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Physics ILecture03-Motion in one dimension-Iη°‘η΄ζΏ±εη«δΊ€ι倧εΈηεΈι’ι»εη©ηη³»
CONTENTS
1. Position, Velocity and Speed
2. Instantaneous Velocity and Speed
3. Motion with Constant Velocity
4. Acceleration
5. Motion Diagram
6. Motion with Constant Acceleration
7. Freely Falling Object
8. Kinmatic Equations & Calculus
1. POSITION, VELOCITY AND SPEED
Scalar: real number π₯, Vector: real number with direction
π₯ ΖΈπ, where the number is just the length of the vector
Position β a vector to note the direction and distance from the origin
0 1 2 3 4β1β2β3
ΰ·π₯
3ΰ·π₯ or 3 ΖΈπβ1.2ΰ·π₯ or β 1.2 ΖΈπ
Displacement β a vector, variation of the position
Notation - β Τ¦π₯ = Τ¦π₯π β Τ¦π₯π, where Τ¦π₯πand Τ¦π₯π are initial and final position
β Τ¦π₯ = Τ¦π₯π β Τ¦π₯π = 4.2 ΖΈπ β 10 ΖΈπ = β5.8 ΖΈπ
Example: The initial position of an object is Τ¦π₯π = 10 ΖΈπ and its final position is Τ¦π₯π = 4.2 ΖΈπ.
What is the displacement?
1. POSITION, VELOCITY AND SPEED
Distance β a scalar corresponding to the displacement
Notation - β Τ¦π₯ = Τ¦π₯π β Τ¦π₯π
β Τ¦π₯ = Τ¦π₯π β Τ¦π₯π = 4.2 ΖΈπ β 10 ΖΈπ = β5.8 ΖΈπ = 5.8
average Velocity β a vector, The displacement divides by the period of time.
Notation - Τ¦π£ππ£π =Τ¦π₯πβ Τ¦π₯π
βπ‘
average Speed β a scalar, different from the average velocity.
Notation - π£ππ£π =π‘ππ‘ππ πππ π‘ππππ π‘πππ£ππππ
βπ‘
Example: The initial position of an object is Τ¦π₯π = 10 ΖΈπ and its final position is Τ¦π₯π = 4.2 ΖΈπ.
What is the distance of its movement?
1. POSITION, VELOCITY AND SPEED
0 10 20 30 40β10β20β30
ΰ·π₯
0 π 10 π 20 π 30 π 40 π 50 π
-40
-30
-20
-10
0
10
20
30
40
50
0 10 20 30 40 50 60
x (
m)
time (s)
t x
0 25
10 40
20 30
30 0
40 -20
50 -30
average velocity between t=0 and t=50:
Τ¦π£ππ£π =Τ¦π₯ 50 β Τ¦π₯ 0
50=
β30 ΖΈπβ25 ΖΈπ
50= β
55
50ΖΈπ (m/s)
average speed between t=0 and t=50:
π£ππ£π =40β25 +(40β(β30))
50=
85
50(m/s)
1. POSITION, VELOCITY AND SPEED
Displacement: β Τ¦π₯ = Τ¦π₯π β Τ¦π₯π = 2 β 12 ΖΈπ = β10 ΖΈπ (m)
Distance: β Τ¦π₯ = β10 ΖΈπ = 10 (n)
average Velocity: Τ¦π£ππ£π = ΰ΅β Τ¦π₯βπ‘ =
β10 ΖΈπ
4β1= β
10
3ΖΈπ (m/s)
Example: A particle is moving along the x-axis. Its initial position is Τ¦π₯π = 12 ΖΈπ (m) at
time π‘π = 1 (s) and its final position is Τ¦π₯π = 2 ΖΈπ (m) at time π‘π = 4 (s). Find out its
displacement and average velocity during the time interval.
2. INSTANTANEOUS VELOCITY AND SPEED
Velocity β a vector, The infinitesimal displacement divides by the
infinitesimal period of time.
Notation - Τ¦π£ = limβπ‘β0
Τ¦π₯πβ Τ¦π₯π
βπ‘= lim
βπ‘β0
π₯ π‘π βπ₯ π‘π
βπ‘ΖΈπ = lim
βπ‘β0
π₯ π‘+βπ‘ βπ₯ π‘
βπ‘ΖΈπ
Speed β a scalar, The norm of the average velocity.
Notation - π£ = Τ¦π£ =ππ₯ π‘
ππ‘ΖΈπ =
ππ₯ π‘
ππ‘
=ππ₯ π‘
ππ‘ΖΈπ
http://588ku.com/image/qicheyibiaoban.html
2. INSTANTANEOUS VELOCITY AND SPEED
The velocity: Τ¦π£ π‘ = limβπ‘β0
Τ¦π₯ π‘+βπ‘ β Τ¦π₯ π‘
βπ‘
= limβπ‘β0
3 π‘+βπ‘ 2+2 π‘+βπ‘ β 3π‘2+2π‘
βπ‘ΖΈπ = 6π‘ + 2 ΖΈπ (m/s)
The average velocity: Τ¦π£ππ£π π‘ =Τ¦π₯ 3 β Τ¦π₯ 1
3β1=
33β5
2ΖΈπ = 14 ΖΈπ (m/s)
Compared with Τ¦π£ 1 = 8 ΖΈπ (m/s), Τ¦π£ 2 = 14 ΖΈπ (m/s),
Τ¦π£ 3 = 20 ΖΈπ (m/s)
Example: The position of an object moving on the x-axis varies in time according to
the equation Τ¦π₯ π‘ = 3π‘2 + 2π‘ ΖΈπ, where π₯ is in meters and π‘ is in seconds. (a) Find the
velocity as a function of time. (b) Find the average velocity in the intervals between
π‘ = 1 and π‘ = 3 s.
3. MOTION WITH CONSTANT VELOCITY
Object in constant velocity motion, its instantaneous velocity is Τ¦π£ π‘ = π£0 ΖΈπ. As you
know the velocity, you can find out its position as a function of time by integration with a specified constant of Τ¦π₯ π‘ = 0 = Τ¦π₯ 0 = Τ¦π₯0.
Τ¦π£ π‘ = π£0 ΖΈπ
π Τ¦π₯ π‘
ππ‘= Τ¦π£ π‘ = π£0 ΖΈπ π Τ¦π₯ π‘ = π£0ππ‘ ΖΈπ
ΰΆ±π Τ¦π₯ π‘ = ΰΆ±π£0ππ‘ ΖΈπ ΰΆ±Τ¦π₯0
Τ¦π₯ π‘
π Τ¦π₯ π‘β² = ΰΆ±0
π‘
π£0π π‘β² ΖΈπ
Τ¦π₯ π‘ = Τ¦π₯0 + π£0π‘ ΖΈπ
Τ¦π₯ π‘ β Τ¦π₯0π‘ β 0
= Τ¦π£ππ£π = π£0 ΖΈπ Τ¦π₯ π‘ β Τ¦π₯0 = π£0π‘ ΖΈπ Τ¦π₯ π‘ = Τ¦π₯0 + π£0π‘ ΖΈπ
3. MOTION WITH CONSTANT VELOCITY
π Τ¦π₯ π‘
ππ‘= Τ¦π£ π‘ = 5.0 ΖΈπ
ΰΆ±10 ΖΈπ
Τ¦π₯ π‘
π Τ¦π₯ = ΖΈπ ΰΆ±2.0
π‘
5.0ππ‘
Τ¦π₯ π‘ = 10 + 5.0 π‘ β 2.0 ΖΈπ = 5.0π‘ ΖΈπ (m)
Τ¦π₯ 10 = 50 ΖΈπ (m)
Example: A particle moves with a constant velocity Τ¦π£ π‘ = 5.0 ΖΈπ (m/s).
The position is Τ¦π₯ 2.0 = 10 ΖΈπ (m) at π‘ = 2.0 (s).
(a) Please find the position as a function of time.
(b) Please find its position at π‘ = 10 (s).
4. ACCELERATION
average Acceleration β a vector, The velocity variation
divides by the period of time.
Notation - Τ¦πππ£π =π£πβπ£π
βπ‘
Acceleration β a vector, The infinitesimal velocity variation divides by
the infinitesimal period of time.
Notation - Τ¦π = limβπ‘β0
π£πβπ£π
βπ‘= lim
βπ‘β0
π£ π‘π βπ£ π‘π
βπ‘= lim
βπ‘β0
π£ π‘+βπ‘ βπ£ π‘
βπ‘
Derivation - Τ¦π =ππ£ π‘
ππ‘=
π2 Τ¦π₯ π‘
ππ‘2
Τ¦π£ π‘ = 3π‘2 β 27 ΖΈπ (m/s)
Τ¦π π‘ = 6π‘ ΖΈπ (m/s2)
Example: A particle moves according to the expression Τ¦π₯ π‘ = 4 β 27π‘ + π‘3 ΖΈπ, where π₯ is in
meters and π‘ is in seconds. Please find its velocity and acceleration as a function of time.
4. ACCELERATION
https://www.pdhpe.net/the-body-in-motion/how-do-biomechanical-principles-influence-movement/motion/acceleration/
Τ¦π =π Τ¦π£ π‘
ππ‘=π2 Τ¦π₯ π‘
ππ‘2
CONTENTS
1. Position, Velocity and Speed
2. Instantaneous Velocity and Speed
3. Motion with Constant Velocity
4. Acceleration
5. Motion Diagram
6. Motion with Constant Acceleration
7. Freely Falling Object
8. Kinmatic Equations & Calculus
5. MOTION DIAGRAM
car at rest
car in motion with constant velocity
car in motion with constant acceleration
car in motion with constant deceleration
https://giphy.com/gifs/dog-cartoons-UKm1AF0UrCkb6
6. MOTION WITH CONSTANT ACCELERATIONObject in constant acceleration motion, its instantaneous acceleration is Τ¦π π‘ = π0 ΖΈπ. As you know the acceleration, you can find out its velocity and position as a function of time
by integration with two specified constants of
Τ¦π₯ π‘ = 0 = Τ¦π₯ 0 = Τ¦π₯0 and Τ¦π£ π‘ = 0 = Τ¦π£ 0 = Τ¦π£0 = π£0 ΖΈπ.
Τ¦π π‘ = π0 ΖΈπ = Τ¦πππ£π =Τ¦π£ π‘ β Τ¦π£0π‘ β 0
Τ¦π£ π‘ β Τ¦π£0 = π0π‘ ΖΈπ
Τ¦π£ π‘ = Τ¦π£0 + π0π‘ ΖΈπ = π£0 ΖΈπ + π0π‘ ΖΈπ
ππ£ π‘
ππ‘= π0 ΖΈπ π Τ¦π£ = π0 ΖΈπππ‘ ΰΆ±
π£0
π£ π‘
π Τ¦π£ = ΰΆ±0
π‘
π0 ΖΈπππ‘β²
Τ¦π£ π£0
π£ π‘= π0 ΖΈπ π‘
β²0π‘ Τ¦π£ π‘ β Τ¦π£0 = π0π‘ ΖΈπ Τ¦π£ π‘ = Τ¦π£0 + π0π‘ ΖΈπ
6. MOTION WITH CONSTANT ACCELERATIONObject in constant acceleration motion, its instantaneous acceleration
is Τ¦π π‘ = π0 ΖΈπ. As you know the acceleration, you can find out its velocity
and position as a function of time by integration with two specified
constants of Τ¦π₯ π‘ = 0 = Τ¦π₯ 0 = Τ¦π₯0 and Τ¦π£ π‘ = 0 = Τ¦π£ 0 = Τ¦π£0 = π£0 ΖΈπ.
Τ¦π£ π‘ = π£0 ΖΈπ + π0π‘ ΖΈπ
t
v
π‘0
π£ π‘ = π£0 + π0π‘
π£0
π£0 + π0π‘0
Τ¦π£ 0 = π£0 ΖΈπ, Τ¦π£ π‘0 = π£0 ΖΈπ + π0π‘0 ΖΈπ
The area in v-t graph:
π₯ π‘0 β π₯ 0 =π£0 + π£0 + π0π‘0
2π‘0 = π£0π‘0 +
π0π‘02
2
π₯ π‘ = π₯0 + π£0π‘ +π0π‘
2
2
Τ¦π₯ π‘ = Τ¦π₯0 + π£0π‘ ΖΈπ +π0π‘
2
2ΖΈπ
6. MOTION WITH CONSTANT ACCELERATIONObject in constant acceleration motion, its instantaneous acceleration
is Τ¦π π‘ = π0 ΖΈπ. As you know the acceleration, you can find out its velocity
and position as a function of time by integration with two specified
constants of Τ¦π₯ π‘ = 0 = Τ¦π₯ 0 = Τ¦π₯0 and Τ¦π£ π‘ = 0 = Τ¦π£ 0 = Τ¦π£0 = π£0 ΖΈπ.
π Τ¦π₯
ππ‘= Τ¦π£ π‘ = π£0 ΖΈπ + π0π‘ ΖΈπ π Τ¦π₯ = π£0 + π0π‘ ππ‘ ΖΈπ
ΰΆ±Τ¦π₯0
Τ¦π₯ π‘
π Τ¦π₯ = ΰΆ±0
π‘
π£0 + π0π‘ ππ‘ ΖΈπ
Τ¦π₯ π‘ = Τ¦π₯0 + π£0π‘ ΖΈπ +π0π‘
2
2ΖΈπ
6. MOTION WITH CONSTANT ACCELERATION
Three Equations:Τ¦π₯ = Τ¦π₯0 + π£0π‘ ΖΈπ +
π0π‘2
2ΖΈπ
Τ¦π£ = π£0 ΖΈπ + π0π‘ ΖΈπ 1st equation
2nd equation
Τ¦π£ β π£0 ΖΈπ = π0π‘ ΖΈπ
Τ¦π₯ β Τ¦π₯0 = π£0π‘ ΖΈπ +π0π‘ ΖΈπ
2π‘ Τ¦π₯ β Τ¦π₯0 π0 ΖΈπ = π£0 ΖΈπ β π0π‘ ΖΈπ +
π0π‘ ΖΈπ
2β π0π‘ ΖΈπ
β Τ¦π₯ β Τ¦π = π£0 ΖΈπ β π£ ΖΈπ β π£0 ΖΈπ +1
2π£ ΖΈπ β π£0 ΖΈπ β π£ ΖΈπ β π£0 ΖΈπ
π£2 = π£02 + 2 Τ¦π β β Τ¦π₯ 3rd equation
π£2 = π£02 + 2ππ
6. MOTION WITH CONSTANT ACCELERATION
π£π = 108ππ
βΓ
1000π
1ππΓ
1β
3600π = 30 (m/s)
π£π = 72ππ
βΓ
1000π
1ππΓ
1β
3600π = 20 (m/s)
π =100 (m)
pick up the right equation: π£π2 = π£0
2 + 2ππ
400 = 900 + 2π Γ 100 π = β2.5 (m/s2)
pick up the right equation: π£π = π£0 + ππ‘
20 = 30 β 2.5 Γ π‘ π‘ = 4 (s)
Example: You start to brake your car from a speed of 108 to 72 km/h when spotting a police car.
The traveled distance is 100 m. Assume that the car is in constant acceleration motion, please
calculate its acceleration and the time required for the decrease in speed.
6. MOTION WITH CONSTANT ACCELERATION
π£π = 3 Γ 104 (m/s)
π£π = 5 Γ 106 (m/s)
π = 2 (cm) = 21π
100ππ= 0.02 (m)
pick up the right equation: π£π2 = π£0
2 + 2ππ
2.5 Γ 1013 = 9 Γ 108 + 2π Γ 0.02 π β 2.5 Γ1013
0.04= 6.25 Γ 1014 (m/s2)
pick up the right equation: π£π = π£0 + ππ‘
5 Γ 106 = 3 Γ 104 + 6.25 Γ 1014 Γ π‘ π‘ β 8 Γ 10β9 (s)
Example: An electron in the cathode-ray tube of a television set enters a region in which it
accelerates uniformly in a straight line from a speed of 3 Γ 104 m/s to a speed of 5 Γ 106 m/s in
a distance of 2 cm. How long is the electron in constant acceleration?
6. MOTION WITH CONSTANT ACCELERATION
45 Γ 2 + 45π‘ =5
2π‘2
Example: A mortorcycle traveling at a constant speed of 45 m/s passes a trooper on a car hidden
behind a billboard. 2 second after the speeding mortorcycle passes the billboard, the trooper sets
out from the billboard to catch the mortorcycle, accelerating at a constant rate of 5.00 m/s2.
How long does it take her to overtake the mortorcycle?
t
vπ£ π‘ = π0π‘ , π0=5
v=45
t=2
https://giphy.com/gifs/drive-fast-adrenaline-14hVD0HODUPY08
7. FREELY FALLING OBJECT
π‘ = 0π‘ = 1
π‘ = 2
π‘ = 3
π‘ = 4
t y(t) (m) vy(t) (m/s) a(t) (m/s2)
0 0 0 -9.8
1 -4.9 -9.8 -9.8
2 -19.6 -19.6 -9.8
3 -44.1 -29.4 -9.8
4 -78.4 -39.2 -9.8
-100 -9.8
https://media.giphy.com/media/cXqQCO1bWp5tK/giphy.mp4
7. FREELY FALLING OBJECT
t y(t) (m) vy(t) (m/s) a(t) (m/s2)
0 -19.6 19.6 -9.8
1 -4.9 9.8 -9.8
2 0 0 -9.8
3 -4.9 -9.8 -9.8
4 -19.6 -19.6 -9.8
5 -44.1 -29.4 -9.8
6 -78.4 -39.2 -9.8
-100 -9.8
8. KINEMATIC EQUATIONS & CALCULUS
π‘
π£ π‘
π‘1 π‘2 π‘3 π‘4 π‘5 π‘6 π‘7 π‘8
βπ‘2 βπ‘4
π£ π‘6
π₯ = limβπ‘πβ0
π=1
π
π£ π‘π βπ‘π
π₯ π‘π β π₯ 0 = ΰΆ±0
π‘π
π£ π‘ ππ‘
ACKNOWLEDGEMENT
εη«δΊ€ι倧εΈηεΈι’
θͺδΈ»ζεΈηΏθ¨η«
γη§ζι¨θ£ε©γ
