Linear motion present

(BB101 ) ENGINEERING SCIENCE

BY

NUR ELYANI BINTI MUSAJMSK

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OBJECTIVE

Define scalar and vector quantities. Define linear motion Define uniform and non-uniform motion. Describe distance, displacement, speed,

velocity, average velocity, instantaneous velocity, acceleration and deceleration.

Apply the concept of linear motion in solving the related problems by using formula

Illustrade velocity – time graph. Determine the velocity, acceleration and

displacement from the graphs.

CHAPTER 2

SCALAR QUANTITY & VECTOR QUANTITY

Definition Quantity which has magnitude

only.

Example Distance, speed, Length, Time,

density

SCALAR QUANTITYObjective……Define scalar quantity and vector quantity

CHAPTER 2

SCALAR QUANTITY & VECTOR QUANTITYVECTOR QUANTITY

Definition- Quantity which have both

magnitude and direction.

Example Displacement, velocity, Force,

Acceleration

Objective……Define scalar and vector quantity

LINEAR MOTION

Linear motion is the motion of an object, moving in a straight line or path.

The movement of any object depend on frequent questions:

How far ! How fast ! Which direction !

Objective……Define linear Motion

CHAPTER 2

LINEAR MOTION

Definition Uniform motion is reaching and maintaining its maximum speed on a straight line.

Example A falling coconut A car moving in same speed in

a straight line

Objective……Define uniform motionDefine non – linear motion

Uniform Motion

LINEAR MOTION

DefinitionNon linear motion that undergoes a change in velocity, either by changing speed or changing direction.

Example A roller coaster ride A snake crawling

Non Linear MotionObjective……Define uniform motionDefine non – linear motion

Distance & Displacement

Distance (How far)Definition is total route taken from

one place into another place. Scalar Quantity because it has

magnitude only SI unit is meterDisplacement Definition is distance along with

direction Vector quantity because it has both

magnitude & direction. SI unit is meter (m)

Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration

To understand about …….

Distance Displacement

An aeroplane is flying from Airport A to Airport B with a long distance about 350km without direction.

But if the aeroplane use a direction move towards Airport B, its displacement is 100km.

Speed and Velocity (How Fast)

Speed Velocity

Speed is the rate of change of distance.

Speed is scalar quantity. Speed = Distance, m Time, s SI unit is meter per

second (ms-1).

Velocity is the rate of change of displacement.

Velocity is vector quantity.

Velocity = Displacement, m Time, s

SI unit is meter per second (ms-1).

CHAPTER 2


Average Velocity & Instantaneous Velocity.

Average Velocity Instantaneous Velocity- Average velocity is

for velocity of entire trip.

- Instantaneous Velocity is velocity of an object at the particular instant or time.

CHAPTER 2


Acceleration and Deceleration.

Acceleration Deceleration.

- Acceleration is the rate change of velocity.

- Acceleration is vector quantity. Acceleration = v - u t Where : v = Final velocity (last high value) u = Initial velocity(start to moving) t = time SI unit is ms-2.

Example : A car starts to move fast From rest or low speed.

- Deceleration is the rate change of velocity that has negative value. (Slowing down)

Acceleration = v - u tWhere : v = Final velocity (last value) u = Initial velocity(start to slowing down) t = time

Example : A car starts to move slowly after max speed.

- SI unit is ms-2.*Formula same as acceleration.


EXAMPLE 1

A student walks to the south for 30 m, then to the east for 40 m and finally to the north for 30 m before he stops to rest. The total time taken by the student for the whole journey is 40 seconds.

a) What is the total distance travelled?b) What is the displacement of the student?c) What is the average speed of the student and his velocity?

Solution Total distance =

30 m + 40 m + 30 m

= 100 m Displacement =

40 m to the east.

Average speed = 100 m

40

= 2.5ms -1

Velocity = 40 m = 1.0 ms¯ ¹ 40 s

Displacement

Starting Point Rest place

30 m30 m

40 m


Example 2

In a car test, the velocity of a car increases uniformly from rest to 30 ms-1 in 15 seconds. What is the acceleration of the car? Solution:Acceleration = Final Velocity, v – Initial Velocity, u

Time taken, t

= 30 ms -1 – 0 ms-1 15 s = 2 ms-2


The Equation of motion

Quantity Symbol Unit

Distance/displacement s m

Time t s

Initial Velocity u ms¯¹

Final Velocity v ms¯¹

Acceleration/deceleration

a ms¯²

Physic Quantity Equation Unit

Displacement s=ut+ s= ( u + v ) t

M(Metre)

Velocity V = u + at ms¯¹(metre/second)

Acceleration A= v – u t

ms¯²(metre/second)


Example 3

A driver accelerates his car from 20 ms¯². What is his velocity after 8 seconds ? Solution :

U = 20ms¯¹ a = 2 ms¯² t = 8s v = ?

? xFirst trial v = u +2 a s

? Second trial v = u + a t v = 20 + 2 (8) = 36 ms¯¹

3 quantities are known

Unable to calculate, because only two

quantity are known


The velocity time graph

t (s)

v, m sˉ¹

t (s)

v, m sˉ¹

t (s)

v, m sˉ¹

Increasing velocity Constant/Uniform

acceleration

Constant/Uniform velocity Zero acceleration

Decreasing velocity Constant/Uniform

deceleration

Illustrade velocity – time graph.

Question & Answer

ExerciseA. A cow moves 3 meter to the

east and then 4 meter to the north. Find the :

i. Total distance moved by the cow

ii. Displacement of the cow

B. A runner runs from the starting line and achieves a velocity of 18m/s in 3 seconds. Calculate his acceleration.


solution

A cow moves 3 meter to the east and then 4 meter to the north. Find the :i) Total distance moved by the cowii) Displacement of the cowSolution :

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

AC = √AB²+ BC² = √3²+ 4² = √9+ 16 = √25 = 5m

3m

4m

A

C

B

3m

4m

B

C

A

5m


Solution no 2

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

Solution u = o m/s v = 18 m/s t = 3s a = v – u t a = 18 – 0 3 a = 6 m/s²


Final Exam (Dis 2011)

A. A runner runs from the starting line and achieves a velocity of 20 m/s¹ in 2.6 seconds. (CLO3)

i. Calculate his acceleration (2 marks)

ii. If the acceleration is constant, calculate the displacement traveled in 6.5 second. ( 2 marks)


Q1(d)

Solution 3


i. Calculate his acceleration (2 marks)

v = u + at 20 = 0 + a(2.6) …………..(1 m) a = 20 2.6 a = 7.69 m/s² …………(1 m)


Continue………….


ii If the acceleration is constant, calculate the displacement traveled in 6.5 second. (2 marks) s = u t + 1 at² 2 s = 0 + 1 ( 7.69) (6.5) ² ………( 1 m) 2 s = 162.45 m …………..( 1 m)


Final exam (June 2012)

A car starting from rest and accelerates at a constant acceleration of 5 m/s² for 10 s. The car then travels at a constant velocity for 15 s. the brakes are then applied and the car stops in 5 s.i. Sketch a velocity- time graph for the

whole journey


Q1(d)

0

50

a = v – u t

5 = v – 0 10v = 50

m/s

v (m/s)

t (s)10 25 30

CONCLUSION

Define scalar and vector quantities. Define linear motion Define uniform and non-uniform motion. Describe distance, displacement, speed,

velocity, average velocity, instantaneous velocity, acceleration and deceleration.


Illustrade velocity – time graph. Determine the velocity, acceleration and

displacement from the graphs.

Achieved the objective

The fastest train in the world is Japanese maglev 581 km/h.

compared to

bullet train at Malysia (Ipoh KL) 150km/h

Summary

shows a person move with 2ms-1 along a circular path. The magnitude of the quickness is the same, but the direction keeps changing. Hence we say that the velocity of this person keep changing when

he moves through the path.

References

www.youtube.comModul BB101 – Engineering Science (JMSK)

http://www.youtube.com/

Linear motion present

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