The motion of a trolley moving along a straight runway is studied.
A paper tape is attached to the tail of the trolley
The paper tape passes through a ticker tape timer
Ticker
The timer prints dots on the tape at regular time intervals Tape
Timer
Trolley Paper tape
attached to the
trolley
As the trolley moves, the paper passes under through the timer
and dots are printed at regular time intervals on the tape.
Ticker
Tape
Timer
Dot printed by the
ticker tape timer on
the paper tape
Dots are printed on the tape with fixed time intervals Dt.
If the trolley is moving with uniform velocity, the dots
are evenly spaced.
Dots are printed on the tape with fixed time intervals Dt.
If the trolley is moving with increasing speed, the
spacing of the dots increases.
first dot last dot
Dots are printed on the tape with fixed time intervals Dt.
If the trolley is moving with decreasing speed, the
spacing of the dots decreases.
first dot last dot
The frequency of the ticker tape timer = fs
fs is the number of dots that the timer prints in one second.
fs is measured in Hz (Hertz).
If fs = 50 Hz , 50 dots are printed in 1 s.
If fs = 100 Hz , 100 dots are printed in 1 s.
The time interval Dt between printing two adjacent dots
is given by
1
Dt
fs
If fs is in Hz, Dt is in s.
If fs = 50 Hz , Dt = 0.02 s.
One dot is printed every 0.02 s.
If fs = 100 Hz , Dt = 0.01 s.
One dot is printed every 0.01 s.
The time interval between printing two adjacent dots is
also known as a ‘tick’.
For a ticker tape timer of 50 Hz,
a one-tick interval is 0.02 s ,
a two-tick interval is 0.04 s ,
a three-tick interval is 0.06 s ,
a five-tick interval is 0.1 s .
The time interval between printing two adjacent dots is
also known as a ‘tick’.
For a ticker tape timer of 100 Hz,
a one-tick interval is 0.01 s ,
a two-tick interval is 0.02 s ,
a three-tick interval is 0.03 s ,
a five-tick interval is 0.05 s .
A tape can be analyzed in one-tick intervals.
A section of the tape is selected for analysis.
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
first dot last dot
The instant when the first dot is printed is assigned t = 0.
The instant when the second dot is printed is assigned t = Dt.
The instant when the third dot is printed is assigned t = 2Dt.
The instant when the fourth dot is printed is assigned t = 3Dt.
A tape can be analyzed in one-tick intervals.
A section of the tape is selected for analysis.
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
The distance travelled by the object is l1 from t = 0 to t = Dt
l1
The average velocity in this time interval is V1
Dt
V1 can be assumed to be the instantaneous velocity of the object at
the instant t = 0.5 Dt
A tape can be analyzed in one-tick intervals.
A section of the tape is selected for analysis.
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
The distance travelled by the object is l2 from t = Dt to t =2Dt
l2
The average velocity in this time interval is V2
Dt
V2 can be assumed to be the instantaneous velocity of the object at
the instant t = 1.5 Dt
A tape can be analyzed in one-tick intervals.
A section of the tape is selected for analysis.
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
The distance travelled by the object is l3 from t = 2Dt to t =3Dt
l3
The average velocity in this time interval is V3
Dt
V3 can be assumed to be the instantaneous velocity of the object at
the instant t = 2.5 Dt
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
A velocity-time graph is now available.
time velocity Velocity
t1 = 0.5 Dt V1 = l1Dt
t2 = 1.5 Dt V2 = l2Dt V3
t3 = 2.5 Dt V3 = l3Dt V2
Note that V1
V1 : V2 : V3 = l1 : l2 : l3 Time
0 Dt Dt Dt
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
time velocity The average acceleration of the object
t1 = 0.5 Dt V1 = l1Dt in the interval t1 < t < t2 is given by
t2 = 1.5 Dt V2 = l2Dt
V2 V1
t3 = 2.5 Dt V3 = l3Dt a12
t 2 t1
l 2 l1
Dt Dt
1.5Dt 0.5Dt
l 2 l1
(Dt ) 2
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
time velocity The average acceleration of the object
t1 = 0.5 Dt V1 = l1Dt in the interval t2 < t < t3 is given by
t2 = 1.5 Dt V2 = l2Dt
V3 V2
t3 = 2.5 Dt V3 = l3Dt a23
t3 t 2
l3 l 2
Dt Dt
2.5Dt 1.5Dt
l3 l 2
(Dt ) 2
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
If the object is moving with uniform acceleration, the velocity-
time graph is a straight line.
Velocity
V3 V2 = V2 – V1
l3 - l2 = l2 – l1 V3
V2
V1 ,V2 , V3 form an
V1
arithmetic progression and
so do l1 ,l2 , l3 Time
0 Dt Dt Dt
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
If the object is moving with uniform acceleration, the velocity-
time graph is a straight line.
a12 = a23 Velocity
l2 l1 l3 l 2 V3
(Dt ) 2 (Dt ) 2 V2
V1
Time
0 Dt Dt Dt
t=0 t=Dt t=2Dt t=3Dt
l1 l2 l3
If the object is moving with uniform acceleration, the velocity-
time graph is a straight line.
The uniform acceleration
is also given by Velocity
V3 V1
a
t 3 t1 V3
l3 l 2
V2
D t Dt
2 .5 Dt 0 .5 Dt V1
l3 l1 Time
0 Dt Dt Dt
2( Dt ) 2
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 50 Hz. The time interval for one tick is
0.02 s
first dot 8 mm 11 mm 14 mm 17 mm
last dot
The sequence has a common difference
of 3 mm. Therefore the object is
moving with uniform acceleration.
0.011 m 0.008 m
a 0.02 s 0.02 s 0.011 m 0.008 m 7.5 m s - 2
0.03 s 0.01 s (0.02 s) 2
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 50 Hz. The time interval for one tick is
0.02 s
first dot 8 mm 11 mm 14 mm 17 mm
last dot
The sequence has a common difference
of 3 mm. Therefore the object is
moving with uniform acceleration.
0.017 m 0.014 m
a 0.02 s 0.02 s 0.017 m 0.014 m 7.5 m s - 2
0.07 s 0.05 s (0.02 s) 2
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 50 Hz. The time interval for one tick is
0.02 s
first dot 8 mm 11 mm 14 mm 17 mm
last dot
The sequence has a common difference
of 3 mm. Therefore the object is
moving with uniform acceleration.
0.017 m 0.011 m
a 0.02 s 0.02 s 0.017 m 0.011 m 7.5 m s - 2
0.07 s 0.03 s 2(0.02 s) 2
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 50 Hz. The time interval for one tick is
0.02 s
first dot 8 mm 11 mm 14 mm 17 mm
last dot
The sequence has a common difference
of 3 mm. Therefore the object is
moving with uniform acceleration.
0.017 m 0.008 m
a 0.02 s 0.02 s 0.017 m 0.008 m 7.5 m s - 2
0.07 s 0.01 s 3(0.02 s) 2
Very often, a ticker tape is analyzed in intervals longer
than one tick. This is because
• the length is very short for one tick. There is
large percentage error in measuring the length
corresponding to one tick.
• the spacing between the dots may vary in an
irregular pattern. The reason may be that the
timer does not print the dots in exactly the same
time interval. Analyzing the paper tape at
intervals of more than one tick may help to
averaging the errors.
A tape can be analyzed in five-tick intervals.
The time interval for one-tick is Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
The instant when the first dot is printed is assigned t = 0.
The instant when the sixth dot is printed is assigned t = 5Dt.
The instant when the eleventh dot is printed is assigned t = 10Dt.
The instant when the sixteenth dot is printed is assigned t = 15Dt.
A tape can be analyzed in five-tick intervals.
The time interval for one-tick is Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
The distance travelled by the object is l1 from t = 0 to t = 5Dt
l1
The average velocity in this time interval is V1
5Dt
V1 can be assumed to be the instantaneous velocity of the object at
the instant t = 2.5 Dt
A tape can be analyzed in five-tick intervals.
The time interval for one-tick is Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
The distance travelled by the object is l2 from t = 5Dt to t = 10Dt
l2
The average velocity in this time interval is V2
5Dt
V2 can be assumed to be the instantaneous velocity of the object at
the instant t = 7.5 Dt
A tape can be analyzed in five-tick intervals.
The time interval for one-tick is Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
The distance travelled by the object is l3 from t = 10Dt to t = 15Dt
l3
The average velocity in this time interval is V3
5Dt
V3 can be assumed to be the instantaneous velocity of the object at
the instant t = 12.5 Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
A velocity-time graph is now available.
Velocity
time velocity
t1 = 2.5 Dt V1 = l1Dt
t2 = 7.5 Dt V2 = l2Dt V3
t3 = 12.5 Dt V3 = l3Dt V2
Note that V1
V1 : V2 : V3 = l1 : l2 : l3 Time
0 Dt Dt Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
The average acceleration of the object
time velocity
in the interval t1 < t < t2 is given by
t1 = 2.5 Dt V1 = l1Dt
t2 = 7.5 Dt V2 = l2Dt V2 V1
a12
t 2 t1
t3 = 12.5 Dt V3 = l3Dt
l2 l1
5 Dt 5Dt
7.5Dt 2.5Dt
l 2 l1
(5Dt ) 2
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
The average acceleration of the object
time velocity
in the interval t2 < t < t3 is given by
t1 = 2.5 Dt V1 = l1Dt
t2 = 7.5 Dt V2 = l2Dt V3 V2
a23
t3 t 2
t3 = 12.5 Dt V3 = l3Dt
l3 l2
5Dt 5Dt
12.5Dt 7.5Dt
l3 l 2
(5Dt ) 2
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
If the object is moving with uniform acceleration, the velocity-
time graph is a straight line.
Velocity
V3 V2 = V2 – V1
l3 - l2 = l2 – l1 V3
V2
V1 ,V2 , V3 form an
arithmetic progression and V1
so do l1 ,l2 , l3
Time
0 Dt Dt Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
If the object is moving with uniform acceleration, the velocity-
time graph is a straight line.
Velocity
a12 = a23
V3
l 2 l1 l3 l 2 V2
(5Dt ) 2 (5Dt ) 2
V1
Time
0 Dt Dt Dt
t=0 t=5Dt t=10Dt t=15Dt
first dot last dot
l1 l2 l3
The uniform acceleration is also
given by Velocity
V3 V1
a
t3 t1
V3
l3 l2
5Dt 5Dt
V2
12.5Dt 2.5Dt
V1
l3 l1
Time
2(5Dt ) 2 0 Dt Dt Dt
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 100 Hz. The time interval for one tick is
0.01 s
first dot 28 mm 34 mm 40 mm 46 mm last dot
The sequence has a common difference
of 6 mm. Therefore the object is
moving with uniform acceleration.
0.034 m 0.028 m
a 0.05 s 0.05 s 0.034 m 0.028 m 2.4 m s - 2
0.075 s 0.025 s (0.05 s) 2
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 100 Hz. The time interval for one tick is
0.01 s
first dot 28 mm 34 mm 40 mm 46 mm last dot
The sequence has a common difference
of 6 mm. Therefore the object is
moving with uniform acceleration.
0.046 m 0.040 m
a 0.05 s 0.05 s 0.046 m 0.040 m 2.4 m s - 2
0.175 s 0.125 s (0.05 s) 2
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 100 Hz. The time interval for one tick is
0.01 s
first dot 28 mm 34 mm 40 mm 46 mm last dot
The sequence has a common difference
of 6 mm. Therefore the object is
moving with uniform acceleration.
0.046 m 0.034 m
a 0.05 s 0.05 s 0.046 m 0.034 m 2.4 m s - 2
0.175 s 0.075 s 2(0.05 s) 2
Consider the section of the paper tape below, printed by a ticker
tape timer of frequency 100 Hz. The time interval for one tick is
0.01 s
first dot 28 mm 34 mm 40 mm 46 mm last dot
The sequence has a common difference
of 6 mm. Therefore the object is
moving with uniform acceleration.
0.046 m 0.028 m
a 0.05 s 0.05 s 0.046 m 0.028 m 2.4 m s - 2
0.175 s 0.025 s 3(0.05 s) 2
The disadvantages of using a ticker tape timer are
• The timer prints dots by hammering on the tape. As a
result the object experiences a frictional force which
opposes its motion. The motion of the object is affected.
For good results, the timer has to be adjusted so that the
hammering force on the tape has to be minimized but a
weak hammering force will print dots which are not
clear enough.
• The timer can only study the motion of an object moving
along a straight line in one direction only. The method
fails if the object reverses its direction of motion.