Begin with the surprising result: a falling object can stop accelerating
A light object is released, gains speed, and later moves downward at a roughly constant speed. At first this seems odd: gravity has not vanished, and the object has not stopped falling. The explanation is that the upward resistive force can grow until it balances the downward weight. With zero resultant force, the object no longer changes its velocity. That limiting steady falling speed is called terminal velocity.
This is an important next step after learning free-body diagrams. In Volume 0450 you followed the forces on a trolley and chose the direction of acceleration. Here the body is falling through air, so the opposing force is not necessarily constant. The same discipline applies: select the body, identify external forces on it, calculate the resultant, and interpret its sign.
A good examination answer distinguishes three stories: what the object is doing, what forces act, and how those forces change over time. The object may be travelling down while its acceleration becomes smaller. Later it can continue down with zero acceleration. Neither motion requires gravity to weaken.
Quick route through the lesson
First label down as the positive direction. Second identify weight downward and drag upward. Third use weight = mass multiplied by gravitational field strength. Fourth write downward resultant force = weight minus drag. Fifth use resultant force = mass multiplied by acceleration. Sixth explain how drag changes as speed changes, and recognise that terminal velocity means constant velocity, not rest.
The transferable skill is to separate a measured or supplied speed from a force calculation. A table of speeds does not by itself supply air resistance; a table of resistive forces does not automatically supply exact positions or travel times. You need the relationship and the conditions of the question before making either inference.
What the 2027 G3 SEC syllabus actually expects
The first Singapore-Cambridge Secondary Education Certificate examination is in 2027, replacing the former N- and O-Level examinations under Full Subject-Based Banding. The 2027 SEAB G3 syllabus directory, updated on 26 October 2026, identifies standalone Physics as K323, corresponding to previous reference code 6091. Combined Science options have separate codes. Check the registered subject rather than treating every G3 Science candidate as a standalone Physics candidate.
The official K323 Physics syllabus includes Newtonian mechanics, descriptions of motion with and without air resistance, and terminal velocity. The learning outcomes include interpreting displacement-time and velocity-time graphs, understanding resultant force and acceleration, and describing falling bodies in a uniform gravitational field. The syllabus calls for applying Newton’s laws rather than reciting their statements word for word.
For standalone K323, Paper 1 is one hour, 40 multiple-choice marks and 30 per cent; Paper 2 is one hour forty-five minutes, 80 marks and 50 per cent; Paper 3 is a one-hour-fifty-minute practical examination, 40 marks and 20 per cent. These details come from the 2027 subject document. The worked examples below are original practice, not official paper questions or a guarantee of question selection.
A free-body diagram begins with one object
Imagine a small falling test body, not a person. Draw a dot or rectangle to represent it. The force of gravity acting on the body is downward and is labelled weight, W. Air resistance acts upward when the body moves downward relative to the air and is labelled D, or drag. If a problem states that only these two vertical forces matter, do not add an invented support force.
Force arrows represent forces acting on the selected body. They are not arrows showing its travel path. A falling body can move downward even if the resultant force points upward, because its current velocity and acceleration need not point the same way. Readers often lose marks when they confuse these two different physical quantities.
Use a consistent vertical sign convention. Here downward is positive, so W is positive and the upward drag is negative. In one dimension F_resultant = W − D. Choosing upward as positive is equally valid, but then both the equation and reported signs must change consistently.
Weight has units of force, not mass
Mass tells us how much matter an object contains and is measured in kilograms. Weight is the gravitational force acting on that mass and is measured in newtons. In a simple near-Earth model, W = mg. The K323 syllabus uses approximately 10 m/s² for free-fall acceleration near Earth in the stated context. Numerically the gravitational field strength is approximately 10 N/kg.
For a fictional 0.20 kg model body, weight is 0.20 multiplied by 10 = 2.0 N downward. You must not say it weighs 0.20 N. The numbers 0.20 kg and 2.0 N refer to different quantities with different units, although they describe the same body.
The weight remains close to 2.0 N throughout a short classroom-height descent if the problem assumes a uniform gravitational field. Changes in acceleration are therefore explained through changes in the opposing force, not through the disappearance of weight as the body falls.
Why air resistance is usually not constant during a fall
An object moving through air interacts with air. That interaction produces a resistive force that typically changes with relative speed as well as shape, orientation and properties of the air. For the simple falling-body questions in this lesson, assume the upward drag grows as the object’s downward speed increases. That is an assumption about the model, not a universal numerical law for every object.
Do not invent a drag equation unless one is supplied. Real drag can depend on several factors and may follow different approximations in different flow conditions. At this level, a careful qualitative answer is often more useful than an unsupported calculation. Say that greater downward speed produces greater opposing air resistance under the stated conditions.
The phrase ‘air resistance increases with time’ is a weaker explanation because time itself is not the direct reason. The object may eventually reach a stable speed; then drag need not keep increasing. State the relevant link between the speed through the air and the resistive force.
The first instant: gravity dominates
Suppose the model body begins from rest and the fictional exercise takes its initial drag as negligible. The weight is 2.0 N downward. The downward resultant is therefore approximately 2.0 N. With mass 0.20 kg, the initial acceleration is 2.0 divided by 0.20 = 10 m/s² downward.
The object gains downward speed at first. The acceleration of 10 m/s² is not a permanent promise once air resistance grows. The ‘free-fall’ idealisation that ignores resistance is distinct from a real or modelled fall where drag matters. A strong answer explains which situation the question describes.
If the task asks why the acceleration is greatest near release, compare the forces. Near release, drag is least in the simple model. As downward speed grows, greater drag reduces the downward resultant while weight remains unchanged.
A full worked table of forces and acceleration
Take the same fictional 0.20 kg falling body with W = 2.0 N. Consider four snapshots. At one snapshot drag is 0 N, so the downward resultant is 2.0 N and downward acceleration is 10 m/s². At another, drag is 0.6 N, so the resultant is 1.4 N and acceleration is 7.0 m/s². At a third, drag is 1.4 N, giving a 0.6 N resultant and 3.0 m/s² acceleration.
At the fourth snapshot, drag is 2.0 N. Resultant force = 2.0 − 2.0 = 0 N, so acceleration is zero. The values are illustrative force states rather than measurements taken at equal time intervals. It would be wrong to infer the travel distance by treating them as a timed speed record.
A useful check is that the drag values increase while the calculated downward acceleration falls. The signs, magnitudes and physical description agree. If you obtained a larger downward acceleration when upward drag increased, inspect the subtraction and sign convention.
Terminal velocity is not zero velocity
When drag equals weight, the two forces on the falling body balance. The resultant force is zero and the acceleration is zero. If the body was already moving downward, it continues moving downward at constant velocity under these idealised conditions. The phrase ‘terminal’ describes a steady limit to the speed, not the end of the motion.
One of the simplest mistakes is to write ‘the object stops because forces are balanced’. Balanced forces mean no change in velocity; they do not require the existing velocity to be zero. An object at rest can remain at rest under balanced forces, but a moving object can also continue at constant velocity.
The correct explanatory chain is: downward speed has increased; drag has risen; eventually upward drag matches downward weight; resultant force becomes zero; acceleration becomes zero; downward speed stays constant. This sequence is clearer than a vague sentence such as ‘gravity and air cancel out’.
How to reason when drag exceeds weight
Consider a different snapshot in which our fictional body still travels downward but the upward drag is 2.4 N and weight is 2.0 N. With down positive, the resultant is 2.0 − 2.4 = −0.4 N. Acceleration equals −0.4 divided by 0.20 = −2.0 m/s². The negative sign tells us acceleration is upward relative to our chosen axis.
An upward acceleration does not instantly reverse a downward velocity. The body can still be moving downward while slowing. This is analogous to a vehicle travelling forward while braking creates a backward acceleration. Direction of travel is read from velocity; direction of acceleration comes from the resultant force.
This case becomes particularly useful when a canopy opens or the body’s drag conditions change. Avoid assuming a new terminal speed is reached instantly. The body must pass through a transition in which speed and the resistive force adjust.
A canopy change introduces a new phase
Imagine a model that falls for a while, then opens a broad paper canopy. Before opening, the body may have reached one steady downward speed. Immediately after opening, its speed has not jumped to the new lower speed, but the canopy changes the drag behaviour. At the same downward speed, drag can become larger than weight, so resultant force points upward and the body slows.
As speed falls, the drag generally reduces again until it balances weight at a new lower terminal velocity in the simplified model. This explanation uses a sequence rather than pretending the body instantly switches between two constant speeds. There is a brief phase of upward acceleration while downward motion continues.
Do not infer safe descent of a person from a paper model. This is a classroom conceptual example, not a design standard, safety instruction, or prediction of full-scale equipment. The physical habit being tested is identifying a force model in each phase.
A force-time story and a velocity-time story are not the same graph
A velocity-time graph describes the object’s motion. For a body released from rest and approaching terminal velocity, downward velocity typically rises and then levels off. The slope of that graph, which represents acceleration, becomes smaller and approaches zero in the simplified account. A force-time graph would instead describe how weight and drag change or stay constant.
The downward weight might be a horizontal line in a weight-versus-time graph. The upward drag might rise toward that line as speed increases. Their difference is the resultant. A learner should not label the plateau of a velocity graph ‘zero speed’. The plateau indicates a constant speed if its height is positive.
When reading any figure, first identify the vertical axis. If it is ‘speed’, the graph does not show direction. If it is ‘velocity, downward positive’, negative values would represent upward motion. If it is ‘resultant force’, the gradient no longer directly represents acceleration.
Understand what the gradient tells you
Suppose a graph gives downward velocity at two times, 0.8 m/s at 0.4 s and 1.4 m/s at 0.8 s. The average acceleration across that interval is (1.4 − 0.8) divided by (0.8 − 0.4) = 1.5 m/s² downward. This is an interval average. It is not automatically the instantaneous acceleration at either endpoint.
In a changing-drag problem acceleration is not generally uniform throughout the fall. Do not use one average gradient for an entire curved graph and call it the acceleration at every point. If a tangent is supplied, its gradient can estimate acceleration at a chosen instant. If only a rough curve is drawn, qualitative reasoning may be appropriate.
The vocabulary matters: ‘speed rises more slowly’ means the slope of the speed-time graph is decreasing. It does not mean that the speed has begun to decrease. The distinction can be expressed in a single calm sentence.
Recognise an approaching plateau without overclaiming
A curve that becomes flatter suggests the magnitude of acceleration is becoming smaller. If it reaches a clearly horizontal section above zero, that section represents constant nonzero velocity. If a few noisy readings appear similar, however, it may be premature to declare exact terminal velocity. Measurement limits and short observation windows can make a rising curve look flat.
Look for the question’s wording. ‘Approaches terminal velocity’ is supported by a steadily decreasing gradient in an appropriate model. ‘Has reached terminal velocity’ usually requires an explicit constant-velocity condition or enough evidence that the resultant is zero. Do not silently replace one with the other.
In practice you should ask what is measured: height, displacement, average speed over short intervals, or instantaneous speed. Those quantities can support different strengths of conclusion. Recording a drop distance alone is not enough to reconstruct every instant of acceleration.
Newton’s third-law partners act on different bodies
Air pushes upward on the falling body through drag. The falling body exerts an equal and opposite interaction force on the air. Those two forces form a third-law pair and act on different bodies. They do not cancel on the falling body’s free-body diagram.
Weight is the gravitational force of Earth on the body. The corresponding third-law partner is the gravitational force of the body on Earth. It is not the air resistance upward on the body. Both weight and drag act on the falling body and can contribute to its resultant; they are not third-law partners of each other.
This explains why identifying the object is essential. If a question asks for the action and reaction, give the interacting pair of bodies and directions. If it asks whether the forces on the falling body are balanced, compare weight and drag acting on that one selected body.
A safe, small-scale visual observation
A school may demonstrate falling objects with light paper cups or small paper filters under supervision in a clear area. Keep the drop short, avoid sharp or heavy objects, and prevent anyone from standing below an elevated apparatus. A phone video can be used only where school rules and permissions allow. No student should attempt a high drop or test a human-carrying device.
Two differently shaped lightweight paper objects can fall at different rates because their drag conditions differ. But observing that one reaches the floor later does not, by itself, establish exact terminal velocities. Their masses, initial conditions and motion through the air may also differ.
Keep the focus on the question’s purpose. For a qualitative explanation, compare the relevant forces. For a measurement question, define the measured quantity, record time or position at a sensible precision, and say what data can actually support.
Practice question one: find acceleration from a force reading
A fictional instrument reports that a 0.25 kg falling object has weight 2.5 N downward and resistive force 1.75 N upward. Find the acceleration. With downward positive, resultant force is 2.5 − 1.75 = 0.75 N downward. Acceleration is 0.75 divided by 0.25 = 3.0 m/s² downward.
The explanation should include the direction. A bare number 3.0 without a direction is incomplete when forces are presented as vectors. Write why the value is less than 10 m/s²: upward resistance has reduced the net downward force below the full weight.
Do a reverse check: mass multiplied by acceleration is 0.25 × 3.0 = 0.75 N, which agrees with the resultant. If your answer were 17 m/s², that would signal that you had added drag instead of subtracting it.
Practice question two: recover a resistive force
A 0.40 kg model body falls downward with acceleration 6.0 m/s² while gravitational field strength is taken as 10 N/kg. What upward resistance is consistent with these conditions? Its weight is 4.0 N. Its downward resultant is 0.40 × 6.0 = 2.4 N. Therefore upward drag is 4.0 − 2.4 = 1.6 N.
The common error is to report 2.4 N as drag. That quantity is the resultant, not the resistance. Label W, D and F_resultant separately before rearranging. The model says W − D = ma, so D = W − ma for a downward positive acceleration.
The answer also makes physical sense. Drag of 1.6 N is positive and smaller than weight, so the body is accelerating downward. If the calculated drag had been negative in a conventional falling-body scenario, the stated assumptions would need checking.
Practice question three: identify the steady-speed condition
A question states that a 0.12 kg model is falling at constant downward velocity in uniform gravity, using g = 10 N/kg, and that its only vertical forces are weight and air resistance. What is air resistance? Weight is 1.2 N downward. Since acceleration is zero, the resultant is zero. Upward air resistance is 1.2 N.
Notice what was not needed. No velocity value was supplied, because the balancing-force condition is sufficient to find drag in this idealised question. But you cannot use weight alone to calculate the numerical terminal speed; that would require an additional drag-versus-speed relationship or suitable motion data.
This difference between a force at terminal velocity and the speed itself is one of the most important checks. The equality of forces determines zero acceleration. It does not supply a unique speed unless further physics is provided.
Practice question four: motion and acceleration have opposite signs
A falling object is moving downward at 3.0 m/s when an enlarged canopy makes its acceleration 1.5 m/s² upward for the next 0.4 s, taken as constant solely for this calculation. With down positive, new velocity is 3.0 − (1.5 × 0.4) = 2.4 m/s downward.
The body has not started rising. It is still travelling downward, but more slowly. If a candidate writes ‘it moves upward because acceleration is upward’, the answer confuses acceleration with velocity. Always state the old velocity, signed acceleration, time interval and new velocity.
The constant-acceleration simplification is an exercise condition. A real canopy transition may have varying acceleration, so one snapshot of force is not automatically enough to predict a whole interval.
Practice question five: read a descriptive graph
Imagine a downward velocity-time graph that rises steeply for the first second, rises more gently during the second, and then approaches a horizontal line. Describe the forces. At first weight exceeds drag substantially, giving sizeable downward acceleration. As speed rises, drag grows and resultant force shrinks. The curve becomes flatter because acceleration decreases.
A graph that levels off is compatible with the approach to terminal velocity. It does not show the body stopping. If asked for the final motion, say constant downward velocity once forces balance. If asked for the initial acceleration, do not calculate a number unless a gradient or sufficient numerical readings are available.
This style of answer earns clarity through order: observation from graph, force mechanism, acceleration consequence, motion interpretation. Avoid repeating ‘gets faster’ four times without explaining why the rate of change changes.
Practice question six: distinguish two falling shapes
Two objects have the same mass but different shapes. One presents a larger area to the moving air. It may experience more air resistance at the same downward speed under otherwise comparable conditions, and may approach a lower terminal speed. This is a qualitative expectation, not a guarantee for every possible shape and flow arrangement.
The strongest answer names the controlled conditions before making the comparison. If mass, shape, orientation and air conditions all differ, the observation cannot isolate projected area as the cause. Do not claim that greater area always gives a precise percentage reduction in terminal speed.
The same discipline appears in other scientific topics: change one relevant factor and keep others as similar as practical. When controls are missing, say that the explanation is plausible rather than proven.
Why ‘heavier always falls faster’ is not a safe rule
In the ideal model without air resistance, objects near Earth share approximately the same free-fall acceleration regardless of mass. With air resistance, mass and shape matter to how weight compares with drag. Two objects with different masses can have different motion histories, but the conclusion cannot be made from mass alone without knowing how resistance behaves.
Do not answer by repeating an everyday impression. First ask whether air resistance is ignored. If it is, acceleration is approximately g for both. If it is included, identify what is known about resistance, shape and initial conditions. A short explanation grounded in assumptions is better than a dramatic slogan.
This is also a reminder not to interchange weight and acceleration. An object with twice the mass has twice the weight in the same gravitational field, but the mass also changes its resistance to acceleration.
Units and precision should support the idea
Weight and drag use newtons. Mass uses kilograms. Acceleration uses metres per second squared. Speed uses metres per second. Distances use metres, and time uses seconds. A learner who labels drag as m/s or terminal speed as N has crossed the line between different concepts, even if the arithmetic seems tidy.
Check units as you rearrange F = ma. Force divided by mass gives N/kg, equivalent to m/s² in the SI system. Weight divided by gravitational field strength gives kilograms. A slope on a velocity-time graph has units of m/s², not metres.
Do not quote more precision than the measurements support. If a question specifies a different rounding instruction, follow it. Otherwise present useful precision and keep intermediate calculations adequately accurate before rounding.
A comparison table can be explained in sentences
Suppose a record shows four force states: W = 2 N every time, with upward D = 0.2, 0.8, 1.6 and 2.0 N respectively. The corresponding downward resultants are 1.8, 1.2, 0.4 and 0 N. With a 0.20 kg mass the accelerations are 9.0, 6.0, 2.0 and 0 m/s².
A clear two-sentence summary is: ‘As air resistance rises toward the constant weight, the net downward force and acceleration become smaller. When air resistance equals weight, the resultant and acceleration are zero, even though the body may still be descending.’
Do not treat a list of independent force states as if it supplies an equal-time trajectory. To calculate distance travelled you would also need velocity or the relevant time history. The same numerical table cannot answer every kind of motion question.
What a diagram cannot prove
A free-body diagram is an idealised representation of selected forces. It helps you reason, but it does not independently verify that a particular real object has a particular drag value. If the question provides a force diagram, use its information. If you are investigating a physical drop, you need observations or a validated model to establish values.
A tidy arrow drawing is not a substitute for a stated assumption. The arrows should be labelled with known quantities and plausible directions. Where lengths are intended to show relative forces, make the comparison consistent with the problem. When the drawing is schematic rather than to scale, do not measure arrow lengths as if they were recorded newtons.
Good physics writing often separates ‘the diagram assumes’ from ‘the experiment recorded’. This is a habit that prevents a simple classroom model from sounding like a certified full-scale prediction.
How to answer a ‘describe’ question
A ‘describe’ prompt usually asks you to report what happens or how it changes. For a falling object approaching terminal velocity, say that its downward speed increases from rest, the rate of increase becomes smaller, and eventually its speed becomes constant in the simplified model.
Use quantities carefully. If the object has already reached a terminal velocity, it travels at constant speed in a straight downward path. Do not confuse this with a body that remains stationary under balanced forces. Motion and change in motion are separate.
Descriptions should follow the chronology or graph sections in the question. Avoid inserting an unsupported numerical calculation merely to sound advanced. A clean, relevant answer is stronger.
How to answer an ‘explain’ question
An ‘explain’ prompt asks for the causal chain. Begin with weight acting downward. When the object has a low downward speed, resistance is smaller than weight. Resultant force is downward, so the body accelerates downward. As it gains speed, resistance increases and the downward resultant falls. At the terminal condition, drag balances weight and acceleration is zero.
This chain is not merely a description of a line becoming flat; it connects the line to forces. Where the question includes a canopy change, add the relevant phase: drag becomes greater than weight at the current speed, producing upward acceleration and slower downward travel until a new balance develops.
Do not write ‘no forces act at terminal velocity’. Both weight and drag still act. What vanishes is their resultant under the model assumptions.
How to answer an ‘evaluate’ question
An ‘evaluate’ prompt may ask whether evidence supports a claim. Suppose a student says a few similar average speeds prove exact terminal velocity. The useful response identifies what the evidence does show—a late-stage reduction in speed change—and why it may not establish an exact steady state: short intervals, measurement precision, and uncertainty about conditions.
Suggest a proportionate improvement such as collecting more readings over comparable intervals with the same object and release method, or using a more precise measurement method. Do not propose unsafe heights or imply that more data guarantees a particular result. The improved method should address the named weakness.
Evaluation is not an invitation to dismiss the whole investigation. Give credit to the limited observation while withholding a stronger claim that the record cannot support.
Exam craft: the two-mark explanation
For a short two-mark style question about falling speed, a compact answer might be: ‘As downward speed increases, upward air resistance increases, so the resultant downward force and acceleration decrease. When resistance equals weight, the resultant is zero and the object continues falling at constant velocity.’
That model response is concise because each clause performs a distinct function. It does not use technical terms for decoration. Mark schemes vary by actual question, so learn the underlying reasoning rather than memorising a single sentence as if it were guaranteed to secure particular marks.
Exam craft: show working rather than only the result
In a calculation question, write the known weight, the drag and the chosen positive direction. Then state the resultant equation and substitute values. Example: W − D = ma; 3.0 − 1.2 = 0.30a; therefore a = 6.0 m/s² downward. This makes every step inspectable.
Never write 3.0 + 1.2 = 4.2 N simply because two forces appear in the question. Forces are vectors and direction matters. If a question includes another force such as tension, include it only where it is genuinely acting on the selected body.
Read the actual diagram and words before assigning directions. The single-object rule protects your force inventory, and the sign convention protects your algebra.
Exam craft: choice questions with plausible traps
A multiple-choice question may offer ‘zero speed’, ‘zero acceleration’, ‘zero weight’, and ‘zero air resistance’ at terminal velocity. Under the usual model, zero acceleration is correct. The other options confuse balancing with absence of motion or absence of forces.
Another item may show a downward-moving object with an upward resultant. The correct conclusion is that it is accelerating upward; the object can be slowing while still travelling down. A distractor that says it must be moving upward relies on the same confusion.
To choose efficiently, name the physical quantity being asked about. If it says acceleration, use the resultant. If it says current velocity, use the motion evidence. If it asks for reaction force, identify the interacting second body.
Exam craft: structured data-response questions
A data-based question may combine a graph, short narrative and several calculations. Read the axes before applying a formula. Find the intervals in which velocity rises, levels or falls. Use the slope for acceleration when the graph is numerical. Connect the slope’s sign to the force result, and express the outcome in words.
Do not assume a line in one graph automatically proves a causal mechanism. A velocity-time curve shows motion, not direct measurement of weight or drag. If the question supplies a physical model of resistance, use it. If it does not, distinguish an observed trend from an inferred explanation.
After calculating, return to the exact question. A prompt requesting an explanation of the plateau needs the force balance, not just a gradient of zero.
A short comparison with the trolley from Volume 0453
The trolley case in Volume 0453 used an assumed constant opposing force in a fitted model. A falling-body model with drag increasing with speed makes a different assumption. Both use the same resultant-force principle, but the behaviour can differ markedly because the resistance term behaves differently.
This is why copying an equation from one story into another without checking assumptions is risky. In the trolley worksheet, recovering a single constant resistance parameter was the mathematical task. For a falling object, a terminal condition may be solved without knowing a complete drag law, but predicting the exact terminal speed generally needs additional information.
The transferable skill is not a formula collection. It is the habit of asking which body, which forces, which conditions, and which observations justify the model.
Practice routine for a learner who often rushes
Work through five short phases. Spend one reading pass locating the body and direction. On scratch paper draw the force arrows. Write the resultant with its sign. If numbers are supplied, calculate using units. Then finish with a motion sentence stating whether the body speeds up, slows down or keeps constant velocity.
Review the same idea in three forms: a diagram, a small numerical force question, and a verbal graph interpretation. If you get one wrong, identify the precise misunderstanding. Did you use weight as mass, add opposing forces, mistake zero acceleration for rest, or confuse acceleration direction with motion direction?
Practise again with a changed mass or a sudden canopy opening. The ability to transfer reasoning to a new scenario shows stronger learning than repeating one completed exercise.
Self-test: predict before checking
Question A: A falling object has weight 5 N and upward drag 3 N. Is its resultant up or down? Answer: 2 N downward. Question B: The same object has upward drag 5 N. What is its acceleration? Answer: zero, assuming no other vertical forces. Question C: Does that mean it must be at rest? Answer: no; it could continue at constant downward velocity.
Question D: A downward-moving object has 6 N upward drag and 5 N downward weight. What is its acceleration direction? Answer: upward; it can still be moving downward while slowing. Question E: Which forces are a third-law pair: weight and drag on the same body, or air on body and body on air? Answer: the latter.
If any answer feels surprising, draw the body and annotate the directions instead of memorising the correct letter.
A second self-test with arithmetic
Let mass be 0.50 kg and use g = 10 N/kg. Weight is 5.0 N. With drag 2.0 N upward, resultant is 3.0 N downward and acceleration 6.0 m/s² downward. With drag 4.0 N, resultant is 1.0 N down and acceleration 2.0 m/s² down. With drag 5.0 N, acceleration is zero.
Now suppose drag is 5.5 N while the body is still moving downward. The resultant is 0.5 N upward; taking down as positive gives −1.0 m/s². The body slows its descent at that instant. Writing the negative sign in a chosen coordinate system and translating it into ‘upward’ is a useful two-step check.
These forces are fictional snapshots, not a timed experimental table. Do not infer that 0.50 kg always implies a particular terminal speed.
How primary-school science reasoning transfers
The PSLE Science Learning Guide builds habits of stating a claim and supporting it with relevant evidence. The same habits remain valuable here, but secondary physics adds signed forces, numerical models and more precise definitions. The progression is from noticing that resistance matters to describing exactly how force balance changes acceleration.
Students can revisit the PSLE English, Mathematics and Science route if they need help reading a question, measuring a quantity or explaining cause and effect. Revisiting a foundation is not a step backward when it repairs a missing link that would otherwise make a harder physics problem unreliable.
Where to continue after this volume
For the broader series, start at G3 SEC Examinations: The Complete EMS Learner’s Guide. For subject-specific science preparation, explore the Science Hub and the Secondary 4 Science guide. They serve as subject pathways rather than duplicate owners of this terminal-velocity lesson.
The next mathematical step is to read graphs and calculate distances with the correct graph assumptions. The next English step is to turn observed visual movement into a precise spoken explanation. When these skills meet, a learner becomes able to tell a physical story, perform a numerical check and avoid claims beyond the evidence.
Final understanding check
Terminal velocity is the constant speed reached when the resistive force balances the weight in an appropriate falling-body model. It is not zero speed. Before that condition, if weight exceeds drag, the body accelerates downward. If drag temporarily exceeds weight while the body still descends, it accelerates upward and slows. Direction of velocity and direction of acceleration must be treated separately.
The dependable method is simple: choose one body; draw only forces acting on it; choose a positive direction; find the resultant; calculate acceleration where possible; and explain the motion without inventing missing measurements. Learn the physics behind the sentence. Then the graph, equation and words begin to agree.
Official references
Check the SEAB SEC overview, the 2027 G3 syllabus directory and the 2027 K323 Physics syllabus for authoritative subject scope and paper arrangements. Published examination details may be amended by SEAB, so candidates should also follow the school and official examination instructions.
This learner’s guide uses original fictional values and explanations. It is not an official SEAB practice paper, a real-world parachute design guide or a report of a measured performance.
