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Ticker Tape Timer (Motion Practical)

Key idea: Use a ticker tape timer to measure speed and estimate acceleration from dot spacing, with common mistakes and worked examples (O Level Physics).

  • SEC G3 Physics 2027
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Learning objectives

  • State what speed means
  • State what velocity means, including its direction
  • Calculate average speed from total distance and total time
  • Calculate acceleration as change in velocity divided by time taken
  • State what uniform acceleration means
  • Interpret examples of non-uniform acceleration
  • Plot and interpret displacement–time and velocity–time graphs in one dimension
  • Deduce rest and uniform or non-uniform velocity from a displacement–time graph
  • Deduce rest, uniform velocity and uniform or non-uniform acceleration from a velocity–time graph
  • Use signed area under a velocity–time graph to determine displacement
  • Recall constant free-fall acceleration near Earth as approximately 10 m/s²

1. Definition

A ticker tape timer makes dots on a paper tape at equal time intervals, Δ t.

If you attach the tape to a moving object, the spacing between dots shows the object’s speed and whether it is accelerating.

Where this fits

The ticker-tape timer is not on the Paper 3 apparatus list, but analysing its tape is in the syllabus: it uses the same kinematics ideas of average speed, acceleration and motion graphs.

2. Key Ideas

  • Time between dots is fixed: Δ t (s).
  • Number of time intervals = number of gaps, not number of dots.
  • Average speed over a section:
    • v = s/t = s/(NΔ t) where N is the number of gaps in that section.
  • Bigger gaps → higher speed.
  • Equal spacing → constant speed.
  • Gaps increasing → acceleration.
  • Gaps decreasing → deceleration.

3. Detailed Explanations

A. How a ticker timer works

Paper tape attached to a moving object passes under a vibrating marker. The marker makes dots at equal time intervals; the tape's distance moved during each interval determines the gap between dots.
The timer fixes the time between marks, Δt. The object's speed determines how far the tape moves—and therefore the gap—during each interval.

You will be told the time interval between dots (e.g. Δ t = 0.02 s).

Sometimes the frequency is given instead (e.g. 50 Hz). Then:

Δ t = 1/f

B. What the tape shows

Ticker-tape spacing for three types of motionThree ticker tapes are marked at equal time intervals. Equal dot spacing shows constant speed, increasing spacing shows speeding up, and decreasing spacing shows slowing down. Each row contains six dots and therefore five time intervals.Dots are made at equal time intervalsConstant speedSpeeding upSlowing down6 dots = 5 gaps = 5Δt
Scroll diagram horizontally to read all labels.
The timer controls the time between dots. Motion controls the spacing: equal, increasing or decreasing. Six dots contain five time intervals.

If the tape has dots at equal time intervals:

  • Tape X: dots are evenly spaced → constant speed.
  • Tape Y: spacing increases → accelerating.
  • Tape Z: spacing decreases → decelerating.

C. Calculating speed from a tape

Pick a section of tape containing a known number of gaps (for example, 5 gaps between 6 dots).

  1. Measure the distance s covered over that section.
  2. Work out the time t for that section: t = (number of intervals) × (time per interval).
  3. Calculate speed: v = s / t.

Tip: Using multiple intervals reduces percentage uncertainty (better than using one gap only).

D. Estimating acceleration (using speeds from sections)

If you calculate speeds for consecutive equal-time sections, you can estimate the (average) acceleration:

a = (Δ v)/(Δ t)

This connects directly to:

4. Common Mistakes

  • Counting dots instead of gaps (time intervals = gaps).
  • Using the wrong Δ t (you must be told its value, or find it from frequency).
  • Using only one gap (large percentage uncertainty).
  • Not converting units (e.g. cm to m).
  • Measuring along a curved tape or using inconsistent endpoints.

5. Exam Tips

  • If asked “describe the motion”, state constant speed / accelerating / decelerating based on spacing.
  • If asked to “calculate speed”, show v = s / t with units.
  • If asked about acceleration, compare how the spacing changes (uniform vs non-uniform).
  • For time-interval skills, see: Measurement Of Time.

6. Worked Examples

Modelled example 1

Calculate speed from a section

Core

Problem

A ticker timer makes dots every Δ t = 0.02 s. A section of tape has 10 gaps and length 18 cm. Find its average speed.
Study the worked solution

  1. Find the section time

    Method

    Multiply the number of gaps by the time per gap.

    Reason

    Each gap represents one complete time interval.

    Working

    t = 10(0.02) = 0.20 s

  2. Convert the distance

    Method

    Express 18 cm in metres.

    Reason

    Speed in m s⁻¹ requires metres and seconds.

    Working

    s = 18 cm = 0.18 m

  3. Calculate average speed

    Method

    Divide section distance by section time.

    Reason

    Average speed is total distance per total time over the selected section.

    Working

    v = 0.18/0.20 = 0.90 m s⁻¹

Guided practice 2

Find the time interval from frequency

About 6 min

Problem

A ticker timer runs at f = 50 Hz. A section has 5 gaps and length 12 cm. Find its average speed.

Derive interval, time and speed

Unit: s
Unit: m s^-1

Hints

Hint 1: invert frequency
A frequency of 50 Hz means 50 equal intervals each second.
Hint 2: scale to five gaps
After finding Δ t, multiply by five and use v = s/t.
View solution step by step

  1. Find the time per gap

    Method

    Take the reciprocal of frequency.

    Reason

    Frequency counts intervals per second, so its reciprocal is seconds per interval.

    Working

    Δ t = 1/50 = 0.020 s

  2. Find the section time and distance

    Method

    Multiply by five gaps and convert centimetres to metres.

    Reason

    The calculation must describe the same selected tape section in SI units.

    Working

    t = 5(0.020) = 0.10 s, s = 0.12 m

  3. Calculate the speed

    Method

    Divide section distance by section time.

    Reason

    This is the average speed over the five-gap interval.

    Working

    v = 0.12/0.10 = 1.2 m s⁻¹

Common misconception 3

Dots vs gaps (time intervals)

Find and correct the mistake

Learner response

A section has 6 dots across 15 cm and Δ t = 0.02 s. A learner uses 6 time intervals and reports 1.25 m s⁻¹. Locate the first error and calculate the corrected speed.

Diagnose before viewing the correction

First error
Unit: m s^-1

View solution step by step

  1. Count intervals correctly

    Method

    Subtract one from the number of dots.

    Reason

    Time elapses between consecutive dots.

    Working

    Six dots contain five gaps, so t = 5(0.02) = 0.10 s.

  2. Calculate the corrected speed

    Method

    Convert 15 cm to 0.15 m and divide by the corrected time.

    Reason

    The learner’s extra interval made the time too large and speed too small.

    Working

    v = 0.15/0.10 = 1.5 m s⁻¹

Examiner practice 4

Estimate acceleration from two consecutive sections

5 marks

Examination question

Dots are 0.10 s apart. Two consecutive sections each have 5 gaps; their lengths are 0.25 m and 0.35 m. Estimate the average acceleration between the sections. [5 marks]

Show section time, both speeds and acceleration

View solution step by step

  1. Find the equal section duration

    1 mark

    Method

    Multiply five gaps by the interval.

    Reason

    Each speed represents motion over one five-gap section.

    Working

    t_section = 5(0.10) = 0.50 s

  2. Calculate both section speeds

    2 marks

    Method

    Divide each section length by the common duration.

    Reason

    These average speeds estimate the velocities at the section midpoints.

    Working

    v₁ = 0.25/0.50 = 0.50 m s⁻¹, v₂ = 0.35/0.50 = 0.70 m s⁻¹

  3. Use midpoint-to-midpoint time

    2 marks

    Method

    Divide the speed change by 0.50 s.

    Reason

    Consecutive equal sections have midpoints separated by one section duration.

    Working

    a ≈ (0.70-0.50)/0.50 = 0.40 m s⁻²

Challenge 5

Find the length of tape from speed

Minimal support

Reverse calculation

Dots are 0.02 s apart. Over 20 gaps, the average speed is 1.8 m s⁻¹. Find the tape-section length.

Work backwards without the model

Unit: m

Hints

Hint 1: find the section time first
t = 20Δ t.
Hint 2: reverse the speed relationship
From v = s/t, use s = vt.
View solution step by step

  1. Find the total time

    Method

    Multiply 20 gaps by the interval.

    Reason

    The quoted speed applies over the complete selected section.

    Working

    t = 20(0.02) = 0.40 s

  2. Find the distance

    Method

    Rearrange average speed to s = vt.

    Reason

    Distance is speed multiplied by elapsed time.

    Working

    s = (1.8)(0.40) = 0.72 m

7. Mind Stretchers

Mind stretcher 1: Uniform or non-uniform acceleration?Extension

A ticker timer makes dots every Δ t = 0.10 s. The tape is split into three consecutive sections, each with 5 gaps:

  • section 1 length: 0.30 m
  • section 2 length: 0.42 m
  • section 3 length: 0.63 m

(a) Estimate the speed in each section.
(b) Is the acceleration uniform?

Show Answer

Time for each section:

t_section = 5(0.10) = 0.50 s

Speeds:

v₁ = 0.30/0.50 = 0.60 m s⁻¹ v₂ = 0.42/0.50 = 0.84 m s⁻¹ v₃ = 0.63/0.50 = 1.26 m s⁻¹

Accelerations between sections (midpoint-to-midpoint time is 0.50 s):

a₁₂ = (0.84-0.60)/0.50 = 0.48 m s⁻² a₂₃ = (1.26-0.84)/0.50 = 0.84 m s⁻²

Acceleration is not uniform because the values are different.

Mind stretcher 2: Work backwards (find the frequency)Extension

A section of tape has 24 gaps and length 1.20 m. The object’s speed over that section is 2.0 m s⁻¹.

Estimate the ticker timer frequency.

Show Answer

Total time for the section:

t = s/v = 1.20/2.0 = 0.60 s

Time per gap:

Δ t = t/24 = 0.60/24 = 0.025 s

Frequency:

f = 1/(Δ t) = 1/0.025 = 40 Hz

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027