Terminal Speed and Air Resistance Revision Notes | A Level AQA Physics

1. The Hook: Why Don't All Objects Fall at the Same Rate?

Drop a sheet of paper and a textbook from the same height. The textbook hits the floor first — every time. Gravity pulls them both downward with the same acceleration of $9.8 \text{ m s}^{-2}$, so why the difference? The answer is air resistance, and understanding it unlocks two of the most practically important ideas in A Level mechanics.

This lesson covers two specification requirements: terminal speed, and the qualitative effect of air resistance on projectile trajectories and vehicle maximum speed. These topics appear regularly in AQA exam papers, often in the context of skydivers, cars, or thrown balls — so the real-world anchors here are genuinely useful.

2. Forces on a Falling Object: Building Towards Terminal Speed

Picture a skydiver the moment they step out of a plane. At that instant, their velocity is zero, so air resistance is also zero. The only significant force acting is their weight, $W = mg$, directed downward. The resultant force is therefore equal to $mg$, and by Newton's second law, the skydiver accelerates downward.

Here is where it gets interesting. As the skydiver speeds up, the drag force — air resistance — increases. Drag depends on speed: the faster you move through air, the greater the resistive force pushing back against you. This means the resultant force (weight minus drag) gets smaller as speed increases, so the acceleration decreases over time.

Eventually, the drag force grows large enough to exactly equal the weight. At that point, the resultant force is zero. By Newton's first law, the skydiver continues moving at a constant velocity — this is terminal speed. The acceleration has dropped to zero, not because any force has switched off, but because two forces are now perfectly balanced.

A common point of confusion: terminal speed is not a special property of the object — it is a dynamic equilibrium between forces. Change the object's shape or mass, and the terminal speed changes too. A skydiver in a spread-eagle position has a larger cross-sectional area, so drag increases more rapidly, and terminal speed is lower (around 55 m/s) compared to a head-down dive position (around 90 m/s).

The velocity-time graph above shows the characteristic shape for terminal speed. The gradient (acceleration) starts steep and decreases continuously, eventually becoming zero when the curve flattens. The flat section is terminal speed. Examiners frequently ask you to describe or explain this shape, so make sure you can link each part of the curve to the force balance at that moment.

3. The Parachute Moment: A Change in Terminal Speed

The skydiver scenario becomes even richer when the parachute opens. At the moment of deployment, the skydiver is already moving at terminal speed. Opening the parachute dramatically increases the cross-sectional area, which causes drag to surge well above the weight. The resultant force is now upward, so the skydiver decelerates.

As the skydiver slows down, drag decreases again. The deceleration continues until drag once more equals weight — but now at a much lower speed, because the parachute keeps drag high even at low velocities. A new, lower terminal speed is reached, safe for landing.

On a velocity-time graph, this appears as a sudden drop in velocity after the parachute opens, followed by the curve levelling off again at a lower terminal speed. Examiners love this scenario because it tests whether you understand that terminal speed is not fixed — it depends on the balance of forces, which can change.

4. Factors Affecting Terminal Speed

Terminal speed is reached when drag equals the driving force. For a falling object, the driving force is weight ($W = mg$). For a vehicle, the driving force is the engine thrust. Anything that changes either of these quantities will change the terminal speed.

• Mass: A heavier object has greater weight, so drag must be larger to balance it. Since drag increases with speed, a heavier object reaches a higher terminal speed.
• Cross-sectional area: A larger area increases drag at any given speed, so the drag-weight balance is reached at a lower speed — lower terminal speed.
• Shape and streamlining: A streamlined shape reduces the drag coefficient, meaning less drag at the same speed. This raises the terminal speed.
• Fluid density: Denser fluids (e.g. water vs air) produce greater drag at the same speed, lowering terminal speed. This is why objects fall more slowly in water.
• Surface texture: Rougher surfaces increase drag, lowering terminal speed.

For a vehicle, the same logic applies. The maximum speed of a car is its terminal speed — reached when the driving force from the engine equals the total resistive forces (air resistance plus friction from the road and drivetrain). A more powerful engine can sustain a larger driving force, so it reaches balance at a higher speed. A more aerodynamic body shape reduces drag, again allowing a higher maximum speed for the same engine output.

5. Air Resistance and Projectile Trajectories

In the idealised projectile model you may have met at GCSE, air resistance is ignored. The horizontal velocity stays constant throughout the flight, and the vertical motion is governed purely by gravity. The resulting path is a perfect parabola. At A Level, you need to understand qualitatively how air resistance changes this picture.

Air resistance always acts in the direction opposite to the velocity vector. For a projectile, the velocity vector changes direction continuously — it points forward and downward at varying angles. This means air resistance has both a horizontal component (opposing forward motion) and a vertical component (opposing downward motion on the way down, opposing upward motion on the way up).

The practical consequences are significant. The horizontal velocity decreases throughout the flight rather than staying constant, so the projectile travels a shorter horizontal range. The peak height is also reduced because the upward component of velocity is reduced more quickly. The trajectory is no longer a symmetric parabola — the descent is steeper than the ascent, and the landing point is closer to the launch point than the idealised model predicts.

The asymmetry of the trajectory is a detail that examiners test directly. Many students assume the path is still symmetric — it is not. The object spends more time descending than ascending because the horizontal velocity has been reduced, and the descent is steeper. If asked to sketch a realistic trajectory alongside an idealised one, always show the realistic path falling short and curving more steeply on the way down.

6. Worked Analysis: Interpreting a Velocity-Time Graph for Terminal Speed

Exam questions on terminal speed frequently present a velocity-time graph and ask you to describe or explain specific features. The step-by-step below walks through a typical analysis question so you can see exactly how to structure a full-mark response.

Worked example: Analysing a Velocity-Time Graph for a Falling Object

A ball bearing is dropped from rest into a tall cylinder of oil. The velocity-time graph shows the ball's speed increasing from 0, curving upward with decreasing gradient, and eventually becoming horizontal at $v = 0.12 \text{ m/s}$. Describe and explain the motion shown by the graph. [4 marks]

1. At $t = 0$: velocity $= 0$, so drag $= 0$. Resultant force $= W - 0 = W = mg$ (downward). — The ball starts from rest. With no velocity, there is no drag force. The full weight acts as the resultant force. [B1] — stating initial resultant force equals weight.
2. As $v$ increases, drag increases. Resultant force $= W - F_{\text{drag}}$ decreases. By $F = ma$, acceleration $a = \frac{W - F_{\text{drag}}}{m}$ decreases. — The gradient of a velocity-time graph equals acceleration. A decreasing gradient means decreasing acceleration — caused by increasing drag reducing the resultant force. [M1] — linking gradient to acceleration and drag to resultant force.
3. When $v = 0.12 \text{ m/s}$: $F_{\text{drag}} = W$. Resultant force $= 0$. Therefore $a = 0$. — A horizontal line on a velocity-time graph means zero acceleration — constant velocity. This is terminal speed. The drag force has grown to exactly equal the weight. [A1] — correctly identifying terminal speed and the force balance.
4. Terminal speed $= 0.12 \text{ m/s}$. — Read directly from the graph as the constant final velocity. Always state the value with units when reading from a graph. [B1] — correct value with units.

Answer: The ball accelerates from rest because the resultant force equals its weight (drag is zero initially). As speed increases, drag increases, reducing the resultant force and therefore the acceleration. The gradient of the graph decreases continuously. When drag equals weight, the resultant force is zero, acceleration is zero, and the ball moves at constant terminal speed of 0.12 m/s.

7. Exam Technique: Getting Full Marks on Terminal Speed Questions

Terminal speed questions are reliable mark-scorers if you follow a consistent structure. The most common command words are 'describe', 'explain', and 'sketch'. Each demands a different approach, and mixing them up is one of the most frequent reasons students lose marks.

For 'describe' questions, state what happens in sequence — do not give reasons. For 'explain' questions, you must give the scientific cause using force language: always name the forces, state their relative magnitudes, and state the resultant force. For 'sketch' questions on trajectories, always draw the realistic path falling short of and below the idealised parabola, with a steeper descent.

8. Bringing It Together: A Synthesis Question

The two specification requirements in this lesson — terminal speed and the effect of air resistance on projectiles and vehicle maximum speed — are connected by a single underlying idea: drag is a velocity-dependent resistive force that changes the motion of any object moving through a fluid. Whether you are analysing a skydiver, a thrown ball, or a car on a motorway, the same force-balance reasoning applies.

For the exam, the highest-mark questions will ask you to apply this reasoning to unfamiliar contexts. A question might describe a cyclist, a raindrop, or a spacecraft re-entering the atmosphere. The approach is always the same: identify the driving force, identify the drag force, consider how drag changes with speed, and describe or explain the motion in terms of the resultant force.