Free Pearson Edexcel International A Level Physics lesson

Constant acceleration and free fall

Choose an interval, keep the signs and test the model.

What you’ll learn

  • Select and rearrange constant-acceleration equations with consistent signs.
  • Apply a constant gravitational field to upward and downward motion.
  • Plan and interpret a free-fall measurement of g with realistic uncertainty.

Choose an equation for one constant-acceleration interval

The symbols s, u, v, a and t describe displacement, initial velocity, final velocity, acceleration and elapsed time within one chosen interval. The four familiar equations apply to one-dimensional motion with constant acceleration. Write known values with signs before choosing a formula. Elapsed time is positive; displacement, velocity and acceleration depend on the positive direction. A braking acceleration is not automatically negative unless the initial motion is positive.

Use v = u + at when displacement is absent; s = (u + v)t/2 when acceleration is absent; s = ut + ½at² when final velocity is absent; and v² = u² + 2as when time is absent. The average of initial and final velocities works here because velocity changes linearly in time. Squaring velocity removes its sign, so a square-root result needs direction supplied by the physical situation. Quadratic equations for time can have two positive solutions when a position is passed twice.

Separate stages when acceleration changes. The final velocity of the first stage becomes the initial velocity of the next, while elapsed time restarts for that stage. Add signed displacements, and split at reversals if total distance is required. Do not extrapolate a braking model beyond the stop unless a continuing force would actually reverse the object. Check your result against a velocity–time sketch: its gradient should equal a and its signed area should equal s.

v = u + at s = ½(u + v)t s = ut + ½at² v² = u² + 2as

Symbols and units

  • u and v are initial/final velocity in m s⁻¹; a is constant acceleration in m s⁻².
  • s is displacement in m; t is elapsed time in s. All signs follow one chosen axis.

How the equation explains the idea

  1. Start with v = u + at. The straight velocity–time line has area s = (u + v)t/2, giving s = ut + at²/2 after substitution.
  2. Eliminate t instead: multiply s = (u + v)t/2 by 2a and use at = v − u. Then 2as = (u + v)(v − u) = v² − u².
  3. These equations require constant acceleration on the selected interval; a changing drag force generally breaks that assumption.

Worked example

A two-stage delivery run

  1. A cart starts at +2.0 m s⁻¹ and accelerates at +1.5 m s⁻² for 4.0 s. Its final velocity is 2.0 + 1.5 × 4.0 = 8.0 m s⁻¹.
  2. First displacement = 2.0 × 4.0 + ½ × 1.5 × 4.0² = 20 m.
  3. It then brakes uniformly to rest in 2.0 s. Second displacement = ½(8.0 + 0) × 2.0 = 8.0 m.

Total displacement is +28 m over 6.0 s; one constant acceleration does not describe both stages.

Keep this idea. Use one constant-acceleration interval at a time and recover direction after squaring a velocity.

Free fall is an acceleration condition

A freely falling object moves under gravity alone; it may initially be travelling upward, sideways or downward. Near Earth’s surface, use a uniform downward gravitational acceleration g = 9.81 m s⁻² unless a different value is stated. Weight is mg and field strength is force per unit mass, g = W/m in N kg⁻¹. Combining W = mg with ΣF = ma shows that the free-fall acceleration does not depend on mass when other forces are negligible.

With upward positive, a = −g throughout an unobstructed flight, including its highest point. At that instant vertical velocity is zero but acceleration and weight are not. A drop from rest has u = 0; a downward throw does not. Use the displacement from the launch position, not an absolute height above an unrelated reference. The same equations describe a fall with downward positive if every velocity and displacement sign is changed consistently.

Ignoring air resistance is a physical approximation, not a property of every falling object. Compact dense objects over modest distances often make a useful approximation; feathers, parachutes and long falls may not. A falling object approaching terminal speed is not accelerating at g because drag reduces the resultant force. A result such as a calculated time should be checked against the flight geometry and the chosen positive direction before accepting a root.

W = mg g = W/m With upward positive: v = u − gt; y = ut − ½gt²

Symbols and units

  • W is weight in N; m is mass in kg; g is downward gravitational acceleration magnitude in m s⁻².
  • u and v are upward-signed velocities in m s⁻¹; y is upward displacement in m; t is time in s.

How the equation explains the idea

  1. Free fall neglects forces other than gravity, so W = mg and a = −g with up positive.
  2. Insert a = −g into constant-acceleration relations: v = u − gt and y = ut − gt²/2. At the highest point v = 0 but a remains −g.
  3. Mass cancels in a = −W/m = −g; this conclusion does not include air resistance.

Worked example

An upward throw from a raised platform

  1. A ball is launched upward at 8.0 m s⁻¹ from a platform 6.0 m above the ground. Use g = 10 m s⁻² and ignore drag.
  2. Ground is y = −6.0 m, so −6.0 = 8.0t − 5.0t².
  3. The roots are t = (8 ± √184)/10. The physical time is 2.16 s; the negative root is before launch.
  4. v = 8.0 − 10 × 2.16 ≈ −13.6 m s⁻¹.

The ball reaches the lower ground after about 2.16 s while moving downward.

Keep this idea. Gravity fixes the acceleration, not the velocity; state the sign convention and whether drag is negligible.

Core practical 1: determine g from free fall

A useful method releases a small dense ball from rest using an electromagnet or other reproducible release, with electronic timing to a contact pad. Measure the vertical fall distance using consistent points on the ball and pad; the ball’s initial lower surface to first contact is one defensible geometry. Use a stable low-height setup over a catch tray, keep hands and feet out of the fall path, secure apparatus and follow the supervising teacher’s risk assessment. The activity described here is a plan, not evidence that a learner has performed it.

Measure times at several heights and repeat each height. Inspect release delays and contact triggering; a highly precise timer does not guarantee that its start coincides with release. With u = 0 and constant g, s = ½gt², so plot s vertically against the square of the measured time horizontally. A straight-line gradient gives g/2. Use a range large enough to reveal curvature or an intercept, and do not force the origin if the data do not support it. A height offset can cause an intercept; a constant time delay generally changes t² nonlinearly, so the same intercept correction cannot fix both.

Alternatively, two light gates can estimate speeds from a short flag length divided by each blocking time, then estimate g from (v₂ − v₁)/the elapsed time between the corresponding speed instants. The speed estimate belongs approximately to the middle of each blockage, not its leading edge; a long flag weakens an instantaneous-speed approximation. Record gate timing definitions and alignment. For either method, compare g with 9.81 m s⁻² within a stated uncertainty, discuss drag and release bias, and explain which improvement addresses the dominant cause.

For release from rest: s = ½gt² Gradient of s against t² = g/2

Symbols and units

  • s is downward displacement from release in m; t is fall time in s; g is acceleration in m s⁻².
  • The horizontal coordinate t² has units s²; the s-versus-t² gradient has units m s⁻².

How the equation explains the idea

  1. For release from rest with negligible air resistance, s = gt²/2. Let X = t², so s = (g/2)X is a straight-line model.
  2. Measure a best-fit gradient k = Δs/Δ(t²), then g = 2k. A nonzero release velocity or timing offset can invalidate the through-origin model.
Plotting against time squared reveals g as a gradient

Conditions. Computed rest-release model s = ½gt² with g = 9.81 m s⁻². Air resistance and timing offsets are neglected; these points are model values rather than experimental readings.

Time squared / s²; Fall distance / m.

The line slope is 4.905 m s⁻². Multiplying by two gives g = 9.81 m s⁻². The horizontal coordinate is t², not t, so the straight line does not mean constant falling speed.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Read g from a transformed graph

  1. Two well-separated points on the best-fit s-versus-t² line are (0.0400 s², 0.198 m) and (0.1600 s², 0.786 m).
  2. Gradient = (0.786 − 0.198)/(0.1600 − 0.0400) = 4.90 m s⁻².
  3. g = 2 × 4.90 = 9.80 m s⁻². A small nonzero intercept should be interpreted rather than silently discarded.

The factor of two belongs to the graph gradient: g is 9.80 m s⁻², not 4.90 m s⁻².

Keep this idea. Define release and contact carefully, use repeats across several heights, and obtain g from a justified gradient.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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