2.1 Linaer Motion FIZIK

SMK TAT BENGSMK TAT BENG

PHYSICS PHYSICS FORM 4FORM 4

2.1 Linear Motion 2.1 Linear Motion

Linear Motion (Pergerakan Linear) Linear motion is motion in a straight line. Examples of linear motion are:-

Examples of non-linear motion are:-

A car moving in a straight line A falling coconut

A train moving in a straight line A moving bullet

A snake crawling A roller-coaster ride

Linear Motion (Pergerakan Linear) Dynamics is the study of the motion of an object due to a

force. Kinematics is the study of the motion of an object without

being concerned with the forces which cause the motion.

All physical quantities can be grouped as scalar quantities and vector quantities.

Linear motion

Kinematics – without consideration of force

Dynamics – force are importance

Linear Motion (Pergerakan Linear) A scalar quantity is a quantity which has magnitude only. A vector quantity is a quantity which has both magnitude

and direction.

(a) Ali moves three steps.

(b) Ali move three step to the right.

Example Scalarquantity

Vectorquantity

Magnitude

three

Unit

Magnitude

three

UnitDirection

right

Distance and Displacement (Sesaran)Distance Distance is the total length of the path travelled by an object. Distance does not take into account the direction travelled by the

object. Distance is thus a scalar quantity.

Displacement The displacement of an object is the distance of its final position

from its initial position in a specified direction. (jarak yang dilalui dalam arah yang tertentu)

Displacement is the length of the straight line connecting the two locations, in a specified direction.

Displacement is a vector quantity. Both distance and displacement have the same SI unit which is

metre, m.

Distance and Displacement (Sesaran)Example

Figure above shows a school bus travelling from location A tolocation H to pick up some students. Find the...(a) total distance travelled by the bus,(b) Displacement of the bus

Distance and Displacement (Sesaran)Solution (a) Distance is defined as the total length of the path travelled from

one location to another.The total distance travelled by the bus using the path A-B-C-D-E-F-G-H = 160 m + 160 m + 160 m + 160 m + 240 m + 140 m + 80 m = 1100 m.

(b) Displacement is the length of the straight line connecting the two locations, in a specified direction. For the bus travelling from A to

H, its displacement is determined by measuring the length of the straight line connecting A to H and the direction of H from A as shown in Figure 2.3 Hence, the displacement is 180 m, in the north direction of A.

Distance and Displacement (Sesaran) Distance is a scalar quantity while displacement is a

vector quantity.

Examples of scalar quantity and vector quantity

Scalar quantity Vector quantity

Distance

Area

Speed

Displacement

Acceleration

Velocity

Displacement, distance, acceleration, area, velocity, speed

Distance and Displacement (Sesaran)THE DIFFERENCES BETWEEN DISTANCE AND

DISPLACEMENT

Aspect Distance

Distance taken with consideration of

Direction.

Total route taken bya motion.

Vector quantity, direction & magnitude

are importance.

Scalar quantity,with magnitude only.

Definition

Type of quantity

SI unit Metre (m)

Displacement

Example 1Example 1

Figure above shows a car moving round a big roundabout which has a radius of 70 m. Calculate...(a) the distance moved by the car,(b) the displacement of the car.

Example 1Example 1

Solution (a) The distance moved by the car is the circumference of

circle with radius 70 m. Circumference = 2πr

= 2 x (22/7) x 70m = 440m

(b) • Consider a car moves from point A and makes one complete circle.

• This means that point A is also the end point of the motion. Hence the displacement of the car is zero.

Example 2Example 2A cow moves 3 m to the east and then 4 m to the north. Find the...(a) total distance moved by the cow,(b) Displacement of the cow

Solution(a)

(b)

The total distance moved by the cow = Length of AB + Length of BC= 3 m + 4 m= 7 m

Let the starting point of the cow be A and the end point be C. The displacement is calculated from A to C, in the direction of AC. The displacement is 5 m in the direction of AC.

Example 3Example 3

Every morning Amirul walks to Boon Hong's house which is situated 80 m to the east of Amirul's house. They then walk towards their school which is 60 m to the south of Boon Hong's house. What is the distance travelled by Amirul and his displacement from his house?

Solution

Example 3Example 3SolutionDistance travelled by Amirul = Total length of the path travelled = PQ + QR = 80 m+ 60 m = 140 m

Displacement of Amirul= Distance of Amirul's final position, R from his initial position,P== 100mtan θ = 60 / 80 = 36.9o

The displacement of Amirul is 100 m on a bearing of 90° + 36.9° = 126.9° from his house.

m22 6080

Example 4Example 4If a boy takes 10 s to finish each of the following paths AB, find the total distance travelled, and the displacement in each case. (Take π = 3.14)

Solution:(a) (i) Total distance travelled = (2 + 4 +6+8 + 6 +2)m

= 28 m(ii) Displacement = the shortest distance between A and B = (4 + 8) m

= 12 m

(b)

Example 4Example 4Solution:(b) (i) Total distance travelled = (π x 2m) + (π x 4m)

= 18.84m(ii) Displacement = the shortest distance between A and B = (2 + 2 + 4 + 4) m

= 12 m

Speed and Velocity Speed and velocity are physical quantities which measure

how fast an object is moving. The speed limit on the highway in Malaysia is 110kmh-1.

We cannot travel more than 110km in an hour. But we can travel in any direction.

“An athlete is running towards the checkpoint at a velocity of 24ms-1. When we talk about velocity, we should indicate the direction (towards the checkpoint) and also the magnitude (24ms-1).

Speed is a scalar quantity while velocity is a vector quantity.

Speed is the rate of change of distance (kadar perubahan jarak terhadap masa) while velocity is the rate of change of displacement (kadar perubahan sesaran terhadap masa).

Speed and Velocity In mathematical writing:

Both speed and velocity have the same SI unit. They are measured in ms-1 or metres per second.

)(,)(,

sttakentimeTotalmstravelleddisanceTotalSpeed

)(,)(,

sttakentimeTotalmsntdisplacemeTotal

Velocity

Speed and Velocity

The car moves forward at a speed of 2.2 ms-1 or at a velocity of 2.2 ms-1.

The car moves backwards at a speed of 2.2 ms-1 or at a velocity of -2.2 m s-1

The car is moving round a circle with an average speed of 44 ms-1,

but the velocity is 0 ms-1.

Comparison Between Speed and Velocity

Aspect Speed

Rate of change of displacement

Rate of change of distanceDefinition

Type of quantity

Velocity

Vector quantityScalar quantity

Formula

SI unit ms-1ms-1

)(,)(,

sttakentimeTotalmsntdisplacemeTotal

Velocity

)(,

)(,sttakentimeTotalmstravelleddisanceTotal

Speed

Example 5Example 5A man running in a race covers 60 m in 12 s.(a) What is his average speed in :- (i) ms-1, (ii) kmh-1

(b) If he takes 40 s to complete the race, what is his distance covered?(c) Another man runs with a speed of 7.5 m s-1, how long did he take to complete the race?

Solution

151260

tan.)(

ms

takentimetravelledcedistotalspeedaverageia

Example 5Example 5Solution

11860601

10005

.)(

kmhx

speedAverageiia

mx

timexspeedceDistime

cedisSpeedib

20045

tan

tan.)(

s

speedcedistime

timecedisSpeediib

7.265.7

200

tan

tan.)(

Example 6Example 6In a 200 m swimming competition. a swimmer takes 12.2 s for the first lap, 12.6 s for the second lap, 12.8 s for the third lap and 12.5 s for the fourth lap. One lap is 50 m long. Find the average speed of the swimmer.

Solution

199.35.126.128.122.12

200

tan.)(

ms

takentimetotaltravelledcedistotalspeedAverageiia

Example 6Example 6A cow walked along a curved path from P andended at Q which is 70 m away from P. Q is atthe south-west of P. The distance travelled bycow is 240 m and the time taken is 160 s

Calculate the(a) Average speed(b) Average velocity Of the cow when moving from P to Q

Example 6Example 6SolutionTotal distance travelled = 240mDisplacement = 70mTime taken = 160s

15.1160240

tan)(

ms

takentimetotaltravelledcedistotalspeedAverageia

Pofdirectionwestsouththeinmssm

takentimetotaltravelledntdisplacemetotalvelocityAverageb

144.016070

)(

Acceleration and Deceleration

Acceleration is the rate of change of velocity.

The SI unit of acceleration is ms-2

Acceleration is a vector quantity. Physically, when an object moves faster and faster it is

moving with acceleration. Object that moves slower and slower is moving with

deceleration or negative acceleration.

)()()()(

ttakenTimeuvelocityInitialvvelocityFinalaonAccelerati

Example 7A runner runs from the starting line and achieves a velocity of 18 m s-1 in 3 seconds. Calculate his acceleration.

Solution

2

1

63)018(

)()()()(

mssms


Example 8A van accelerated uniformly from a velocity of 15 ms-1 to 20 ms-1 in 2.5 s. What was the acceleration of the van?

Solution Initial velocity, u = 15 ms-1

Final velocity, v = 20 ms-1

Time take = 2.5 s

2

1

0.25.2)1520(

)()()()(

mssms


Example 8A car travelling at 24 m s-1 slowed down when the traffic light turned red. After undergoing uniform deceleration for 4 s, it stopped in front of the traffic light. Calculate the acceleration of the car.Solution Initial velocity, u = 24 ms-1

Final velocity, v = 0 ms-1

Time take = 4 s

2

1

0.64)240(

)()()()(

mssms


A negative value indicates deceleration

Example 9A car increases its velocity steadily from 72 kmh-1 to 108 kmh-1 in 5 s. What is its acceleration in m s-2?

Solution

1

1

20

6060100072

72

ms

xxkmhu

1

1

30

60601000108

108

ms

xxkmhv

2

1

25)2030(

)()()()(

mssms


Example 10

An object moves from rest with a uniform acceleration of 2 m s-2. What is the velocity of the object after 30 s?


Time take = 30 sAcceleration = 2 m s-2

160

)302(0

)()()()(

ms

xatuv


Example 11A car moving at a constant velocity of 30 cm s-1 came to a stop 6 s after its brakes were applied. What was the deceleration of the car?


final velocity, v = 0 ms-1

Time take = 6s

2

2

1

0.5

0.56)300(

)()()()(

msonDecelerati

mssms


Study of Motion with the Ticker Timer A ticker timer is used A ticker timer is used to study the motion of an object for a short to study the motion of an object for a short

period of timeperiod of time.. Figure below shows the ticker timerFigure below shows the ticker timer

When the terminals are connected to the a.c. power supply, the When the terminals are connected to the a.c. power supply, the vibrating steel strip (hitter) will vibrate vibrating steel strip (hitter) will vibrate 50 times every second50 times every second and and make 50 dots a second on a tape being pulled through it.make 50 dots a second on a tape being pulled through it.

Study of Motion with the Ticker Timer The time interval between two consecutive dots (time for The time interval between two consecutive dots (time for

one tick) is one tick) is 1 / 50 s1 / 50 s or or 0.02 s0.02 s..

Method To Calculate timeMethod To Calculate time1 dot 1 dot

B C

1 tick

(2 dot form 1 tick)

Time taken from B to C = 1 tick = 1 x 0.02s = 0.02 s

Method To Calculate timeMethod To Calculate time

Time taken from A to E = 4 tick = 4 x 0.02s = 0.08 s

B C DA E

4 tick (formed by 5 dots)

B D FA C E G

Time taken from A to E = 6 tick = 6 x 0.02s = 0.12 s

6tick (formed by 7 dots)

Method To Calculate timeMethod To Calculate time

Time interval = number of dot space x 0.02 s

The ticker timer can be used to determine the...(a) time interval of motion,(b) displacement of the object,(c) velocity of the object,(d) acceleration of the object, (e) type of motion of the object.

Types of Motion (Patterns)

.The distance between the dots is the same. It shows that the body is moving with constant speed.

Direction of motion

Initial Final


.The distance between the dots is increasing. It shows that the body is moving faster.

Direction of motion

Initial Final


.The distance between the dots is decreasing. It shows that the body is decreasing.

Direction of motion

Initial Final


.The distance between the dots is short. It shows that the body is moving very slowly.

Direction of motion

Initial Final


.The distance between the dots is long. It shows that the body is moving very fast.

Direction of motion

Initial Final

Determination of the Average Speed of a Trolley by Using the Ticker Timer

A B C D E F G H JI

18cm

Solution stepGuide

Time taken = number of ticks x 0.02 = 9 x 0.02s = 0.18s

1. Determine the time (the number of ticks)

Total distance travelled is 18cm2. Determine the distance travelled

Speed = 18cm / 0.18s = 100cms-13. Speed = Distance / time taken

Determination of the Acceleration of a Trolley by Using the Ticker Timer

A B C D E F G

5cm1cm

Direction of motion

Solution stepGuide

One part = 2 ticks = 2 x 0.02s = 0.04s

1. Determine the time of one part

u = 1cm / 0.04s = 25cms-1

2. Find initial velocity, u = S initial / t initial

Determination of the Acceleration of a Trolley by Using the Ticker Timer

Solution stepGuide

v = 5cm / 0.04s = 125cms-1

3. Find final velocity, u = S final / t final

Total time, t = (6 - 1) x 0.04s = 5 x 0.04s = 0.2s

4. Determine the total time Total time = (total parts -1) x time of one part

Acceleration, a = (v – u ) / t = 125 – 25 / 0.2 = 500cms-1

5. Applying the formula, a = (v – u) / t

Example 12Example 12

Based on the ticker tape shown in Figure above, find...(a) the acceleration.(b) the average speed.

Example 12Example 12Solution(a) Time taken for 1 part = 1 tick = 0.02 s

Initial velocity, u = 4 cm / 0.02 s = 200 cm s-1

Final velocity, v = 6 cm / 0.02 s = 300 cm s-1

t = (3 - 1) x 0.02 s = 0.04s

Acceleration, a = (300cms-1 – 200cms-1) / 0.04 = 2500 cms-2 / 25 ms-2

Example 12Example 12Solution(b) Total distance = 4 cm + 5 cm + 6 cm = 15 cm

Total time = 3 parts x 0.02 = 0.06 s

Average speed, v = s / t = 15cm / 0.06s = 250cms-1


Figure above shows a tape chart consisting of six consecutive sections of a ticker tape which are glued side by side to show the motion of a trolley down a sloping runway. Every section of the ticker tape has ten ticks. The frequency of the a.c supply is 50 Hz. Find...(a) the total displacement,(b) the average velocity,(c) the acceleration.

Example 13Example 13Solution(a) Total displacement = 35 + 30 + 25 + 20 + 15 + 10 = 135 cm

(b) Total time = 6 x 10 x 0.02 s = 1.2 s

Average velocity = Total displacement / total time = 135 cm / 1.2 s = 112.5 cms-1

6

Sections of 6 tapes

10

Each section have 10 ticks

0.02sTime for 1 ticks

Example 13Example 13Solution

(c) Initial velocity, u = 35 cm / (10 x 0.02 s) = 175.0 cms-1

Final velocity, v = 10 cm / (10 x 0.02) s = 50.0 cms-1

Total time, t = (6 -1) strips x (0.02 x 10) = 5 x (0.02 x 10)

= 1.0 s

Acceleration, a = (v – u) / t = 50.0 cms-1 -175.0 cms-1) / 1.0 s = -125.0 cms-1


Figure above shows a section of a ticker-tape that was pulled through a ticker-timer by a trolley as it moved down a raised runway. What is the acceleration of the trolley if the ticker-timer was connected to a 50 Hz alternating current power supply?.

Example 14Example 14Solution(a) Initial velocity, u = 0.4 cm / 0.02 s = 20.0 cms-1

Final velocity, v = 2.0 cm / 0.02 s = 100.0 cms-1

Time interval, t = (6 - 1) x 0.02 s = 0.1s

Acceleration, a = (100 cms-1 – 20 cms-1) / 0.1 s = 800 cms-2 / 8.0 ms-2


Figure above shows a strip of ticker tape depicting the motion of a toy car with uniform acceleration. Determine the acceleration of the toy car.

Example 15Example 15Solution(a) Initial velocity, u = 0.2 cm / 0.02 s = 10.0 cms-1

Final velocity, v = 1.6 cm / 0.02 s = 80.0 cms-1

Time interval, t = (8 - 1) x 0.02 s = 0.14 s

Acceleration, a = (80 cms-1 – 100 cms-1) / 0.14 s = 500 cms-2 / 5.0 ms-2


Figure above shows a chart representing the movement of a trolley with uniform deceleration. Determine its acceleration

Example 16Example 16SolutionTime for each 10 tick strip = 10 x 0.02 s = 0.2 s

u = 7.2 cm / 0.2 s = 36.0 cms-1

v = 1.2 cm / 0.2 s = 6.0 cms-1

Total time, t = (6 - 1) x 0.02 s x 10 = 1.0 s

Acceleration, a = (6.0 cms-1 – 36.0 cms-1) / 1.0 s = - 30.0 cms-2

Equations of Motion Problems on linear motion with uniform acceleration can

often be solved quickly using the equations of motion. The following symbols are used in the equations of motion: u = initial velocity v = final velocity t = time a = acceleration s = displacement

First equation of motion a = (v - u) / t at = v – u v = u + at ----------------- (1)

Equations of Motion Second equation of motion Average velocity = (u + v ) / 2 Displacement = average velocity x time S = [( v + u / 2 )] t = [( u + u + at ) / 2] t S = ut + ½ at2 ---------------- (2)

Third equation of motion s = average velocity x time s = [(u + v) / 2] t Substituting t = (v – u) / a s = [(v + u) / 2] x [(v – u) / a] = (v2 – u2) / 2a 2as = v2 – u2

v2 = u2 + 2as ---------------- (3)

uv = u + at

a = (v – u) / t

Equations of Motion Fourth equation of motion s = average velocity x time = [(u + v ) / 2] x t s = ½ (u + v) t ----------------- (4)

Example 17Example 17A school bus accelerates with an acceleration of 4.0 ms-2 after picking up some students at a bus stop. Calculate the(a) velocity,(b) distancetravelled by the bus after 5 s.

Solution (a) Initial velocity, u = 0 ms-1

Acceleration , a = 4.0 ms-2

Time, t = 5 s Using equation (1) v = u + at = 0 + (4.0 x 5) = 20.0 ms-1

Example 17Example 17Solution (b) Using equation (2) s = ut + ½ at2

= (0 x 5) + ( ½ x 4.0 x 52) = 50.0 m

Example 18Example 18In the long jump event, Jerak was running at a velocity of 5ms-1 towards the long jump pit. He needed to achieve a velocity of 10ms-1 s after covering a distance of 4.5 m before lifting himself off the ground from the jumping board.

(a) Calculate the required acceleration for Jerak to do so.(b) Calculate the time taken for him to cover the horizontal distance of 4.5 m.

Example 18Example 18Solution (a) Initial velocity, u = 5 ms-1

Final Velocity, v = 10 ms-1

s = 4.5 m Using equation (3) v2 = u2 + 2as 102 = 52 + (2 x a x 4.5) a = 8.3 ms-2

(b) Using equation (4) s = [(u + v) / 2] t 4.5 = [(5 + 10) / 2] x t t = 0.6 s

Example 19Example 19A train is moving at a velocity of 25 m s-1, and it brakes with a deceleration of 2 m s-2 before it stops. Find(a) the time taken for the train to stop,(b) the braking displacement of the train.

Solution:Given u = 25 m s-1; v = 0 m s-1; a = –2 m s-2

(a) By v = u + at 0 = 25 + (-2) x t Thus, t = 12.5 s(b) By v2 = u2 + 2as 02 = 252 + 2 x (-2) x s Thus, s = 156.25 m

Example 20Example 20A car is accelerated at 6 ms-2 from an initial velocity of 2 ms-1 for 10 seconds. What is(a) the final velocity, and(b) the distance moved?

SolutionGiven a = 6 ms-2 , u = 2.0 ms-1, t =10 s(a) Final velocity, v = u + at = 2 + (6 x 10) = 62 ms-1

(b) Distance moved, s = ut + ½ at2 =(2 x 10)+ ½ (6 x 102) = 320 m

Example 21Example 21A driver travelling at a velocity of 108 kmh-1 notices a cow in the middle of the road 80 m in front of him. On seeing the cow, the driver instantly applies the brakes and is able to bring the car to a stop after 6 seconds.(a) What is the deceleration of the car?(b) Calculate the distance travelled by the car from the time the driver applies the brakes until it comes to a stop.(c) Is the driver able to avoid knocking the cow?

SolutionGiven u = 30 ms-1 , v = 0 ms-1, t = 6 s(a) a = (v - u) / t = (0 - 30) / 6 = -5 ms-2

Example 21Example 21SolutionGiven u = 30 ms-1 , v = 0 ms-1, t = 6 s(b) s = ut + ½at2

= (30 x 6) + ½ (-5 x62) = 90m

(c) Since the car moved a distance of 90 m before it came to a stop, the driver would not be able to avoid knocking down the cow.

THE ENDTHE END

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2.1 Linaer Motion FIZIK