engineering fundamentals session 8 (3 hours). motion
TRANSCRIPT
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