Laboratory 1:
One Dimensional Motion
Aleksandr Shpiler
PH1004 Section E2
January 26, 2005
Abstract
The objective of this lab was to investigate one dimensional motion in a straight line. The
first part of the experiment analyzes instantaneous and average velocity in one dimensional
motion. The second half of the experiment studies accelerated motion. Instantaneous velocity
was obtained using a movement of an object on virtually frictionless surface over a period of
time. The data from experiment was acquired using photo gates and Data Acquisition, a software
to record the data.
Raw Data Tables and Analysis
Part A
Table 1: Data from part A of the experiment
Trial
1
2
3
4
5
6
7
8
9
10
Part A
xb
xG1
Delta x
107.8
70.2
0.05
tG3 – tG1
2.727
2.705
2.709
2.721
2.729
2.709
2.732
2.705
2.719
2.7
tbG2
0.072
0.074
0.072
0.072
0.074
0.074
0.074
0.074
0.074
0.073
cm
cm
cm
xub
xG3
Delta t
110.2
159.9
0.001
cm
cm
s
m
m
m
xub
xG3
Delta t
1.102
1.599
0.001
m
m
s
Above table converted to meters
Part A
xb
xG1
Delta x
1.078
0.702
0.0005
Table 2: Results of motion with acceleration experiment (original data)
Trial
1
2
3
4
5
6
7
8
9
10
xG1, cm
70.58
78.35
84.45
89.5
97.05
106.85
115.1
123.1
130.85
144
tbG1, s
0.027
0.025
0.026
0.022
0.022
0.019
0.02
0.019
0.017
0.02
Table 2a: Results of motion with acceleration experiment (data in meters)
Trial
1
2
3
4
5
6
7
8
9
10
xG1, m
0.7058
0.7835
0.8445
0.895
0.9705
1.0685
1.151
1.231
1.3085
1.44
tbG1, s
0.027
0.025
0.026
0.022
0.022
0.019
0.02
0.019
0.017
0.02
Table 3: Instantaneous Velocity (original data)
Part B
xinital
Delta x
Delta t
60
0.05
0.001
cm
cm
s
Table 3a: Instantaneous Velocity (data converted to meters)
Part B
xinital
Delta x
Delta t
0.6
0.0005
0.001
m
m
s
Part A
Average Velocity:
1. The distance between the photogates was found by subtracting the position of the third gate
G3 from the position of the first gate G1:
xfi = xG3 – xG1 = 1.599 - 0.702= 0.897 m
2. The uncertainty for xfi was found by adding the two uncertainties xG1 and xG2.
The uncertainty for each is 0.0005m which brings the total to:
xfi = xG1 + xG3 = 0.001 m.
3. The average time it took the glider to move from one photogate to another is the average of all
the times for the ten trials:
t ft = AVG (tG3-tG1) = (2.727+2.705+2.709+2.721+2.729+2.709+2.732+2.705+2.719+2.7) / 10
= 2.7156 s
4. The uncertainty of t13 comes from two sources. One of the sources is statistical fluctuation
which is calculated using standard deviation of the ten trials. Standard deviation:
fi = 0.011423 s.
The other source is the uncertainty of the time measurement by photogates:
tins = 0.001s
The total uncertainty in the time is thus given by:
2
t fi   fi  t ins
=
0.011423 2  0.0012 = 0.011467 s.
2
5. With enough information we can now calculate the average velocity:
v av 
x fi

t fi
0.897m
m
 0.330314
2.7156s
s
6. Now we calculate the part of the uncertainty incurred by the distance measurement.
v av, x 
x ft
t ft
=
0.001
m
0.000368
2.7156
s
Now we calculate the uncertainty in the average velocity due to imprecision in time measure.
vav,t 
x ft
t
2
ft
t ft =
0.897
m
0.011467 = 0.001395
7.3744
s
Finally we calculate the total uncertainty of the average velocity.
v av  x ave, x  x ave,t = 0.000368 + 0.001395 = 0.001763
m
s
Instantaneous velocity:
1. In order to calculate the instantaneous velocity we have to find the width of the flag:
w flag =|xub– xb| = 1.102 – 1.078 = 0.024 m
2. The uncertainty of the flag measurement is the sum of the uncertainties for each measurement,
since there are two measurements and the uncertainty of each is 0.0005 m, the total uncertainty
for the flag is:
w = xub + xb = 0.0010 m
3. The average time for the flag to pass the second photo gate is:
tavg = AVG (tbG2) = (0.072+0.072+0.072+0.074+0.074+0.074+0.074+0.074+0.074+0.073)/10
= 0.0733 s
4. The uncertainty in the time it takes the flag to pass through the photo gate depends on two
uncertainties. The first quantity is t b  fl  standard deviation of the times measured. Standard
deviation is:
 = 0.000949 s.
The other source is the uncertainty of the time measurement by photogates:
t bins  0.001s.
The total uncertainty in time for the instantaneous velocity is:
t b   b2 fl  t b2ins  (0.000949 s) 2  (0.001s) 2 = 0.001378 s
5. The instantaneous velocity is:
v
v=
w
tb
0.024
m
= 0.327422
0.0733
s
Uncertainty due to the flag width measurement is:
v w 
w
0.0010
m
=
= 0.013643
0.001378
s
tb
The uncertainty due to the imprecision in time is:
v t 
w
0.024
m
t b =
0.001378 = 0.006157
2
2
s
tb
(0.001378)
Both components and the total uncertainty are:


v = vw +vt. = 0.013643
m
m
m
+ 0.006157
= 0.0198
s
s
s
Part B
1. The instantaneous velocity of every trial of the experiment in part B is
Calculated by dividing the width of the flag w by the time it took the flag to pass by photo gate
v= w/tbG1
m
s
x = xG1 - xinitial
in meters
v2
m2
s2
0.888889
0.96
0.923077
1.090909
1.090909
1.263158
1.2
1.263158
1.411765
1.2
0.1053
0.183
0.244
0.2945
0.37
0.468
0.5505
0.6305
0.708
0.8395
0.790123
0.9216
0.852071
1.190083
1.190083
1.595568
1.44
1.595568
1.99308
1.44
2. The acceleration of this motion can be found in a graphic way, by plotting the graph of
Instantaneous Velocity squared over the distance traveled (Chart1).
Chart 1:
Instantaneous Velocity Squared vs. Distance Traveled
velocity squared
(m^2/s^2)
2.5
2
y = 1.2557x
1.5
1
0.5
0
0
0.5
1
Distance (m)
1.5
2
3. The equation of the line in the graph is given by the equation:
y = 1.2557x
It is evident from the above equation that the m slope m equal to 1.2557.
Linest Function Results:
1.255672
0.061595
0
#N/A
With the slope being 1.255672 ; y intercept is 0
The intercept of  m = 0.061595
4. Taking the formula:
v xf2  v xi2  2a x ( x f  xi )
and rewriting it in different form we obtain:
vin2  2a x ( x f  xi )
Since the initial velocity is 0 m/s and the final velocity is calculated by the instantaneous velocity
when the flag moves by the photo gate.
If we compare the two equations:
2
vinst
 2a x ( x f  xi )
y = 1.2557x
It is evident that 2a x = 1.2557
Since ( x f  xi ) = x in the equation from the chart, the acceleration is
= 0.62785
m
.
s2
5. The value of the uncertainty in the acceleration, a, comes from two sources—that caused by
statistical fluctuations, and that caused by the instrument imprecision. We assume that the
uncertainty due to instrument imprecision is negligible (but only for this experiment) so that the
total uncertainty in the acceleration is equal to the random error which is determined by the
uncertainty of the slope found using LINEST. The value of a is half the uncertainty of the
slope which gives us
 m = 0.061595
0.5 * 0.061595 = 0.03079
Δ a x = 0.03079
m
.
s2
Results
Part A:
Average Velocity: v = 0.3303  0.0018
Fractional uncertainty: vav / vav =
m
s
0.001763
= 0.00534
0.330314
Instantaneous Velocity: v w = 0.0136  0.0062
Fractional uncertainty: v w /v w =
m
s
0.006157
= 0.4513
0.013643
Part B:
Acceleration: a = 0.628  0.062
m
s2
Fractional uncertainty: Δ a x / a x =
0.061595
m
= 0.0981 2
0.62785
s
Conclusion
The uncertainty in the average velocity due to imprecision in time measure was dominant
over the other uncertainty.
The average velocity and the instantaneous velocities do not have a junction in the
bounds of their answers 0.3303  0.0018
m
m
and 0.0136  0.0062 . Instantaneous and
s
s
average velocities do not overlap in their bounds so that means that the real velocity which the
glider traveled was one given by the average velocity.
For part B, the result that was retrieved was only 0.629
m
. The acceleration in the
s2
horizontal direction was the same as the acceleration in the vertical direction however
acceleration a(0.629
m
m
) was not equal to g(9.8 2 ). All the fractional uncertainties were found
2
s
s
to be reasonably low and therefore the experimental results were close to theoretical.