straight motion regular report by fildia putri
DESCRIPTION
Laporan Lengkap Fisika Dasar Univeristas - Gerak Lurus Beraturan .TRANSCRIPT
STRAIGHT MOTION REGULAR
Fildia Putri, Rafika Sri Rahayu , Nul Lailah , Nurul Wisna Afianti, Nur Rahmah
Marisa Raden.
Departement of Chemistry, Faculity of Mathematics and Natural Science
State University of Makassar 2013
Abstract
Have done experiment titled “Straight Motion Regular”. The purpose of
experiment are understand the difference between distance and displacement,
determine the speed and the average speed, determining the relationship between
displacement (Δx) with mileage time (t) objects moving Straight Motion Regular
(GLB) , and (4) understand the regular rectilinear motion (GLB). Data obtained
from measurements of distance, displacement and travel time by using the three-
track a moving object with a rectangular-shaped objects, each of which varies the
speed, the three objects start at point A and then move to the next point until the
point where the first object moves. Then proceed with data collection by
measuring the distance and travel time in regular “Straight Motion Regular” GLB
tube that hung in stative by taking four samples, each point has a different
distance and any use distance measurements using different also, measurements
were made three times over and to get the maximum results and accurate. Activity
results 1 in lab can be concluded that the faster object moves does it take to get
geared increasingly shorter and the activities 2 can be concluded that the higher an
object is placed, faster the movement of objects from one to the other.
Keywords: GLB, distance, displacement, velocity, pace, travel time, speed, meter,
stative, GLB tube.
I. PURPOSE
1. Students can understand the difference between distance and the
displacement.
2. Students can determine the speed and average speed.
3. Students can investigate the relationship between displacement (Δx)
with time (t) regular moving body straight (GLB).
4. Students can understand the regular rectilinear motion (GLB).
II. EXPERIMENTAL METHODOLOGY
A. A Brief Theory
PositionIn order to describe the motion of an object, you must first be able to
describe its position—where it is at any particular time. More precisely, you need
to specify its position relative to a convenient reference frame. Earth is often used
as a reference frame, and we often describe the position of an object as it relates to
stationary objects in that reference frame. For example, a rocket launch would be
described in terms of the position of the rocket with respect to the Earth as a
whole, while a professor’s position could be described in terms of where she is in
relation to the nearby white.
Displacement
If an object moves relative to a reference frame (for example, if a
professor moves to the right relative to a white board or a passenger moves toward
the rear of an airplane), then the object’s position changes. This change in position
is known as displacement. The word “displacement” implies that an object has
moved, or has been displaced.
Displacement is the change in position of an object:
Δx = xf − x0 where Δx is displacement, xf is the final position, and x0 is the initial
position. In this text the upper case Greek letter Δ (delta) always means “change
in” whatever quantity follows it; thus, Δx means change in position.
Always solve for displacement by subtracting initial position x0 from final
position xf.
Distance
Although displacement is described in terms of direction, distance is not.
Distance is defined to be the magnitude or size of displacement between two
positions. Note that the distance between two positions is not the same as the
distance traveled between them. Distance traveled is the total length of the path
traveled between two positions. Distance has no direction and, thus, no sign. For
example, the distance the professor walks is 2.0
m. The distance the airplane passenger walks is 4.0 m.
It is important to note that the distance traveled, however, can be greater
than the magnitude of the displacement (by magnitude, we mean just the size of
the displacement without regard to its direction; that is, just a number with a unit).
For example, the professor could pace back and forth many times, perhaps
walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the
right of her starting point. In this case her displacement would be +2.0 m, the
magnitude of her displacement would be 2.0 m, but the distance she traveled
would be 150 m. In kinematics we nearly always deal with displacement and
magnitude of displacement, and almost never with distance traveled. One way to
think about this is to assume you marked the start of the motion and the end of the
motion. The displacement is simply the difference in the position of the two marks
and is independent of the path taken in traveling between the two marks. The
distance traveled, however, is the total length of the path taken between the two
marks.
Time, Velocity, and Speed.
Time
As discussed in Physical Quantities and Units, the most fundamental
physical quantities are defined by how they are measured. This is the case with
time. Every measurement of time involves measuring a change in some physical
quantity. It may be a number on a digital clock, a heartbeat, or the position of the
Sun in the sky. In physics, the definition of time is simple— time is change, or
the interval over which change occurs. It is impossible to know that time has
passed unless something changes. The amount of time or change is calibrated by
comparison with a standard. The SI unit for time is the second, abbreviated s. We
might, for example, observe that a certain pendulum makes one full swing every
0.75 s. We could then use the pendulum to measure time by counting its swings
or, of course, by connecting the pendulum to a clock mechanism that registers
time on a dial. This allows us to not only measure the amount of time, but also to
determine a sequence of events.
How does time relate to motion? We are usually interested in elapsed time
for a particular motion, such as how long it takes an airplane passenger to get
from his seat to the back of the plane. To find elapsed time, we note the time at
the beginning and end of the motion and subtract the two.
Velocity
Your notion of velocity is probably the same as its scientific definition.
You know that if you have a large displacement in a small amount of time you
have a large velocity, and that velocity has units of distance divided by time, such
as miles per hour or kilometers per hour.
Average Velocity
Average velocity is displacement (change in position) divided by the time of
travel,
(2.5) v - = Δx
Δt = xf − x0
tf − t0
, where v - is the average (indicated by the bar over the v) velocity, Δx is the
change in position (or displacement), and xf and x0 are the final and beginning
positions at times tf and t0, respectively. If the starting time t0 is taken to be zero,
then the average velocity is simply
(2.6) v - = Δxt.
Notice that this definition indicates that velocity is a vector because displacement
is a vector. It has both magnitude and direction. The SI unit for velocity is meters
per second or m/s, but many other units, such as km/h, mi/h (also written as mph),
and cm/s, are in common use. Suppose, for example, an airplane passenger took 5
seconds to move −4 m (the negative sign indicates that displacement is toward the
back of the plane). His average velocity would be
(2.7) v - = Δx
t = −4 m
5 s = − 0.8 m/s.
Speed
In everyday language, most people use the terms “speed” and “velocity”
interchangeably. In physics, however, they do not have the same meaning and
they are distinct concepts. One major difference is that speed has no direction.
Thus speed is a scalar. Just as we need to distinguish between instantaneous
velocity and average velocity, we also need to distinguish between instantaneous
speed and average speed.
Instantaneous speed is the magnitude of instantaneous velocity. For
example, suppose the airplane passenger at one instant had an instantaneous
velocity of −3.0 m/s (the minus meaning toward the rear of the plane). At that
same time his instantaneous speed was 3.0 m/s. Or suppose that at one time during
a shopping trip your instantaneous velocity is 40 km/h due north. Your
instantaneous speed at that instant would be 40 km/h—the same magnitude but
without a direction. Average speed, however, is very different from average
velocity. Average speed is the distance traveled divided by elapsed time.
We have noted that distance traveled can be greater than displacement. So average
speed can be greater than average velocity, which is displacement divided by
time. For example, if you drive to a store and return home in half an hour, and
your car’s odometer shows the total distance traveled was 6 km, then your average
speed was 12 km/h. Your average velocity, however, was zero, because your
displacement for the round trip is zero. (Displacement is change in position and,
thus, is zero for a round trip.) Thus average speed is not simply the magnitude of
average velocity.
Said moving object if the object changed position to a reference point.
Objects will move through a trajectory with a certain length of time. The total
length of the path traversed is called the distance, while the change in the
position of the object from the initial position to the final position is called the
displacement. Distance is a scalar quantity, whereas the displacement is a vector
quantity.
Said object moving regular straight (GLB) if the object is moving on a
straight path and move at a constant speed or no speed change with time,
so the acceleration is zero. Speed is defined as the change in position at
any time or in the form of written mathematical;
v⃗=∆ xt
(1.1)
while the pace is great mileage per unit of time or in the form of written
mathematical
v= xt
(1.2)
Description :
v⃗ : Speeds (m/s)
Δx : Change the position or the displacement (m)
t : The time interval (s)
v : Speed of (m/s)
x : Distance (m)
B. Tool and Material
1. Meter
2. Stopwatch
3. GLB tube
4. Stative
5. Tool written write
C. Identification Variables
Activity 1
1. Variable Control
As for the control variables of activity 1 is always
constant and fixed value is the distance.
2. Variable Manipulation
1 can be seen in the activities that are being manipulated
variable displacement because it causes changes in lab
activities.
3. Variable Response
Response variable in this practicum is the travel time
from the object.
Activity 2
1. Variable Control
Control variables in the second practicum is mileage.
2. Variable Manipulation
In the second activity, the height of which is variable
manipulation.
3. Variable Response
2 activity response variable is the travel time from titk
one particle to another point.
D. Definition Operational Variables
Activity 1
1. Variable Control
Distance is the total length of the path traversed. In the
first activity within each track together and fixed.
2. Variable Manipulation
Displacement is a big change of body position from the
initial position to the final position. In the first movement of
each activity is different trajectories result of the independent
variable.
3. Variable Response
Travel time is faster than an object to get to a certain
distance. Travel time changes due to the distance through which
the object trajectory.
Activity 2
1. Variable Control
Mileage distance is the total length of the path traversed.
In the second activity within each track together and fixed.
2. Variable Manipulation
Height is the vertical position of an object from a
specific point. Height in this activity is affected by distance
trajectory.
3. Variable Response
Travel time travel time is faster than an object to get to a
certain distance. Travel time changes due to the distance through
which the object trajectory.
E. Work Procedures
a. Activity 1
1. Making tracks in the rectangular space, then measure the length of
each side.
2. Provide the code on each corner with code A, B, C, and D.
3. Setting up your friends, as the object is moving at different speeds.
4. The first person standing at point A, then move toward point B, and
then measure the time it used to take the path from point A to point
B (try moving with constant velocity). Proceed to the second and
third then record the results in the table of observations.
5. Perform step 4 with different trajectory for example from point A
to point B and then to point C. Followed by some other path, record
the results in the table of observations
b. Activity 2
1. Take the tube GLB and Statif to pocket one end of the tube
2. Mark of at least 4 points as points A, B, C, and D on the tube (try
having the same interval).
3. Determine / measure the path length of the bottom of the tube (0
cm) to point A, to point B, to point C, and to point D.
4. Hanging one end of the tube at a certain height stative, start of
height about 5 cm from the bottom / base.
5. Lifting the other end of the tube, so that the bubble in the tube is in
the raised end.
6. Lowering the tip had reached the base / pedestal so that the bubble
will move up, measure the time it takes a bubble to reach point A
(start the stopwatch when the bubble right across the 0 cm position
on the tube), repeated retrieval of data as much as 3 times.
7. Repeat steps 4, 5 and 6, with different mileage (to point B, to C,
and to point C) record the results in the table of observations.
III. EXPERIMENTAL RESULTS AND DATA ANALYSIS
A. Observations
D 1,55 C
3,7 2,15
A B
2,13
1. Activity 1
Tabel 1.1.Results Measurement Distance, Displacement and Travel
Time.
NST Stopwatch : 0,1 s
NST Metered : 0,1 cm
No Trajectory Distance Shift Travel Time
(m) (m) (s)
1 From point A to B
1. 2,13
2. 2,13
3. 2,13
1. 2,13
2. 2,13
3. 2,13
1. 8,00
2. 2,50
3. 2,3
2From point A to B
and to C
1. 4,28
2. 4,28
3. 4,28
1. 3,02
2. 3,02
3. 3,02
1. 13,60
2. 5,70
3. 4,80
3 From point A to D
1. 3,70
2. 3,70
3. 3,70
1. 1,53
2. 1,53
3. 1,53
1. 16,00
2. 10,90
3. 7,00
4From point A to B
to C to D
1. 5,83
2. 5,83
3. 5,83
1. 2,13
2. 2,13
3. 2,13
1. 4,61
2. 4,61
3. 4,61
2. Activity 2
40302010
DCBA
cm
Tabel 1.2.Results Measurement Mileage and Travel Time Regular
Straight In Motion.
NST Stopwatch : 0,1 s
NST GLB Tube : 0,1 cm
No Height (cm) Mileage (cm) Travel Time (s)
1 5
10
1. 2,00
2. 2,00
3. 2,30
20
1. 4,00
2. 4,00
3. 4,40
30
1. 6,50
2. 6,70
3. 6,50
40
1. 8,70
2. 8,70
3. 8,40
2 10 10 1. 1,30
2. 1,30
3. 1,20
20
1. 2,60
2. 2,50
3. 2,70
30
1. 3,60
2. 3,60
3. 3,70
40
1. 5,10
2. 5,00
3. 5,00
3 15
10
1. 1,00
2. 1,00
3. 1,00
20
1. 1,90
2. 2,00
3. 2,00
30 1. 2,90
2. 3,00
3. 2,90
40
1. 4,00
2. 3,90
3. 3,80
B. Data Analysis
1. Activity 1
Based on the observations / measurements, great speed and
average speed of each person on each track are as follows:
a. Speed
From point A to B
First Person:
v1=Δ xt
=2,138,00
=0,26 m /s
Second Person:
v2=Δ xt
=2,132,50
=0,30 m / s
Third Person:
v3=Δ xt
=2,132,30
=0,43 m/ s
From point A to B to C
First Person :
v1=Δ xt
= 3,0213,60
=0,22 m /s
Second Person :
v2=Δ xt
=3,025,70
=0,53 m / s
Third Person :
v3=Δ xt
=3,024,80
=0,63 m /s
From point A to D
First Person :
v1=Δ xt
= 1,5316,00
=0,09m /s
Second Person :
v2=Δ xt
= 1,5310,90
=0,14 m /s
Third Person :
v3=Δ xt
=1,537,00
=0,22 m /s
From point A to B to C to D
First Person :
v1=Δ xt
=2,134,61
=0,46 m /s
Second Person :
v2=Δ xt
=2,134,61
=0,46 m /s
Third Person :
v3=Δ xt
=2,134,61
=0,46 m /s
b. Speed of
From point A to B
First Person :
v1=xt=2,13
8,00=0,29 m /s
Second Person :
v2=xt=2,13
2,50=0,85 m /s
Third Person
v3=xt=2,13
2,30=1,00 m /s
From point A to B to C
First Person :
v1=xt= 4,28
13,60=0,31 m /s
Second Person :
v2=xt=4,28
5,70=0,75 m / s
Third Person :
v3=xt=4,28
4,80=0,89m / s
From point A to D
First Person :
v1=xt= 3,70
16,00=0,23 m /s
Second Person :
v2=xt= 3,70
10,90=0,34 m / s
Third Person :
v3=xt=3,70
7,00=0,53 m /s
From point A to B to C to D
First Person :
v1=xt=5,83
4,61=1,26 m /s
Second Person :
v2=xt=5,83
4,61=1,26 m /s
Third Person :
v3=xt=5,83
4,61=1,26 m /s
c. Errors Analysis
v= st
→ v=s t−1
∆ v=|δvδs|∆ s+|δv
δt |∆ t
¿|δs t−1
δs |∆ s+|δs t−1
δt |∆ t
∆ vv
=∆ stv
+ s ∆ t
t2
v
∆ vv
=|∆ ss |+|∆ t
t |∆ v=|∆ s
s+ ∆ t
t |v
Speed
From point A to B
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,102,13
+ 0,058,00|0,26
¿|0,10 x8,00+2,13 x 0,052,13 x 8,00 |0,26
¿|0,80+0,1117,04 |0,26
¿| 0,9117,04|0,26
¿0,05 x 0,26
¿0,013m /s
KR=∆ v1
v1
x100 %
¿ 0,0130,26
x 100 %
¿5,9 %
KR=|0,260 ±0,013|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,102,13
+ 0,052,50|0,30
¿|0,10 x2,50+2,30 x 0,052,13 x2,50 |0,30
¿|0,25+0,015,33 |0,30
¿|0,265,33|0,30
¿0,048 x 0,30
¿0,014 m/ s
KR=∆ v2
v2
x100 %
¿ 0,0140,30
x 100 %
¿4,8 %
KR=|0,300 ±0,014|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,102,13
+ 0,052,30|0,43
¿|0,10 x2,30+2,13 x 0,052,13 x 2,30 |0,43
¿|0,23+0,114,89 |0,43
¿|0,344,89|0,43
¿0,07 x 0,43
¿0,03 m /s
KR=∆ v3
v3
x100 %
¿ 0,030,43
x100 %
¿6,9 %
KR=|0,430 ±0,030|
From point A to B to C
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,103,02
+ 0,0513,60|0,22
¿|0,10 x13,60+3,02 x 0,053,02 x 13,60 |0,22
¿|1,36+0,1541,07 |0,22
¿| 1,5141,07|0,22
¿0,04 x 0,22
¿0,008m /s
KR=∆ v1
v1
x100 %
¿ 0,0080,22
x 100 %
¿4 %
KR=|0,220 ±0,008|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,103,02
+ 0,055,70|0,53
¿|0,10 x5,70+3,02 x 0,053,02x 5,7 |0,53
¿|0,57+0,1517,21 |0,53
¿| 0,7217,21|0,53
¿0,04 x 0,53
¿0,021 m /s
KR=∆ v2
v2
x100 %
¿ 0,0210,53
x100%
¿3,9 %
KR=|0,530 ±0,021|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,103,02
+ 0,054,80|0,63
¿|0,10 x 4,80+3,02 x 0,053,29 x 8,00 |0,63
¿|0,48+0,1514,49 |0,63
¿| 0,6314,49|0,63
¿0,043 x 0,63
¿0,027 m / s
KR=∆ v3
v3
x100 %
¿ 0,0270,63
x 100 %
¿4,3 %
KR=|0,630 ±0,027|
From point A to D
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,101,53
+ 0,0516,00|0,09
¿|0,10 x16,00+1,53 x 0,051,53 x 16,00 |0,09
¿|1,6+0,0924,48 |0,09
¿| 3,1355,63|0,09
¿0,069 x0,09
¿0,006 m / s
KR=∆ v1
v1
x100 %
¿ 0,0060,09
x 100 %
¿6,60 %
KR=|0,090 ±0,006|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,101,53
+ 0,0510,90|0,14
¿|0,10 x10,90+1,53 x 0,051,53 x 10,90 |0,14
¿|1,09+0,0716,67 |0,14
¿| 1,1616,67|0,14
¿0,069 x0,14
¿0,009 m /s
KR=∆ v2
v2
x100 %
¿ 0,0090,14
x100 %
¿6,4 %
KR=|0,140 ±0,009|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,101,53
+ 0,057,00|0,22
¿|0,10 x7,00+1,53 x 0,051,53 x 7,00 |0,22
¿|0,70+0,0810,70 |0,22
¿| 0,7810,70|0,22
¿0,072 x0,22
¿0,015m /s
KR=∆ v3
v3
x100 %
¿ 0,0150,22
x 100 %
¿6,80 %
KR=|0,220 ±0,015|
From point A to B to C to D
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,102,13
+ 0,054,61|0,46
¿|0,10 x 4,61+2,13 x 0,052,13 x2,61 |0,46
¿|0,046+0,1069,82 |0,46
¿|0,159,82|0,46
¿0,007 m / s
KR=∆ v1
v1
x100 %
¿ 0,0070,46
x 100 %
¿1,50 %
KR=|0,460 ±0,007|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,102,13
+ 0,054,61|0,46
¿|0,10 x 4,61+2,13 x 0,052,13 x 4,61 |0,46
¿|0,46+0,1069,82 |0,46
= |0,509,82|0,46
¿0,005m /s
KR=∆ v2
v2
x100 %
¿ 0,0050,46
x 100 %
¿1 %
KR=|0,460 ±0,005|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,102,13
+ 0,054,61|0,46
¿|0,10 x 4,61+2,13 x 0,052,13 x 4,61 |0,46
¿|0,46+0,1069,82 |0,46
= |0,509,82|0,46
¿0,005m /s
KR=∆ v3
v3
x100 %
¿ 0,0050,46
x 100 %
¿1 %
KR=|0,460 ±0,005|
Speed of
From point A to B
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,102,13
+ 0,058,00|0,29
¿|0,10 x8,00+2,13 x 0,052,13 x 8,00 |0,29
¿|0,80+0,106517,04 |0,29
¿|0,906517,04 |0,29
¿0,05 x 0,29
¿0,001 m /s
KR=∆ v1
v1
x100 %
¿ 0,0010,29
x100 %
¿0,34 %
KR=|0,290 ±0,001|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,102,13
+ 0,052,50|0,85
¿|0,10 x2,50+2,30 x 0,052,13 x2,50 |0,85
¿|0,25+0,015,33 |0,85
¿|0,265,33|0,85
¿0,048 x 0,85
¿0,04 m/ s
KR=∆ v2
v2
x100 %
¿ 0,040,85
x 100 %
¿4,7 %
KR=|0,850 ±0,040|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,102,13
+ 0,052,30|0,43
¿|0,10 x2,30+2,13 x 0,052,13 x 2,30 |1,00
¿|0,23+0,114,89 |1,00
¿|0,344,89|1,00
¿0,07 x 1,00
¿0,07 m / s
KR=∆ v3
v3
x100 %
¿ 0,071,00
x 100 %
¿7 %
KR=|1,000± 0,070|
From point A to B to C
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,104,28
+ 0,0513,60|0,31
¿|0,10 x13,60+4,28 x0,054,28 x13,60 |0,31
¿|1,36+0,21458,208 |0,31
¿| 1,5758,208|0,31
¿0,027 x 0,31
¿0,008m /s
KR=∆ v1
v1
x100 %
¿ 0,0080,31
x 100 %
¿2,5 %
KR=|0,3100 ±0,008|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,104,28
+ 0,055,70|0,75
¿|0,10 x5,7+4,28 x0,054,28 x5,70 |0,75
¿|0,57+0,21424,39 |0,75
¿|0,78424,39|0,75
¿0,032 x0,75
¿0,024 m/ s
KR=∆ v2
v2
x100 %
¿ 0,0240,75
x 100 %
¿3,2 %
KR=|0,750 ±0,024|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,104,28
+ 0,054,80|0,89
¿|0,10 x 4,80+4,28 x 0,054,28 x 4,80 |0,89
¿|0,48+0,21420,54 |0,89
¿|0,69420,54|0,89
¿0,033 x 0,89
¿0,029 m /s
KR=∆ v3
v3
x100 %
¿ 0,0290,89
x100 %
¿3,2 %
KR=|0,890 ±0,029|
From point A to D
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,103,70
+ 0,0516,00|0,23
¿|0,10 x16,00+3,70 x 0,053,70 x 16,00 |0,23
¿|1,60+0,18559,20 |0,23
¿|1,78559,20|0,23
¿0,03 x 0,23
¿0,0069 m /s
KR=∆ v1
v1
x100 %
¿ 0,00690,23
x100 %
¿3 %
KR=|0,230 ±0,007|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,103,70
+ 0,0510,90|0,34
¿|0,10 x10,90+3,70 x 0,053,70 x 10,90 |0,34
¿|1,09+0,18540,33 |0,34
¿|1,27540,33|0,34
¿0,0361 x0,34
¿0,0107 m / s
KR=∆ v2
v2
x100 %
¿ 0,01070,34
x 100 %
¿3,15 %
KR=|0,340 ±0,011|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,103,70
+ 0,057,00|0,53
¿|0,10 x7,00+3,70 x 0,053,70 x 7,00 |0,53
¿|0,70+0,18596,360 |0,53
¿|0,88525,9 |0,53
¿0,034 x 0,53
¿0,018m /s
KR=∆ v3
v3
x100 %
¿ 0,0180,53
x 100 %
¿3,4 %
KR=|0,530 ±0,018|
From point A to B to C to D
First Person :
∆ v1=|∆ ss
+ ∆ tt |v1
¿|0,105,83
+ 0,054,61|1,26
¿|0,10 x 4,61+5,83 x 0,055,83 x 4,61 |0,26
¿|0,46+0,2926,876 |1,26
¿| 0,7526,876|1,26
¿0,0279 x1,26
¿0,035m /s
KR=∆ v1
v1
x100 %
¿ 0,0351,26
x 100 %
¿2,7 %
KR=|1,260± 0,035|
Second Person :
∆ v2=|∆ ss
+ ∆ tt |v2
¿|0,105,83
+ 0,054,61|1,26
¿|0,10 x 4,61+5,83 x 0,055,83 x 4,61 |0,26
¿|0,46+0,2926,876 |1,26
¿| 0,7526,876|1,26
¿0,0279 x1,26
¿0,035m /s
KR=∆ v2
v2
x100 %
¿ 0,0351,26
x 100 %
¿2,7 %
KR=|1,260± 0,035|
Third Person :
∆ v3=|∆ ss
+ ∆ tt |v3
¿|0,105,83
+ 0,054,61|1,26
¿|0,10 x 4,61+5,83 x 0,055,83 x 4,61 |0,26
¿|0,46+0,2926,876 |1,26
¿| 0,7526,876|1,26
¿0,0279 x1,26
¿0,035m /s
KR=∆ v3
v3
x100 %
¿ 0,0351,26
x 100 %
¿2,7 %
KR=|1,260± 0,035|
2. Activity 2
a. Speed
Height of 5 cm
Point A
vA=Δ xt
= 102,10
=4,76 cm /s
Point B
vB=Δ xt
= 204,13
=4,84 cm/ s
Point C
vC=Δ xt
= 306,56
=4,57 cm / s
Point D
vD= Δ xt
= 408,60
=4,65 cm/s
Height of 10 cm
Point A
vA=Δ xt
= 101,26
=7,94 cm / s
Point B
vB=Δ xt
= 202,60
=7,69 cm /s
Point C
vC=Δ xt
= 303,63
=8,26 cm /s
Point D
vD= Δ xt
= 405,03
=7,95 cm / s
Height of 15 cm
Point A
vA=Δ xt
= 101,00
=10,00 cm /s
Point B
vB=Δ xt
= 201,96
=10,20 cm /s
Point C
vC=Δ xt
= 302,93
=10,24 cm /s
Point D
vD= Δ xt
= 403,96
=10,10 m / s
b. Errors Analysis
v= st
→ v=s t−1
∆ v=|δvδs|∆ s+|δv
δt |∆ t
¿|δs t−1
δs |∆ s+|δs t−1
δt |∆ t
∆ vv
=∆ stv
+ s ∆ t
t2
v
∆ vv
=|∆ ss |+|∆ t
t |∆ v=|∆ s
s+ ∆ t
t |v
Height of 5 cm
Point A
∆ v A=|∆ ss
+ ∆ tt |v A
¿|0,1010
+ 0,052,10|4,76
¿|0,10 x2,10+10 x 0,0510 x 2,10 |4,76
¿|0,21+0,521,00 |4,76
¿| 0,7521,00|4,76
¿0,033 x 4,76
¿0,157cm / s
KR=∆ v A
v A
x 100 %
¿ 0,1574,76
x 100 %
¿3,3 %
KR=|4,76 ± 0,16|
Point B
∆ vB=|∆ ss
+ ∆ tt |vB
¿|0,1020
+ 0,054,13|4,84
¿|0,10 x 4,13+20 x0,0520 x 4,13 |4,84
¿|0,413+120 x 4,13|4,84
¿|1,41382,6 |4,84
¿0,017 x 4,84
¿0,0822 cm /s
KR=∆ vB
vB
x 100 %
¿ 0,08224,84
x100 %
¿1,69%
KR=|4,84 ± 0,08|
Point C
∆ vC=|∆ ss
+ ∆ tt |vC
¿|0,1030
+ 0,056,56|4,57
¿|0,10 x6,56+30 x 0,0530 x 6,56 |4,57
¿|0,656+1,5196,80 |4,57
¿| 2,156196,80|4,57
¿0,0109 x 4,57
¿0,0498 cm / s
KR=∆ vC
vC
x100 %
¿ 0,04984,57
x 100 %
¿1,09%
KR=|4,57 ± 0,05|
Point D
∆ vD=|∆ ss
+ ∆ tt |v D
¿|0,1040
+ 0,058,60|4,65
¿|0,10 x8,60+40 x0,0540 x 8,60 |4,65
¿|0,86+2344 |4,65
¿|2,86344 |4,65
¿0,0683 x 4,65
¿0,0385 cm / s
KR=∆ vD
vD
x100 %
¿ 0,03854,65
x 100 %
¿6,83%
KR=|4,65 ± 0,04|
Height of 10 cm
Point A
∆ v A=|∆ ss
+ ∆ tt |v A
¿|0,1010
+ 0,051,26|7,94
¿|0,10 x1,26+10 x 0,0510 x 1,26 |7,94
¿|0,126+0,512,6 |7,94
¿|0,62612,6 |7,94
¿0,0496 x7,94
¿0,394 cm /s
KR=∆ v A
v A
x 100 %
¿ 0,3947,94
x 100 %
¿4,9 %
KR=|7,94 ± 0,40|
Point B
∆ vB=|∆ ss
+ ∆ tt |vB
¿|0,1020
+ 0,052,60|7,69
¿|0,10 x2,60+20 x 0,0520 x 2,60 |7,69
¿|0,26+152 |7,69
¿|1,2652 |7,69
¿0,0242 x7,69
¿0,1860 cm / s
KR=∆ vB
vB
x 100 %
¿ 0,18607,69
x 100 %
¿1,34%
KR=|7,70± 0,86|
Point C
∆ vC=|∆ ss
+ ∆ tt |vC
¿|0,1030
+ 0,053,63|8,26
¿|0,10 x3,63+30 x 0,0530 x 8,33 |8,26
¿|0,363+1,5108,9 |8,26
¿|1,863108,9|8,26
¿0,0171 x8,26
¿0,1412 cm /s
KR=∆ vC
vC
x100 %
¿ 0,14128,26
x100 %
¿1,7 %
KR=|8,26 ± 0,14|
Point D
∆ vD=|∆ ss
+ ∆ tt |v D
¿|0,1040
+ 0,055,03|7,95
¿|0,10 x5,03+40 x0,0540 x12,53 |7,95
¿|0,503+2201,2 |7,95
¿|2,503201,2|7,95
¿0,0124 x 7,95
¿0,0985 cm / s
KR=∆ vD
vD
x100 %
¿ 0,09857,95
x 100 %
¿1,2 %
KR=|7,95± 0,01|
Height of 15 cm
Point A
∆ v A=|∆ ss
+ ∆ tt |v A
¿|0,1010
+ 0,051,00|10
¿|0,10 x1,00+10 x 0,0510 x 1,00 |10
¿|0,1+0,510 |10
¿|0,6010 |10
¿0,06 x10,00
¿0,60 cm / s
KR=∆ v A
v A
x 100 %
¿ 0,6010,00
x100%
¿6 %
KR=|10,00± 0,60|
Point B
∆ vB=|∆ ss
+ ∆ tt |vB
¿|0,1020
+ 0,051,96|10,2
¿|0,10 x1,96+20 x 0,0520 x 1,96 |10,2
¿|0,196+139,2 |10,2
¿|1,19639,2 |10,2
¿0,0305 x10,2
¿0,3111cm /s
KR=∆ vB
vB
x 100 %
¿ 0,311110,2
x100 %
¿3,05 %
KR=|10,20± 0,31|
Point C
∆ vC=|∆ ss
+ ∆ tt |vC
¿|0,1030
+ 0,052,93|10,24
¿|0,10 x2,93+30 x 0,0530 x 2,93 |10,24
¿|0,293+1,587,9 |10,24
¿|1,79387,9 |10,24
¿0,0204 x 10,24
¿0,2089 cm / s
KR=∆ vC
vC
x100 %
¿ 0,208910,24
x100 %
¿2,04 %
KR=|10,2 4± 0,21|
Point D
∆ vD=|∆ ss
+ ∆ tt |v D
¿|0,1040
+ 0,053,96|10,10
¿|0,10 x3,96+40 x0,0540 x3,96 |10,10
¿|0,396+2158,4 |10,10
¿| 2,3961548,4|10,10
¿0,0151 x10,10
¿0,5151 m /s
KR=∆ vD
vD
x100 %
¿ 0,515110,10
x100 %
¿5,1 %
KR=|10,10± 0,16|
c. Graph showing the relationship between mileage and travel time
5 10 15 20 25 30 35 40 450
0.51
1.52
2.53
3.54
4.5
f(x) = 0.0987 xR² = 0.999665467755094
Height of 5 cm
Height of 5 cm Travel Dis-tanceLinear (Height of 5 cm Travel Distance)
Travel Distance (cm)
Trav
el T
ime
(s)
t=ms+c
m= st
→ m= ts= 1
m
¿ 10,219
=4,5662 cm / s
KR=(1−R2 ) x100 %
¿ (1−0,998 ) x 100 %
¿0,2 %
5 10 15 20 25 30 35 40 450
0.51
1.52
2.53
3.54
4.5
f(x) = 0.0987 xR² = 0.999665467755094
Height of 10 cm
Height of 5 cm Travel Dis-tanceLinear (Height of 5 cm Travel Distance)
Travel Distance (cm)
Trav
el T
ime
(s)
t=ms+c
m= st
→ m= ts= 1
m
¿ 10,318
=10,1317 cm /s
KR=(1−R2 ) x100 %
¿ (1−0,9997 ) x 100%
¿0,03 %
5 10 15 20 25 30 35 40 450
0.51
1.52
2.53
3.54
4.5
f(x) = 0.0987 xR² = 0.999665467755094
Height of 15 cm Travel Distance
Height of 5 cm Travel Dis-tanceLinear (Height of 5 cm Travel Distance)
Travel Distance (cm)
Trav
el T
ime
(s)
t=ms+c
m= st
→ m= ts= 1
m
¿ 10,318
=10,1317 cm /s
KR=(1−R2 ) x100 %
¿ (1−0,9997 ) x 100%
¿0,03 %
IV. DISCUSSION
1. Activity 1
In activities 1 have been carried out on the measurement of
distance, displacement and travel time by using stopwacth and
rectangular shaped trajectory. In Activity 1 required 3 models that will
move at different speeds, but the trajectory starting from point A to B,
point A to B to C, point A to B to C to D, and the last track from point
A to B to C to D and back to A.
The observations in the table it can be seen that there are
differences in the travel time of each model, it is because the
difference in speed of each model.
In the data analysis performed summation error analysis it is
intended to see the error at the time of measurement, because each
measurement there must be mistakes. Measurement error can be
caused by several factors, among which kejlihan reading measurement
tools, precision holder with models during the stopwatch stops and
starts, the measuring instrument itself, etc.
2. Activity 2
Activities 2 have been carried out on the measurement of
distance and travel time in uniform rectilinear motion with a
stopwatch, stative, and GLB tube, and meters. On the activities needed
four points with the same distance between one point with another
point. Data collection was performed by three in order to obtain more
accurate data.
On the activities of these measurements, the height of the tube
GLB changed from 5 cm to 15 cm in turn. Measurement results
obtained were different from each trajectory and height of the tubes
GLB. However, in these measurements, there are several errors that
are caused by these factors. The factors that promote the occurrence of
measurement errors tube placement is not appropriate, the reading
results are less precise measurements, there is no measurement tool
that can read the distance between the two scales, the slope of the tube
does not fit, the measuring instrument itself, etc..
V. CONCLUSION
1. Activity 1
Based on the observations that have been made to the
measurements of distance, displacement and travel time of an object
in order to achieve certain goals or to get to the point of one another to
the point that it can be concluded that the faster the object is moving
toward the point purpose the time needed to arrive at the destination is
getting shorter or in other words the slower the object was circling
trajectory the longer the time it takes and vice versa.
2. Activity 2
Based on the observations that have been made to the
measurements of distance and travel time in straight motion regular
can be concluded that the higher the placement of objects then the
faster the object moves left the place of origin or the higher the
position of objects that travel time to the movement of an object or
move for reached the surface only takes a short time.
REFERENCES
Laboratorium Fisika Unit Praktikum Fisika Dasar-FMIPA Universitas Negeri Makassar. 2013. Penuntun Praktikum Fisika Dasar I.
Rice University & OpenStage College .2013. College Physics.
Alonso, Marcello & Edward J. Finn. 1980. Dasar-Dasar Fisika Universitas. Erlangga. Jakarta .
http://biologimediacentre.com/mengenal-macam-variabel-dalam-percobaan/
http://temukanpengertian.blogspot.com/2013/06/pengertian-variabel-kontrol.html
Makassar, November 2013
Assistant, Praktikan,
Muh. Ihsanul Amri Fildia Putri
Nim: 1112040183 Nim:1313440017