engineering fundamentals session 8 (3 hours). motion

Engineering Engineering FundamentalsFundamentalsEngineering Engineering

FundamentalsFundamentals

Session 8 (3 hours)Session 8 (3 hours)

Motion

Distance Vs Displacement

• Distance 距離 is a scalar quantity which refers to "how much ground an object has covered" during its motion.

• Displacement s 位移 is a vector quantity which refers to "how far out of place an object is"; it is the object's change in position.

A teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.Distance = ___________

Displacement = _____________

Exercise

Start here

Exercise

displacement =___________

distance traveled=____________

Answer to previous page: distance=12;displacement=[0,0]

The skier moves from A to B to C to D.

Speed and Velocity

• Average speed 速率 – Average speed = total distance / total time taken– ( or rate of change of distance, or changes in

distance per unit time)– a scalar

• Average velocity v 速度 – v = total displacement/total time taken

– (or rate of change of displacement, or changes in displacement per unit time)

– a vector– Unit: ms-1

Answer to previous page: displacement=140m,distance=42

0m

v = ∆s / ∆t

ExampleA car moves 3 km north for 10 minutes and then

3 km east for 10 minutes. Find its average speed and velocity.

3km

3km

Average speed = 6km / 20 minutes = 6 km / (1/3 hour) = 18km/hr

Average velocity = √18 km/ (1/3 hour) at an angle of 45 。

=12.73 km/hr at 45 。

Exercise

• Which part(s) is the car having a positive velocity? ________

• Which part(s) is the car having a negative velocity?_________

• What is the velocity at part B?____

time

displacement

A

BC

Acceleration 加速度 • Car speeds up velocity increases and there

is an acceleration a• Car slows down velocity decreases and

there is a -ve acceleration, or deceleration.• Average acceleration = changes in velocity /

total time taken• (or rate of change of velocity, or changes of

velocity per unit time)• SI unit: ms-2

Answer to previous page: A, C, 0

Average acceleration a = ∆v / ∆t

ExampleA sports car can go from rest to 100 km

hr-1 within 10 seconds. What is its acceleration?

AnswerChanges in velocity = 100 - 0 km hr-1 =

100 /3600 km s-1 = 100/3.6 m s-1 Average acceleration = (10 / 3.6) /10 ms-2

Exercise• Which part(s) of the curve shows

an acceleration? ____• Which part(s) of the curve shows a

deceleration? _____• Which part(s) of the curve shows a

constant (stable) velocity?_____

time

velocity

A

BC D

Exercise• A car is originally at rest. It accelerates

at 2 ms-2 for 1 second. What is its velocity afterwards? _________

• A car is originally moving at a constant velocity of 1 ms-1. It then accelerates at 2 ms-2 for 0.5 second. What is its velocity afterwards? _________

Instantaneous Velocity• Average velocity 平均速度 = ∆s / ∆t (average

over the time interval ∆t )• Instantaneous velocity 瞬時速度 = velocity at

an instant 瞬閒 of time. (∆t 0)• Instantaneous velocity at a time instant t1 =

slope of tangent line at t1.

t1

t

Displacement s

instantaneous velocity (at time t) = slope of tangent at t

Instantaneous Velocity vs Average Velocity

instantaneous velocity at t=2 is 1 ms-1

Instantaneous velocity at t=3 is 0 ms-1

Instantaneous velocity at t=4 is __________Instantaneous velocity at t=8 is __________Instantaneous velocity at t=2 is undefined since it is different at 2+ (slightly > 2) and 2- (slightly < 2).Average velocity between t=0 and t=2 is 1 ms-1

Average velocity between t=0 and 7 is _____________Average velocity between t=7 and 9 is _____________Average velocity between t=0 and 9 is _____________

2 7 9 T(s)

S(m)

2

Exercises

t

t

v

3 10 13

4

6

Plot the v-t graph below:

1

0.5

1.5

Exercises

t

t1 t2 t3 t4

Time instants at which velocities are positive: _________

Time instants at which velocities are negative: ________

Compare velocity at t1 and velocity at t2:___________

Velocity at t3 = ______________

Realistic Displacement-Time curveDiscontinuous

velocity (not realistic)

Velocity gradually increases (realistic)

s

t

v

t

Red curves are unrealistic since the velocities are discontinuous (implies infinite

acceleration)

Constant Acceleration

v

t

a

Motions Equations for Constant Acceleration

• 5 variables :– t time– u initial velocity– v final velocity– a acceleration (constant)– s displacement

v = u + at

v2 = u2 + 2as

2

2

1atuts

Motion Equations (constant acceleration)velocity

u

v

t

s

ttime

displacement

time

An object moving in a straight line with constant acceleration takes 10 s from rest to cover a distance of 100 m. Determine the acceleration of the object.

By using the equation

2

2

1atuts

• u = 0• t = 10 sec• s = 100 m

(Ans) a = 2 m/s2

Example

A particle with u = 80 m/s and zero acceleration for first 5 sec. The particle is then slowed down with acceleration of -15 m/s2. How far it has travelled after 5 sec more? Find its velocity at that time.

2

2

1atuts

• In last 5 sec, • s = 80(5)+0.5(-15)(5)2 = 212.5 m • But it has travelled 80x5 = 400 m

in the first 5 sec

Example

Example (continued)

• Total distance travelled = 400 + 212.5 =612.5 m

• Velocity at that time• v = u + at• v = 80 + (-15)5• = 5 m/s

Motion under the action of gravity

• The acceleration due to gravity 引力 g is the acceleration of a freely falling object as a result of a gravitational force. For most practical purpose is taken as being 9.81 m/s2 at the surface of the earth.

Why is there Gravity?(this slide will not be tested)

Newton hit by an apple. Why does the apple fall

downwards?

m1

m2

r

Law of Gravity 萬有引力

There is gravitational force between 2 masses

An object is thrown vertically upwards with a velocity of 8 m/s. What will be the maximum height it reaches and the time taken for it to reach that height ?

• u = 8 m/s• a = g = -9.81

m/s2

• v = 0 at the maximum height

Example

Positive direction

g

Apply v2 = u2 + 2as

0 = 82 + 2(-9.81) s

s = 3.26 m ( The maximum height )

Also v = u + at

0 = 8 + (-9.81) t

t = 0.82 sec.

Example (Continued)

Force

• A force cannot be seen, only the effect of a force on a body may be seen.

• Force Units: S.I. Unit ,Newton, (N) or (kN)

• Force is a vector quantity. It has both magnitude and direction.

Newton

•Born 1643

•Newton’s Laws

•Gravitational Force

•Calculus

Newton’s First Law (Law of Inertia)

• First Law First Law – Every body will remain at rest or continue in uniform motion in a straight line until acted upon by an external force.

• Inertia 慣性 : tendency for a body maintains its state of rest or move at constant speed

• The greater the mass 質量 , the larger is the resistance to change.

Life Examples

http://spaceflight.nasa.gov/gallery/video/living/net56/fun_destiny_56.asf

Once the dummy is in motion, it will be in motion

Astronauts in Space Shuttle. Observe instances of Law of Inertia in the following video

clip.

Newton’s Second Law• When an external force is applied to a

body of constant mass. It produces an acceleration which is directly proportional to the force

• The large the mass, the more force it takes to accelerate it.

• The large the force, the larger the acceleration.

• Force (F)= mass (m) x acceleration (a)

Video Example• When a given tension (force) is given to

a slider without friction (with air track). View 1-2 video of acceleration from http://www.doane.edu/Dept_Pages/PHY/PhysicsVideoLibrary/videolibrary.html

(Use the flash version if you do not have quicktime.)

Constant force provided by falling object

A net force of 200 N accelerates an object with a mass of 100 kg. What is

the acceleration?

F = 200 Nm = 100 kgF=ma

a=F/m = 2 m/s2

Example

massmeter

Newton’s Third Law

• Every action produces an equal and opposite reaction.

• Action and Reaction

Action Force = Reaction Force

Life Example

Life Example (continued)

Concept map

Newton's law

Newton'first law

Newton's second law

Newton's third law

Scalars and Vectors

Uniformly accelerated motion

Motion

Linear Motion

v = u + at

v2 = u2 + 2as

2

2

1atuts

Gravity=9.81 m/s2

Force

Action = Reaction

F=maLaw of Inertia