = 0.2 = 10 cm/s v = 1.4 = 70 cm/s
0.02
0.02
t = (5 – 1) x 0.02 = 0.08 s
a = 70 – 10 = 750 cms-2 = 7.5 ms-2
0.08
3.
A ticker tape below contains 5 ticks for every
interval AB,BC,CD and DE. Calculate the
acceleration.
u = 8.0 = 80 cm/s v = 2.0 = 20 cm/s
0.1
0.1
t = (4 – 1) x 0.1 = 0.3 s
a = 20 – 80 = - 200 cms-2 = - 2.0 ms-2
0.3
Making a speed-time graph
Ticker tape gain from the experiment:
3.
The diagram above shows a ticker tape chart
for a moving trolley. The frequency of the
ticker-timer used is 50 Hz. Each section has
10 dots-spacing.
5. To identify the types of motion
Distance between the
dots: equal
Type of motion:
Constant velocity
Distance between the dot
increases Uniformly
The velocity is of the
object is increasing
uniformly
(a)
What is the time between two dots?
0.02 s
(b)
What is the time for one strips.
10 x 0.02 = 0.2 s
(c)
What is the initial velocity
2.0 = 10 cm/s
0.2
(d)
What is the final velocity.
12.0 = 60 cm/s
0.2
The object is moving at a
uniform acceleration
Distance between the
dots decreases uniformly
The velocity of the object
is decreasing uniformly
The object is
(e)
What is the time interval to change from
initial velocity to final velocity?
t = (11 – 1) x 0.2 = 2.0 s
(f)
What is the acceleration of the object?
a = 60 – 10 = 25 cms-2
2.0
4. The figure shows a tape chart.
Calculate
(a) the acceleration
(b) the average velocity
(a) u = 4/0.2 = 20 cm/s
v = 24/0.2 = 120 cm/s
t = (6 – 1) x 0.2 = 1.0
a = 120 – 20 = 100 cms-2 = 1 ms-2
1.0
(b) total distance = 4 + 8 + 12 + 18 + 20 + 24 = 84
Average speed = 84 = 70.0 cm/s
1.2
experiencing uniform
decelaration.
Activity 4: To determine displacement, average
velocity and acceleration
Apparatus: Ticker timer, trolley, 12 V power
supply, runway, ticker tape, ruler
Procedure:
1. Raise one end of runway to a reasonable
height.
2. Pass the ticker tape through the ticker timer
and attach it to a trolley at the top of the
runway.
3. Switch on the ticker timer and release the
trolley.
4. When the trolley comes to a stop, cut the
tape.
5.
6.
Mark and cut the tape into 10-tick strips from
the start of the first clear dot
Paste the 10-tick strips side-by-side on a
paper to make a tape chart.
Find
1st strip
Last
strip
The displacement of
the 10-tick
16.1 cm
39.3 cm
The time covered for
the 10-tick strip
0.2 s
0.2 s
Average velocity
over the 10-tick
Strip
16.1
0.2
= 80.5
cm/s
39.3
0.2
= 196.5
cm/s
Change in velocity
between the two 10tick strip
Time taken for the
change in velocity
196.5 – 80.5
= 116.0 cm/s
(5 – 1) x 0.2 = 0.8 s
Acceleration
116.0 = 145.0 cms-2
0.8
Conclusion:
The acceleration of the trolley is 145.0 cms-2
Activity 5:
Aim: Derive linear motion equations and solve
problems
First linear motion equation
a=v–u
t
v – u = at
v = u + at (1)
Second linear motion equation
Displacement = average velocity x time
1. A car accelerates from rest to 25 m s-1 in 4 s.
Find the acceleration of the car.
u = 0, v = 25, t = 4, a = ?
v = u + at
a = (25 – 0) = 6.25 ms-2
4
2.
When a racing car passes through on a
straight track, its velocity is 40 ms-1. After 3
seconds, the racing car achieves the speed
50 ms-1. Calculate the displacement travelled.
u = 40, v = 50, t = 3, s = ?
s = ½ (u + v)t
= ½ (40 + 50)3 = 135 m
3.
The 100 m men world record holder, Usian
Bolt starts his run from rest and achieved his
maximum velocity after he accelerating
uniformly for 9.58 s. Determine his
acceleration.
s = 100 m u = 0 t = 9.58 s a = ?
s = ut + ½ at2
100 = ½ a(9.58)2
a = 2.18 ms-2
4.
Maria rides a bicycle at a velocity of 8 m s–1.
She brakes suddenly and stops after a
distance of 2 m. What is the acceleration of
Maria and her bicycle?
u = 12 v = 0 s = 2 a = ?
v2 = u2 + 2as
0 = 144 + 2(a)2
4a = - 144
a = -16 ms-2
5.
A car moving along a straight road at a
velocity of 30 m s–1 reduces its velocity at a
constant rate until it stops after 5 s. What is
the acceleration of the car?
u=0 a=4 t=5 v=?
v = u + at = 0 + (4)(5) = 20 cm/s
6.
A boy is cycling
down a hill. His
initial velocity is 4
ms-1. After he moves
35 m, his velocity
becomes 10 m s-1.
(a) What is the time
for the boy to
travel at a distance 35 m?
u = 4, v = 10, s = 35 t = ?
s = (u + v) t
2
35 = (4 + 10) t
2
t=5s
displacement
= [initial velocity + final velocity] x time
2
s = ½ (u + v) t
Third linear motion equation
v = u + at
(1)
s = ½ [u + v]t
(2)
Substitute equation (1) into equation (2)
s = ½ [u + u + at]t
s = ½ [2u + at]t
s = ½ [2ut + at2]
s = ut + ½ ut2
(3)
Fourth linear motion equation
Square equation (1) v = u + at
v2 = (u + at)2
v2 = u2 + 2uat + a2t2
v2 = u2 + 2a(ut + ½ at2)
at2
daripada (3) s = ut + ½
v2 = u2 + 2as (4)
v = u + at
.(1)
Where
s :displacement
s= u+v t
2
.(2)
s = ut + ½ at2
.(3)
u :initial velocity
v :final velocity
a :acceleration
v2 = u2 + 2as
.(4)
t :time
(b) What it is his acceleration?
u = 4, v = 10, t = 5, a = ?
v = u + at
a = (10 – 4)
5
a = 1.2 ms-2
7.
A car accelerates from rest at 3 ms-2 along a
straight road. How far has the car traveled
after 4 s?
u = 0, a = 3, t = 4, s = ?
s = ut + ½ at2
= 0 + ½ (3)(16) = 24 m
8. A car is traveling at 20 m/s along a straight
road. The driver puts the brakes on for 5 s. It
this causes a deceleration of 3 m s-2 , what is
the car’s final velocity?
u = 20, t = 5, a = -3, v = ?
v = u + at = 20 + (-3)(5) = 5 m/s
9.
A car moving with constant velocity of 40
ms-1. The driver saw and obstacle in front
and he immediately stepped on the brake
pedal and managed to stop the car in 8 s.
The distance of the obstacle from the car
when the driver spotted it was 180 m. How
far is the obstacle from the car after it stops.
u = 40, v = 0 , t = 8 , s = ?
s = (40 + 0) 8 = 160 m
2
the obstacle from the car after it stops:
180 – 160 = 20 m
10. You can determine your reaction
time by catching a ruler between
your fingers which is released.
Ready to catch the ruler by
opening your fingers at the ‘zero’
mark of the ruler. Catch the ruler
as soon as it is released. Mark the
position where you catch the ruler.
The ruler falls with acceleration 10
m s-2, calculate your reaction time.
u = 0, a = 10, s = 20 cm = 0.2 m t = ?
s = ut + ½ at2
0.2 = 0 + ½ (10)t2
t2 = 0.2/5 = 0.04
t = 0.2 s
TUTORIAL 2.1
1. Which physical quantity is equal to
displacement?
Time
A. Speed
C. Distance
B. Velocity
D. Acceleration
2. The acceleration of a car moving with a
constant velocity will be
A. increased
B. constant
C. zero
D. decreased
3. Muthu cycles from his house to the shop.
While coming back, he stops at Ahmad’s
house.
What is his displacement from his house?
A. 300 m
B. 400 m
C. 1 100 m
4. Which of the following is true regarding the
motion of an object having zero acceleration?
A. The object is not moving
B. The object is moving with uniform
velocity
C. The object is at rest or moves with
uniform velocity
D. The object is moving with maximum
velocity.
5. A tick from a ticker timer is
A. time interval between two consecutive
dots on a ticker tape.
B. distance between two consecutive dots
on a ticker tape.
C. frequency of vibration from the ticker
timer
D. velocity of vibration from the ticker timer.
6. Which statement is true about the ticker tape
shown below?
A.
B.
C.
D.
Velocity between CD is low.
Velocity between DE and velocity
between AB are equal.
Frequency between BC is higher than DE
Time between AB is equal with the time
between DE
A. 20 cm s-2
B. 200 cm s-2
C. 240 cm s-2
D. 400 cm s-2
u = 4/0.1 = 40 v = 24/0.1 = 240
t = (6 – 1) x 0.1 = 0.5 s
a = (240 – 40)/0.5 = 400 cms-2
7. The diagram shows a ticker tape which is
pulled by a trolley through a ticker-timer of
frequency 50 Hz.
The average speed of the trolley’s motion is
A. 0.2 ms-1
C. 0.3 ms-1
B. 0.4 ms-1
D. 0.5 ms-1
Average speed = 4/0.1 = 40 cms-1
8.
Diagram below shows the path travelled by
a car from P to S.
Average velocity from P to Q in the ticker
tape below is
A.
B.
9.
85 cms-1
C. 170 cms-1
-1
200 cms
D. 240 cms-1
Total distance = 2 + 3 + 4 + 2 + 6 = 17 cm
Total time taken = 5 x 0.02 = 0.1 s
v = 17/0.1 = 170 cms-1
Calculate the acceleration.
A.
B.
10.
11.
- 50 ms-2
C. - 500 ms-2
-2
50 ms
D. 500 ms-2
u = 2.5/0.02 = 125 v = 0.5/0.02 = 25
t = (11 – 1) x 0.02 = 2.0 s
a = (25 – 125)/2.0 = - 50 cms-2 = - 0.5 ms-2
The diagram below shows a tape chart
which is produced by a moving trolley. The
frequency of the ticker-tape timer is 50 Hz
and each stripe of ticker tape contains 5
ticks.
What is the trolley’s acceleration? (1994)
What is the displacement of the car? (2007)
A. 5.0 km
B. 6.8 km
C. 8.2 km
D. 9.0 km
12. Diagram 2 shows Ali stands at O. He walks
towards A, then moves towards B and stops
at B.
What is the displacement of Ali? [2012]
A. 2 m towards west
B. 5 m towards east
C. 7 m towards east
PAPER 2, SECTION A
1. Question 1: Melaka Mid 08
Ali walks to the north for a distance of 300 m. He
then turns west and walks for another distance of
400 m to reach Pak Kassim’s stall.
Diagram 1 shows the path taken by Ali.
(a)
Distance is a ………
Tick (√) the correct answer in the box
below
√
3. SPM 2016
Diagram 1 shows a ticker tape with 5 ticks
obtained from an experiment.
Scalar quantity
Vector quuantity
(b)
(c)
(d)
On Diagram 1, mark an arrow to show the
displacement made by Ali
[1 mark]
What is the total distance travelled by Ali?
300 + 400 = 700
[1 mark]
Ali then walks home for another distance of
500 m. What is the total displacement made
by Ali. [1 mark]
0
2. SPM 2011 Question 2
Diagram 1.1 shows a trolley moving down an
inclined plane. The ticker timer vibrates at
frequency 50 Hz. Diagram 1.2 shows the ticker
tape produced by the motion of the trolley.
(a) One tick is the time taken from P to Q [1 mark]
(b) Complete the following sentence by ticking (√)
the correct answer in the box provided.
The ticker tape can be analysed to determine
……√….
Distance and speed of an object
……….
Force and momentum of an object
[1 mark]
(c) (i) What is the type of motion shown by the
ticker tape in Diagram 1?
Constant velocity
[1 mark]
(ii) Give one reason for your answer in 1(c)(i).
The distance between the ticks are equal.
[1 mark]
4. SPM 2019
Diagram 5.1 and Diagram 5.2 show two ticker
tapes with different lengths produced by the
motion of two trolleys.
Diagram 1.1
Diagram 5.1
Diagram 5.2
Diagram 1.2
(a) Underline the correct answer in the bracket to
complete the sentence below.
The type of current used in the ticker timer is
(direct current, alternating current).
[1 mark]
(b)
(i)
Based on Diagram 1.2,
one tick is represented by the time taken
from point P to point Q
[ 1 mark]
(ii)
Compare the distance between PQ and ST.
ST > PQ
1 mark]
(iii) State the type of motion of the trolley.
Accelerate
[1 mark]
(a) What is the meaning of length?
Distance between two points.
[1 mark]
(b) Based on Diagram 5.1 and Diagram 5.2,
compare
(i) the number of tickes
Equal
(ii) the length of the ticker tapes
5.2 > 5.1
(iii) average speed
5.2 > 5.1
[3 marks]
(c) Using your answer in 5(b), state the
relationship between the length of a ticker
tape and average speed.
Directly proportional
[1 mark]
(d) Diagram 5.3 shows a ticker tape chart
produced from the motion of a trolley.
Diagram 5.3
Based on Diagram 5.3, explain the type of
motion from W to Y.
•
•
•
•
Velocity increase constant // uniform
acceleration, a
Increasing in length uniformly
zero acceleration // constant velocity
uniform length