AshPhys

Velocity and acceleration in the lab

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.7Sections 2.7

Goal: Analyse a ticker tape to find velocities and an acceleration.

01The big picture

A ticker-timer prints dots on a moving tape at 50 per second, so each space between dots is 0.020 s. Cut the tape into strips of 5 spaces: each strip is 0.10 s, and its length ÷ 0.10 s is the average velocity during it.

Stuck side by side, the strips form a velocity–time graph (a tape chart). Strips growing by the same amount each time mean uniform acceleration. Motion sensors and video analysis give the same data automatically.

Ticker-timer [paraphrase]
Marks a dot on a tape every 0.020 s (50 Hz) as the tape is pulled through.
Fig. 1 · A tape chart
A tape chartThree paper strips stand side by side on a time axis, 2.0, 4.0 and 6.0 cm tall, each with six evenly spaced dots. A dashed line through their tops is straight.length / cmtime / s2460.10.20.3
Equal steps in strip length mean equal rises in velocity: uniform acceleration.

02Key equations

Velocity from one strip
v=strip length0.10 sv = \frac{\text{strip length}}{0.10\ \mathrm{s}}
Acceleration
a=ΔvΔta = \frac{\Delta v}{\Delta t}
Symbols and units
  • vvaverage velocity during a stripm s−1\mathrm{m\,s^{-1}}
  • Δt\Delta ttime between strip centress\mathrm{s}

03Sample problems with solutions

Sample 1 · A tape chart
Five-space strips from a 50 Hz ticker-timer are 2.0 cm, 4.0 cm and 6.0 cm long (Fig. 1). Find the velocity during each strip and the acceleration.

  1. Each strip lasts 5 × 0.020 = 0.10 s:

    v=0.0200.10, 0.0400.10, 0.0600.10=0.20, 0.40, 0.60 m s−1v = \frac{0.020}{0.10},\ \frac{0.040}{0.10},\ \frac{0.060}{0.10} = 0.20,\ 0.40,\ 0.60\ \mathrm{m\,s^{-1}}

  2. The velocity rises by 0.20 m s⁻¹ every 0.10 s:

    a=0.200.10=2.0 m s−2a = \frac{0.20}{0.10} = 2.0\ \mathrm{m\,s^{-2}}
Answer: 0.20, 0.40, 0.60 m s⁻¹; uniform acceleration 2.0 m s⁻²
Unit check: m s−1s=m s−2 ✓\frac{\mathrm{m\,s^{-1}}}{\mathrm{s}} = \mathrm{m\,s^{-2}}\ \checkmark

04Practice problems

  1. P1
    How long is 10 spaces on a 50 Hz ticker tape?
    Final answer0.20 s
    Answer: 0.20 s
  2. P2
    A five-space strip is 3.5 cm long. Find the average velocity.
    Final answer0.35 m s⁻¹
    Answer: 0.35 m s⁻¹
  3. P3
    Five-space strips are 1.5, 2.5 and 3.5 cm long. Find the acceleration.
    Final answer1.0 m s⁻²
    Answer: 1.0 m s⁻²
  4. P4
    All the strips on a tape chart are the same length. What does that show?
    Final answerConstant velocity (zero acceleration)
    Answer: Constant velocity (zero acceleration)

05Other useful information

Common mistakes

  • ✗ Wrong: Count the dots instead of the spaces.

    ✓ Right: Time = number of spaces × 0.020 s; 6 dots make 5 spaces.

Exam tips

  • The acceleration uses the time between strip centres: 0.10 s for neighbouring strips.

What the syllabus asks

  • 9702 topic 2: measure velocity and acceleration. Coursebook 2.7.

Explore: Practice questions for this lesson