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Next exam on November 14 – 15, 2026:
All concepts

Uniformly Accelerated Linear Motion

匀变速直线运动

The velocity and displacement equations of uniformly accelerated linear motion and their corollaries, free fall as the special case, and reading x-t and v-t graphs.

17 real-exam questions · 4 sittings34 in mock papers12 practice

Formulas

  • Velocity equation for uniform acceleration

    Formula
    v=v0+atv=v_0+at

    Applies when：Constant acceleration a; take the initial velocity direction as positive

  • Displacement equation for uniform acceleration

    Formula
    x=v0t+12at2x=v_0t+\frac{1}{2}at^2

    Applies when：Constant acceleration a; pair with v²−v₀²=2ax to eliminate time

  • Free-fall displacement

    Formula
    h=12gt2h=\frac{1}{2}gt^2

    Applies when：Released from rest under gravity alone, air resistance neglected

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Physics December 2025 Past Paper
    An object on a smooth horizontal surface begins uniform accelerated linear motion from rest. If its acceleration is a=2 m/s2a = 2 \mathrm {~m} / \mathrm { s } ^ { 2 }, then its velocity at the end of 4 seconds is ().
    • A．4 m/s4 \mathrm {~m} / \mathrm { s }
    • B．16 m/s16 \mathrm {~m} / \mathrm { s }
    • C．8 m/s8 \mathrm {~m} / \mathrm { s }
    • D．2 m/s2 \mathrm {~m} / \mathrm { s }
    Show answer & solution

    Answer: C

    1 Uniform acceleration from rest: v=at\mathrm { v } = \mathrm { a } \mathrm { t }. 2 v=2 m/s2×4 s=8 m/s\mathrm { v } = 2 \mathrm {~m} / \mathrm { s } ^ { 2 } \times 4 \mathrm {~s} = 8 \mathrm {~m} / \mathrm { s }.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    A droplet falls freely from 20 m . Landing speed is ( g=10 m/s2\mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } )
    • A．10 m/s10 \mathrm {~m} / \mathrm { s }
    • B．15 m/s15 \mathrm {~m} / \mathrm { s }
    • C．20 m/s20 \mathrm {~m} / \mathrm { s }
    • D．25 m/s25 \mathrm {~m} / \mathrm { s }
    Show answer & solution

    Answer: C

    v2=2gh=400,v=20 m/s\mathrm { v } ^ { 2 } = 2 \mathrm { gh } = 400 , \mathrm { v } = 20 \mathrm {~m} / \mathrm { s }. Answer: C．
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    An object starts from rest and undergoes uniformly accelerated linear motion with a=3 m/s2a = 3 \mathrm {~m} / \mathrm { s } ^ { 2 }. Its velocity at the end of 5 s is
    • A．10 m/s10 \mathrm {~m} / \mathrm { s }
    • B．5 m/s5 \mathrm {~m} / \mathrm { s }
    • C．20 m/s20 \mathrm {~m} / \mathrm { s }
    • D．15 m/s15 \mathrm {~m} / \mathrm { s }
    Show answer & solution

    Answer: D

    v=v0+at=0+3×5=15 m/sv = v _ { 0 } + a t = 0 + 3 \times 5 = 15 \mathrm {~m} / \mathrm { s }.
    Take the full CSCA Physics April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

17 questions across 4 sittings1 of 4 sittings available to verify free online

34 questions across 6 mock papers — our own papers, so a coverage signal rather than real-exam evidence

12 related practice questions availablePractise free: Kinematics

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Glossary

accelerationdisplacement

Related concepts

Describing MotionVelocity and AccelerationProjectile Motion