Bsme 1 3 G1 E2
Bsme 1 3 G1 E2
Bsme 1 3 G1 E2
DATA:
Figure 5: The total distance traversed by the steel ball for every time interval.
1. From the tabulated distance measurements, compute the distance traversed by the
steel ball in each successive time interval and record these values. The distance
traversed during the first time interval and the distance traversed for the next time
interval.
Trial 1 Trial 2
S1=35 cmS2=68 cm S1=33 cmS2=63.2 cm
Distance traversed during 1 interval (cm): Distance traversed during 1 interval (cm):
S2−S 1=S S2−S 1=S
68 cm−35 cm=33 cm 63.2 cm−33 cm=30.2 cm
Trial 3 Trial 4
S1=34 cmS2=65 cm S1=35.1 cmS2=67.2 cm
Distance traversed during 1 interval (cm): Distance traversed during 1 interval (cm):
S2−S 1=S S2−S 1=S
65 cm−34 cm=31 cm 67.2 cm−35.1 cm=32.1 cm
Trial 5 Trial 6
S1=36 cmS2=71.6 cm S1=37 cmS2=73 cm
Distance traversed during 1 interval (cm): Distance traversed during 1 interval (cm):
S2−S 1=S S2−S 1=S
71.6 cm−36 cm=35.6 cm 73 cm−37 cm=36 cm
2. Using the distance traversed in each time interval and the recorded interval, compute the
average speed during each time interval.
Trial 1:
Given: Formula:
t=1.6 s S1=35 cm s v + v +v
v= v ave = 1 2 3
S2=33 cm S3=32 cm t 3
Solution:
S1 35 cm S3 32cm
v1 = ¿ ¿ 21.9 cm/ s v3 = ¿ ¿ 20 cm /s
t 1.6 s t 1.6 s
S2 33 cm
v 2= ¿ ¿ 20.6 cm/ s 21.9+20.6+ 20
t 1.6 s v ave = v ave =20 .8 cm / s
3
Trial 2:
Given: Formula:
t=1.09 s S1=33 cm s v + v +v
v= v ave = 1 2 3
S2=30.2 cm t 3
S3=36.8 cm
Solution:
S1 33 cm S3 36.8 cm
v1 = ¿ ¿ 30.3 cm/ s v3 = ¿ ¿ 33.8 cm/s
t 1.09 s t 1.09 s
S2 30.2cm
v 2= ¿ 30.3+27.7+ 33.8
t 1.09 s v ave = v ave =30 .6 cm/ s
3
¿ 27.7 cm/s
Trial 3:
Given: Formula:
t=0.9 s S1=34 cm s v + v +v
v= v ave = 1 2 3
S2=31 cmS3=35 cm t 3
Solution:
S1 34 cm S3 35 cm
v1 = ¿ ¿ 37.8 cm /s v3 = ¿ ¿ 38.9 mc/ s
t 0.9 s t 0.9 s
S2 31cm
v 2= ¿ ¿ 34.4 cm/ s 37.8+34.4 +38.9
t 0.9 s v ave = v ave =37 .0 cm/s
3
Trial 4:
Given: Formula:
t=0.72 s s v + v +v
S1=35.1 cm v= v ave = 1 2 3
t 3
S2=32.1 cm
S3=32.8 cm
Solution:
S1 35.1cm S3 32.8 cm
v1 = ¿ v3 = ¿ ¿ 45.6 cm/s
t 0.72 s t 0.72 s
¿ 48.8 cm/ s
Trial 5:
Given: Formula:
t=0.53 s S1=36 cm s v + v +v
v= v ave = 1 2 3
S2=35.6 cm t 3
S3=28.4 cm
Solution:
S1 36 cm S3 28.4 cm
v1 = ¿ v3 = ¿ ¿ 53.58 cm/s
t 0.53 s t 0.53 s
¿ 67.92 cm/s
Trial 6:
Given: Formula:
t=0.37 s S1=37 cm s v + v +v
v= v ave = 1 2 3
S2=36 cmS3=27 cm t 3
Solution:
S1 37 cm S3 27 cm
v1 = ¿ ¿ 100 cm/s v3 = ¿ ¿ 72.97 cm /s
t 0.37 s t 0.37 s
S2 36 cm
v 2= ¿ 100+ 97.30+ 72.97
t 0.37 s v ave = v ave =90 . 09 cm/s
3
¿ 97.30 cm/s
1 2
3. Using equation 4, s=v 1 t + a t , compute the acceleration using the total distance recorded and
2
the corresponding total time. Compute the average value of the acceleration thus obtained.
Formula:
1
s=v 1 t + a t 2
2
Trial 1:
Given:
Time (t) = 4.8s
Distance (s) = 100cm
v1 =0 cm/ s
Required:
a=?
Solution:
1
s=v 1 t + a t 2
2
1
100 cm=( 0 )( 4.8 s )+ ( a )( 4.8 s )
2
2
[ 1 2
]
100 cm= ( a ) ( 4.8 s ) ( 2 )
2
2
200 cm a ( 4.8 s )
200 cm=a ( 4.8 s )2 2
= 2
( 4.8 s ) ( 4.8 s )
2
8.69 cm /s =a
Trial 2:
Given:
Time (t) = 3.27s
Distance (s) = 100cm
v1 =0 cm/ s
Required:
a=?
Solution:
1
s=v 1 t + a t 2
2
1
100 cm=( 0 )( 3.27 s ) + ( a ) ( 3.27 s )
2
2
[ 1 2
]
100 cm= ( a ) ( 3.27 s ) ( 2 )
2
2
200 cm a ( 3.27 s )
200 cm=a ( 3.27 s )2 2
= 2
( 3.27 s ) ( 3.27 s )
2
18.70 cm/s =a
Trial 3:
Given:
Time (t) = 2.70s
Distance (s) = 100cm
v1 =0 cm/ s
Required:
a=?
Solution:
1
s=v 1 t + a t 2
2
1
100 cm=( 0 )( 2.70 s ) + ( a ) ( 2.70 s )
2
2
[ 1
]
100 cm= ( a ) ( 2.70 )2 ( 2 )
2
2
200 cm a ( 2.70 s )
200 cm=a ( 2.70 s )
2
2
= 2
( 2.70 s ) ( 2.70 s )
2
27.43 cm/s =a
Trial 4:
Given:
Time (t) = 2.15s
Distance (s) = 100cm
v1 =0 cm/ s
Required:
a=?
Solution:
1 2
s=v 1 t + a t
2
1
100 cm=( 0 )( 2.15 s ) + ( a ) ( 2.15 s )2
2 [ 1 2
]
100 cm= ( a ) ( 2.15 s ) ( 2 )
2
2
2 200 cm a ( 2.15 s )
200 cm=a ( 2.15 s ) 2
= 2
( 2.15 s ) ( 2.15 s )
2
43.27 cm/s =a
Trial 5:
Given:
Time (t) = 1.60s
Distance (s) = 100cm
v1 =0 cm/ s
Required:
a=?
Solution:
1 2
s=v 1 t + a t
2
1
100 cm=( 0 )( 1.6 s ) + ( a ) ( 1.6 s )
2
2
[ 1
]
100 cm= ( a ) ( 1.6 s )2 ( 2 )
2
2
200 cm a ( 1.6 s )
200 cm=a ( 41.6 )
2
2
= 2
( 1.6 s ) ( 1.6 s )
78.13 cm /s 2=a
Trial 6:
Given:
Time (t) = 1.10s
Distance (s) = 100cm
v1 =0 cm/ s
Required:
a=?
Solution:
1
s=v 1 t + a t 2
2
1
100 cm=( 0 )( 1.1 s )+ ( a ) ( 1.1 s )
2
2
[ 1
]
100 cm= ( a ) ( 1.1 s )2 ( 2 )
2
2
200 cm a ( 1.1 s )
200 cm=a ( 41.6 )
2
2
= 2
( 1.1 s ) (1.1 )
2
165.29 cm/s =a
Average Acceleration:
a1+ a2 +a3 + a4 + a5+ a6 8.68+ 18.70+ 27.43+43.27+ 78.13+ 165.29 2
a ave = a ave = a ave =56 . 92cm/ s
6 6
4. Plot a curve to show the relationship between the total distance traveled and the time
elapsed.
Trial 1 Trial 2
Trial 3 Trial 4
Trial 5 Trial 6
CONCLUSION:
The main purpose of conducting this experiment was to study the uniformly accelerated motion of a
body rolling down an inclined plane. The data in trial 1 was based on the online experiment while the data in
trials 2-6 were recorded during the face-to-face experiment. A condition wherein the angle of elevation of the
inclined plane gets higher per trial was applied. The result of the experiments showed that the first trial had an
acceleration of 8.68 cm/s2, 18.70 cm/s2 for the 2nd trial, 27.43 cm/s2 for the 3rd trial, 43.27 cm/s2 for the 4th trial,
78.13 cm/s2 for the 5th trial and 165.29 cm/s2 for the acceleration of the 6th trial. From the following results, it is
evident that the value of acceleration increases per trial. Thus, we can conclude that the higher the elevation,
the faster the acceleration. We can also conclude that the lower the elevation, the longer the time it takes for
the body to traverse the total distance of the inclined plane. In addition, since the object rolled down in an
inclined surface, the body’s velocity in each trial changes at a constant rate, undergoing a uniform acceleration
regardless of time.
However, relative velocity is the speed of one object B in the rest frame of another object A.
There are times when one or more objects move in a frame that isn't stationary relative to another
observer. For instance, a boat traveling through a river with a certain rate of flow or an airplane flying
through the air while being affected by the wind. It is sometimes defined as the velocity of a body in
reference to another body and average velocity is velocity between certain time periods; it's like a kind
of mean(μ) of two velocities.
Uniform acceleration, on the other hand, refers to a constant acceleration over a period of time.
This means that the acceleration of an object does not change over time, and it has a fixed value. An
example of uniform acceleration is when an object is falling freely due to gravity, as gravity produces a
constant acceleration of 9.8 m/s^2 near the Earth's surface.
4. Describe an experiment other than that which you have performed in the laboratory
which would show how a body with uniformly accelerated motion behaves.
It uses an inclined plane at a given angle then, as a substitute for the steel ball, it uses a cart
attached with a ticker tape timer that will produce dots on the paper which will determine the
acceleration of the cart as the angle of the inclined plane increases. The ruler will determine the
distance traveled by the cart in the form of a dot (e.g., 5 dots = 1.5cm). It first identifies 55 continuous
dots from the ticker tape by counting the 5 dots interval. From that, we get the difference from the 5
dots consecutively that determines the distance traveled. To get the average speed, the distance
traveled by the cart over time, which is 5 dots. For acceleration, final velocity minus the initial velocity all
over time. Most of the results result in the same acceleration, as the cart is moving downwards the
acceleration should be constant and with that it observes the uniform accelerated motion.
5. How is uniform linear acceleration related to acceleration due to gravity? Give their
similarities and/or differences by citing specific examples.
If there is no air resistance and an object is at a particular location on Earth, then it will fall with a
uniform linear acceleration, which is associated with the acceleration due to gravity. An object
experiences acceleration owing to gravity when it descends freely toward the surface of the earth from
a specific height. This acceleration is caused by a change in the object's velocity. The force of gravity
causes objects to fall towards the ground. The force of gravity accelerates objects as they descend to
the ground's surface. It is claimed that an object experiences uniform acceleration when it is moving in
a straight line while accelerating at certain periods of time. Uniform acceleration is demonstrated by an
object free falling. The term “Acceleration” refers to a change in velocity, which is a measurement of
motion's direction and speed.