movement
TRANSCRIPT
MOTION
Anatasari Surya, S.PdJunior High School State 1 Metro1
To understand the concept of movement, work, force, and
energy in our daily life
Standard Competence
2
To analyze experimental data of uniform straight motion and uniformly accelerated straight motion, and its application in everyday life
Basic Competence
3
• To find the characteristics of Uniform Linear Motion• To define acceleration as a speed changes per time
taken• To find the distance based on time taken
Indicator
4
• Motion• Virtual Motion• Relative motion• Point of reference• Distance• Displacement
• Speed• Velocity• Acceleration• Deceleration• Uniform motion• Uniformly
acceleration motion
5
Science Words
Use the ground as reference! What she is move?
Look at the picture!
6
What bird is move..?
Use the ground as reference
7
Mention the objects are move and motionless!!
Use the ground as reference 8
• Motion is a change of position with respect to a point of reference.
• Point of reference is something that is considered fixed and used as a comparison.
• Motion very dependent on point of reference– Satu titik acuan melihat suatu gerak sangat lambat– Titik acuan lain melihatnya sangat cepat– Sedangkan titik acuan yang lain lagi melihatnya diam
Conclude:
9
.....
• Motion dependent on point of reference is relative motion
• Virtual motion is looks like motion but actually it is not motion, example:– Sun rises in the east and sets in the west– When you are inside a moving bus and look
towards the window you will see trees moving away from you.
10
Distance = 200 mDisplacement = 200 m to rigth
200 m
What is differencebetween distance and displacement?
0 1 2 3 4 5 6 7 8 9A C
B
2 134Distance = 13 mDisplacement = 5 m to rigth
m
What is differencebetween distance and displacement?
conclude
• Distance is length of all lines passed through by an objectscalar quantity -> have: magnitude and unit
• Diplacement is position change of an object from the initial point
Vector quantity -> have: magnitude, unit, and direction
Distance = 35 mDisplacement = 15 m to left
What its distance?What its displacement?
10 m
25 m
1.
What difference between a distance and a displacement ?
120 m
150 m
150 m
Distance =
Displacement =
AB
C
270 m
90 m
Exercise!
2.
A
B
30 meters
30 meters
Romi run around from A to B. Calculate the distance and displacement!
Exercise
Distance = 60 mDisplacement = 1800 m
= 30 2 m
3.
4. Initial position of adi is 0 m from the reference of point. He walking as far as 5 m to right and then return to left as far as 9 m. Calculate the distance and displacement!
-4 -3 -2 -1 0 1 2 3 4 5m
Speed and Velocity
0 4
Time (second)
the moving object is identified by its change of position of a point of reference If we want to know how far the position has changed, we must know the concept about velocity
Why the change of position of car is longer than a bicycle?
Speed = the number of velocity. (scalar quantity)
Speed = distance time
Velocity = the change of position of each time. (vector quantity)
Velocity = displacement time
scalar quantity:Speed = magnitude, and unit
vector quantity:Velocity = magnitude, unit and direction
For example:The speedometer of motorcycle shows 50 km/hour to west• Speed = 50 km/hour• Velocity = 50 km/hour to west
What difference between a distance and a displacement ?
200 m
250 m
150 m
Distance =
Displacement =
A B
C
450 m
150 m
Speed and Velocity
If Budi go to C from A in 5 seconds that :
Speed =
Velocity =
90 m/s
30 m/s
A
B C
D
E
500
m
200 m
300 m
400 m
350 m
Exercise:Yuni runs 120 m in 1 minute.What is her running speed?
Known : s = 120 mt = 1 minute = 60 sQuestion : V =….?
Answer : V = s / t
= 120 / 60= 2 m/s
• a friend told you that he could ride his bike at a speed of 18 km/hour. To find out whether you could ride faster, you needed 30 seconds to go as far as 180 m. Who rade faster, your or your friend?
Average speed
Average speed is the total traveled distance divided by the total time needed to travel that distance• Average speed = total distance total time
v = s1 + s2 + s3 ……
t1 + t2 + t3 ….
ExampleBudi rides a bicycle traveling 20 km in 20 minutes. And then he travels 16 km in 10 minutes. Find the Budi’s average speed! State your answer in meter/second.
s = 20 km + 16 km = 36 km = 36,000 meter
t = 20 minutes + 10 minutes = 30 minutes = 1800 seconds
v = s : t
= 36,000 m : 1800 s
= 20 m/s
7
v = s1 + s2 + s3
t1 + t2 + t3
= 5000 + 10000 + 10000 600 + 900 + 300= 25000 1800= 13.89 m/s b. 50 km/hour
Exercise :
1. The velocity of a car when it is moving are as follow: the first 10 minutes, the distance is 5 km. the second 15 minutes, the distance is 10 km the third 5 minutes, the distance is 10 km.
Calculate :a. the average velocityb. express the average velocity in km/hour.
2. Sebuah bus melaju di jalan tol yang lurus. Selama 30 menit pertama bus itu menempuh jarak 45 km, 15 menit selanjutnya menempuh jarak 15 km, dan 15 menit selanjutnya menempuh jarak 20 km. Calculate the average speed that bus!
LINEAR ACCELERATE MOTION (LAM)
It is linear motion whose velocity is changing/varying
regularly
ACCELERATION• Symbol: a• Formula:
acceleration = change of velocity time taken
• SI Unit : m/sec2
• The same formula can also be applied for deceleration, but the value of a is negative
Acceleration• Acceleration denotes the change of velocity
per unit of time. (Vector Quantity) acceleration decleration
• The formula :
a = vt – v0 or a = v/t tt – t0
With : a = acceleration (m/s2) vt = The final velocity (m/s) v0 = the initial velocity (m/s)
velocity acceleration velocity acceleration
Equation of Motion
• Mathematical relations relating motion variables are called equation of motion
• For motion with constant acceleration, the variables are:
• Displacement : s• Initial velocity : v0• Final velocity : vt• Constant acceleration : a• Time taken : t
Examples
1. A car changed its speed from 36 km/hour to 72 km/hour in 10 seconds. Calculate the car’s acceleration.
Known : vo = 36 km/hour = 10 m/s
vt = 72 km/hour = 20 m/s
t = 10 s
Question : a …?
Answer :
• a = vt – vo
• • = 20 – 10
10= 1 m/s2
t
• A car changed its speed from 36 km/hour to 54 km/hour within 10 seconds. Calculate the car acceleration!
• a truck move by velocity as 7 m/s in 1 second. and then in 2 second the velocity becomes 9 m/s. Calculate the acceleration?
Linier Motion
• Linier motion is defined as a motion that has a linier path.
• Linier motion with constant velocity is called a regular linier motion.
• Linier motion with constant acceleration is called a dynamic linier motion.
Displacement-time graph
• A displacement-time graph shows how the displacement of an object changes with time.
• The gradient of a displacement-time graph represents the velocity of the object.
Displacement (m)
Time (s)
Zero gradient – stationary object
Displacement (m)
Time (s)
Fixed gradient-uniform velocity
Displacement (m)
Time (s)
Increasing gradient – increasing velocity
Displacement (m)
Time (s)
Decreasing gradient – decreasing velocity
Graph of Linier Motion
• Graph distance on the y-axis and time on the x-axis
• The velocity is 2 m/s
Slope = rise = distance = speed run time
No
Distance (m)
Time (s)
1 20 10
2 40 20
3 60 30
4 80 40
5 100 50
6 120 60
Distance - Time Graph
• If something is not moving, a horizontal line is drawn.
• If something starts out slow and then speeds up, its change in speed can look like this.
Eg.
• A motorcycle move linier with velocity 60 km/hours. Calculate the distance travelled by motorcycle after 2 hour and ½ hour! And make a graph!
hour distance1 -> 60 km/h2 -> 2 x 60 = 120 km/h½ -> ½ x 60 = 30 km/h
Learning CheckpointThis graph shows several stages of motion:
• Stage 1: 100 m in 10 s
• Stage 2: 50 m in 10 s
• Stage 3: 150 m in 20 s
Calculate the speed as indicated by each of the colors.
Calculate the average speed.
What is the total distance?
What is the displacement?
SolutionStage 1: S= d/ t
100 m/ 10 s= 10 m/sStage 2: S= d/t
50 m/ 10 s= 5 m/sStage 3: S= d/t
150 m/ 20 s= 7.5 m/sAve Speed= Tot d/ Tot t
300 m/ 40 s= 7.5 m/sDistance = 300 metersDisplacement = 0 meters
Consider the following position-time curve.
Graphical Analysis of 1-D Motion
slope of curve is: RUN
RISEm
12
12
t - t
x- x
t
x
Slope of a P-t curve at any time is object’s v at that time.
time (s)
posi
tion
(m)
0
5
10
15
0 1 2 3 4 5 6 7
Analysing motion graph
15
3 9
Displacement (m)
Time (s)
What is the displacement of theObject after 3 seconds?What is the velocity during : - the first 3 seconds - the next 3 seconds?
Displacement = 15 mVelocity : the first 3 s = 5 m/s the next 3 s = 0
Find object’s velocity att = 0 to 3.0 s and at t = 6.0 to 7.0 s.
s 0 - s 3
m 0 - m 15
t
x v
t
x v
s 6 - s 7
m 15 - m 0
s
m 5
s
m 15-
(–) sign indicates that the object is moving opposite to how it started
(which we assumed was the (+) direction).
time (s)
posi
tion
(m)
0
5
10
15
0 1 2 3 4 5 6 7
Analysing motion graphs.
Time (s)
Displacement (m)
10 20 30 40 50 60
120
100
80
60
40
20
What is the displacementafter 25 seconds?What is the velocity during: - the first 25 s? - the next 15 s?
A particle in a magnetic field moves as follows:
Find the velocity for each part of the motion
Velocity-time graph
• A velocity-time graph shows how the velocity of an object changes with time.
• The gradient of a velocity-time graph represents the acceleration of the object.
• The area under a velocity-time graph represents the distance traveled by the object.
Examples
Velocity (m/s)
Time (s)
A B
C
15
20 50 60O
What is the acceleration of the car during the part of the journey represented by: - OA - AB - BC
What is the total distance traveled by theCar?
Calculate the average velocity of the car for its whole journey.
Several examples of velocity-time graph
• Regular / uniform linier motionVelocity (m/s)
0 Time (s)
Uniform velocity – zero acceleration
Accelerated dynamic linier motion
Velocity (m/s)
Time (s)0
Uniform acceleration
Velocity (m/s)
Time (s)
Increasing acceleration
Decelerated dynamic linier motion
Velocity (m/s)
Time (s)
Uniform deceleration
Velocity (m/s)
Time (s)
Decreasing acceleration
Decreasing deceleration
Velocity (m/s)
Time (s)
THANK YOUSEE YOU NEXT TOMORROW