(BB101 ) ENGINEERING SCIENCE BY NUR ELYANI BINTI MUSA JMSK
(BB101 ) ENGINEERING SCIENCE
BY
NUR ELYANI BINTI MUSAJMSK
Hair Raiser Roller Coaster POV Ocean Park Hong Kong B M Floorless On-Ride - 3.wmv
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
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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.
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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)
Apply the concept of linear motion in solving the related problems by using formula
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
Apply the concept of linear motion in solving the related problems by using formula
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.
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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)
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
Q1(d)
Solution 3
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)
v = u + at 20 = 0 + a(2.6) …………..(1 m) a = 20 2.6 a = 7.69 m/s² …………(1 m)
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
Continue………….
A. A runner runs from the starting line and achieves a velocity of 20 m/s¹ in 2.6 seconds. (CLO3)
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)
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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
Objective……Describe distance , displacement, speed, velocity, average velocity, instantaneous velocity, acceleration & deceleration
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.
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.
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.