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Measuring velocity and acceleration

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.6Sections 2.6

Goal: Find a velocity from a light gate and an acceleration from two velocities.

velocity

01The big picture

A light gate times how long a card of known length LL blocks its beam. The velocity is L/tL/t; a short card over a short time gives close to the instantaneous velocity.

Two light gates give two velocities. Dividing their difference by the time taken between the gates gives the acceleration. Motion sensors, ticker-timers and video do the same job by recording position against time.

Light gate [paraphrase]
A beam and sensor connected to a timer; it records how long the beam is blocked.
Fig. 1 · A trolley passing two light gates
A trolley passing two light gatesSide view of a trolley on a level track moving to the right. A card stands on the trolley. Two light gates stand over the track ahead of it; each has a beam at the height of the card.vgate 1gate 2card, length Ltime Δt between gates
Each gate times the card; the timer also records the time between the gates.

02Key equations

Velocity at a gate
v=Ltv = \frac{L}{t}
Acceleration between gates
a=v2−v1Δta = \frac{v_2 - v_1}{\Delta t}
Symbols and units
  • LLlength of the cardm\mathrm{m}
  • tttime the beam is blockeds\mathrm{s}
  • Δt\Delta ttime from gate 1 to gate 2s\mathrm{s}

03Sample problems with solutions

Sample 1 · Two light gates
A 5.0 cm card blocks gate 1 for 0.050 s and gate 2 for 0.025 s. It takes 0.40 s to travel between the gates. Calculate the acceleration.

  1. Velocities at the gates (L=0.050L = 0.050 m):

    v1=0.0500.050=1.0 m s−1,v2=0.0500.025=2.0 m s−1v_1 = \frac{0.050}{0.050} = 1.0\ \mathrm{m\,s^{-1}}, \quad v_2 = \frac{0.050}{0.025} = 2.0\ \mathrm{m\,s^{-1}}
  2. Acceleration:

    a=2.0−1.00.40=2.5 m s−2a = \frac{2.0 - 1.0}{0.40} = 2.5\ \mathrm{m\,s^{-2}}
Answer: a=2.5 m s−2a = 2.5\ \mathrm{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
    A 10 cm card blocks a light gate for 0.20 s. Find the velocity.
    Final answer0.50 m s⁻¹
    Answer: 0.50 m s⁻¹
  2. P2
    A trolley’s velocity rises from 0.50 m s⁻¹ to 1.5 m s⁻¹ in 0.80 s. Find the acceleration.
    Final answer1.25 m s⁻² (1.3 m s⁻²)
    Answer: 1.25 m s⁻² (1.3 m s⁻²)
  3. P3
    Why is a narrow card better for measuring an instantaneous velocity?
    Final answerIt blocks the beam for a shorter time, so the velocity changes less
    Answer: It blocks the beam for a shorter time, so the velocity changes less
  4. P4
    Why use a light gate rather than a hand-held stopwatch?
    Final answerNo human reaction time, so a much smaller timing uncertainty
    Answer: No human reaction time, so a much smaller timing uncertainty

05Other useful information

Common mistakes

  • ✗ Wrong: Leave the card length in cm.

    ✓ Right: Convert to metres first: 5.0 cm = 0.050 m.

Exam tips

  • Δt\Delta t is the time between the gates, not the time either gate is blocked.

What the syllabus asks

  • 9702 topic 2: measure velocity and acceleration. Coursebook 2.6.

Explore: Practice questions for this lesson