Physics › Mechanics › Drag and terminal speed

Drag and terminal speed

Friction and drag are the forces that idealised problems leave out, and this topic puts them back in, qualitatively. The key fact is that drag grows with speed, and from it come terminal speed, the two stages of a skydive, and the top speed of any vehicle.

Year 12AQA 3.4.1.4 CIE 3.2.1, 3.2.2, 3.2.3 OCR A 3.2.2(a), 3.2.2(b), 3.2.2(c), 3.2.2(d)(i)EDEXCEL 17

Builds on Newton's laws and the resultant force and Motion graphs and the SUVAT equations.

IN THIS TOPIC

Describe friction, lift and drag qualitatively, including that air resistance increases with speed.
Explain terminal speed using Newton's laws, and sketch the velocity-time graph it produces.
Apply the same balance argument to a parachutist's two terminal speeds and to a vehicle's maximum speed.

COMMON MISCONCEPTION

Falling objects just keep speeding up.

The resistive cast

Friction acts between solid surfaces and opposes relative motion, or attempted motion, between them. The required treatment is qualitative only, and the distinction between static and dynamic friction is not tested. Drag is the resistive force from moving through a fluid, air or water, and lift is the component of the fluid's force perpendicular to the flow, the force that holds aircraft up.

Everything else in this topic follows from the fact that air resistance increases with speed. Move faster and you sweep more air aside each second, and each parcel of it more violently. Friction, by contrast, does not grow with speed in this way, which is why drag, not friction, sets speed limits.

Terminal speed

Drop an object and follow the forces. At the moment of release the speed is zero, so drag is zero and the resultant is the full weight, and the acceleration is g. As speed builds, drag grows, the resultant shrinks, and the acceleration fades.

FIG. 1Weight never changes; drag grows with speed until the net force falls to zero and the fall settles at the first terminal speed. The parachute multiplies the drag at a stroke: for a moment drag exceeds weight (the coral net arrow flips upward), the fall slows, and a second, lower terminal speed is reached. The curve records all of it. Arrow lengths are capped for scale during the brief spike, when the true drag is many times the weight.

Eventually drag has grown until it equals the weight. The resultant force is zero, and by Newton's first law the object stops accelerating and falls at constant velocity. That speed is the terminal speed. Nothing switched off. Two forces reached a stalemate, and the object is still travelling fast.

FIG. 2The velocity-time graph of the same story: a steep start at gradient g, a shrinking gradient as drag grows, and a flat approach to terminal speed.

Every stage of the argument is visible on the velocity-time graph. Initial gradient g, then curvature as the resultant shrinks, then a horizontal asymptote at terminal speed. Sketching that graph with all three features labelled is a standard question in its own right.

WORKED EXAMPLE

The forces at half terminal speed

An 80 kg skydiver has a terminal speed of 50 m s⁻¹, with drag proportional to speed squared. Find the resultant force and acceleration at 25 m s⁻¹.

At terminal speed the drag equals the weight, 80 × 9.81 = 785 N. That one line calibrates the drag without ever knowing its constant.

Halve the speed and the squared law leaves a quarter of the drag, 785/4 = 196 N.

Resultant = 785 − 196 = 589 N downward, so a = 589/80 = 7.4 m s⁻². Still most of g, and that is why the early part of a fall gathers speed so alarmingly.

The parachutist's second terminal speed

A parachutist runs the argument twice. In freefall the body settles at a high terminal speed. Opening the parachute multiplies the area meeting the air, so at that same speed the drag now vastly exceeds the weight. The resultant swings upward, the parachutist decelerates, and drag falls away with the speed until the two balance again at a much lower and thoroughly survivable terminal speed.

FIG. 3The skydiver's velocity-time graph: a high plateau in freefall, a steep deceleration when the parachute opens, and a second, lower plateau.

Nothing new pulls upward when the canopy opens. The old drag force is simply made enormous by the new area, and it falls back as the speed drops.

The top speed of a vehicle

A car's maximum speed is that stalemate again, in a different setting. The engine provides a driving force, and drag rises with speed until the total resistive force matches it. Resultant zero, acceleration zero, and that is the top speed. Anything that raises the driving force or trims the drag, more power or better streamlining, pushes the stalemate to a higher speed. Sports-car design is that trade, pursued expensively.

GUIDED PRACTICE

A top speed from power and drag

A car's drag is 0.35v² newtons at speed v, and its engine delivers a maximum useful power of 25 kW. Using P = Fv at the balance point, find its top speed.

Show the working

At top speed the driving force equals the drag, so P = Fv becomes 25 000 = 0.35v² × v = 0.35v³.

v³ = 71 400, so v = 41 m s⁻¹, about 150 km h⁻¹. The cube is why doubling a car's top speed needs roughly eight times the power.

ASSESSMENT FOCUS

The terminal-speed explanation is a chain, and every link scores. Drag increases with speed, so the resultant force falls, so the acceleration falls, until drag equals weight, resultant zero, constant velocity. Write it in that order and take all of the marks.
At terminal speed the forces are balanced, and the object is still moving and still has weight. “The forces cancel so it stops” is the error the mark scheme is watching for.
On the skydiver graph, the parachute moment is a steep negative gradient and never a vertical cliff to zero. The speed falls to the new terminal value, not to rest.
For objects of the same shape and size, heavier means a faster terminal speed. More weight needs more drag to balance it, and more drag needs more speed, so the raindrop questions all hinge on this and most students guess the wrong way.
For a vehicle's top speed, one sentence does it. At maximum speed the driving force equals the total resistive force, so the resultant force and the acceleration are both zero.

CHECK YOURSELF

A skydiver falls at a terminal speed of 55 m s⁻¹, then opens a parachute and reaches a new terminal speed of 6 m s⁻¹. Explain, in terms of forces, why the new terminal speed is lower.

Show a hint

The weight has not changed. What has, and what does that do to the speed at which drag can match it?

Show the answer

Opening the parachute greatly increases the area pushing through the air, so at any given speed the drag is far larger than before.

Drag therefore matches the unchanged weight at a much lower speed, and it is at that lower speed that the resultant returns to zero and the fall becomes steady.

Between the two plateaus, drag exceeds weight, the resultant acts upward, and the skydiver decelerates: the steep falling section of the velocity-time graph.

Drag grows with speed.

Terminal speed is where it catches the weight.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the drag and terminal speed questions page.

8 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Exponentials and logarithms is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Describe friction, lift and drag qualitatively, including that air resistance increases with speed.
Explain terminal speed using Newton's laws, and sketch the velocity-time graph it produces.
Apply the same balance argument to a parachutist's two terminal speeds and to a vehicle's maximum speed.

Open the full revision checklist to track your progress across the whole unit.