Variable Force and Acceleration

In A-Level Mechanics the force is usually constant, and SUVAT does all the work. Real forces rarely are: a parachutist meets air resistance that grows with speed, a spring pulls harder the further it is stretched, a rocket's thrust changes in time. When the force — and hence the acceleration — varies, SUVAT is the wrong tool and calculus is the right one. The whole technique turns on choosing the correct form of the acceleration, $a = \dfrac{\mathrm{d}v}{\mathrm{d}t}$a=dtdv​ or $a = v\dfrac{\mathrm{d}v}{\mathrm{d}x}$a=vdxdv​, to match whether the force depends on time, position or velocity, and then setting up and solving the resulting differential equation. This lesson builds that decision systematically and works the three canonical cases, including the air-resistance and inverse-square problems that recur on Paper 3.

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1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M), and it is the most overtly calculus-driven topic in Further Mechanics — it leans directly on the further-calculus compulsory content (separable first-order differential equations, integration by partial fractions and standard forms). It is strongly AO1/AO3: set up the equation of motion correctly (the modelling choice of sign and acceleration form is the crux), then carry the integration to a result, often with an initial condition and a limiting (terminal-velocity) interpretation. Paper 3 weights these AO1 40% / AO2 25% / AO3 35%. The prerequisite is confident integration — including $\int \frac{1}{a+bv}\,\mathrm{d}v$∫a+bv1​dv and $\int \frac{v}{a+v^2}\,\mathrm{d}v$∫a+v2v​dv — from A-Level and Further Maths.

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2 Core theory — the decisive identity and the three cases

The acceleration can be written two ways, both from the chain rule:

$$\boxed{\,a = \frac{\mathrm{d}v}{\mathrm{d}t} = v\frac{\mathrm{d}v}{\mathrm{d}x}\,},$$a=dtdv​=vdxdv​​,

the second because $\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}v}{\mathrm{d}x}\dfrac{\mathrm{d}x}{\mathrm{d}t} = v\dfrac{\mathrm{d}v}{\mathrm{d}x}$dtdv​=dxdv​dtdx​=vdxdv​. A third form, $a = \dfrac{\mathrm{d}}{\mathrm{d}x}\bigl(\tfrac12 v^2\bigr)$a=dxd​(21​v2), is sometimes the cleanest of all when integrating with respect to position, since $v\,\mathrm{d}v = \mathrm{d}\bigl(\tfrac12 v^2\bigr)$vdv=d(21​v2); it makes the appearance of the kinetic-energy term completely transparent. All three are the same acceleration — the rate of change of velocity — merely expressed against different independent variables, and the freedom to choose is precisely what lets us turn an otherwise intractable equation of motion into a separable one. Choosing the right form is the entire art:

Force depends on	Equation of motion	Acceleration form & method
time, $F = f(t)$F=f(t)	$m\dfrac{\mathrm{d}v}{\mathrm{d}t} = f(t)$mdtdv​=f(t)	integrate w.r.t. $t$t for $v(t)$v(t), again for $x(t)$x(t)
position, $F = f(x)$F=f(x)	$mv\dfrac{\mathrm{d}v}{\mathrm{d}x} = f(x)$mvdxdv​=f(x)	integrate $mv\,\mathrm{d}v = f(x)\,\mathrm{d}x$mvdv=f(x)dx (work–energy)
velocity, $F = f(v)$F=f(v)	$m\dfrac{\mathrm{d}v}{\mathrm{d}t} = f(v)$mdtdv​=f(v) or $mv\dfrac{\mathrm{d}v}{\mathrm{d}x} = f(v)$mvdxdv​=f(v)	separate variables (choose $t$t- or $x$x-form to suit the question)

The guiding rule: use $v\,\mathrm{d}v/\mathrm{d}x$vdv/dx whenever the answer or the force is in terms of distance, and $\mathrm{d}v/\mathrm{d}t$dv/dt whenever it is in terms of time. For velocity-dependent forces both work — pick the one that matches what is asked.

Why does the choice matter so much? Because picking the wrong form produces an integral you cannot evaluate. If a force depends on position, say $F = f(x)$F=f(x), and you write $m\,\mathrm{d}v/\mathrm{d}t = f(x)$mdv/dt=f(x), the right-hand side is a function of $x$x while the left integrates over $t$t — you cannot separate the variables without first knowing $x(t)$x(t), which is what you are trying to find. Switching to $mv\,\mathrm{d}v/\mathrm{d}x = f(x)$mvdv/dx=f(x) puts $v$v on the left and $x$x on the right, and the equation separates immediately. The whole skill of this topic is recognising, from the form of the force and from what the question asks for, which version of the acceleration makes the equation separable. A short checklist: force in $t$t → use $\mathrm{d}v/\mathrm{d}t$dv/dt; force in $x$x → use $v\,\mathrm{d}v/\mathrm{d}x$vdv/dx; force in $v$v, time wanted → $\mathrm{d}v/\mathrm{d}t$dv/dt; force in $v$v, distance wanted → $v\,\mathrm{d}v/\mathrm{d}x$vdv/dx.

A general truth worth stating: integrating $mv\,\mathrm{d}v = f(x)\,\mathrm{d}x$mvdv=f(x)dx gives $\tfrac12 mv^2 = \int f\,\mathrm{d}x + C$21​mv2=∫fdx+C — this is the work–energy theorem, the change in kinetic energy equals the work done. Position-dependent forces are therefore always amenable to an energy argument, and conservative ones ($F = -\mathrm{d}U/\mathrm{d}x$F=−dU/dx) give energy conservation directly. This is more than a computational convenience: it tells you that for a position-dependent force the speed at a point is determined by the work done getting there, independent of how long the journey took — which is exactly why the SHM relation $v^2 = \omega^2(a^2 - x^2)$v2=ω2(a2−x2) can be written down without ever solving for time. Whenever a question links a speed to a position and never mentions time, reach first for the energy/$v\,\mathrm{d}v/\mathrm{d}x$vdv/dx approach.

It is also worth noticing the limiting behaviour that velocity-dependent forces produce. When a resistance grows with speed, there is generally a terminal velocity at which the resistance exactly balances the driving force and the acceleration vanishes. Mathematically this is the value of $v$v making $\mathrm{d}v/\mathrm{d}t = 0$dv/dt=0, and it is also the horizontal asymptote of the $v(t)$v(t) graph: the speed climbs (or falls) towards it but never crosses it for motion started on one side. Spotting the terminal velocity before doing any integration — simply by setting the net force to zero — gives you a check on the final answer and often a quick mark in its own right. The whole qualitative shape of resisted motion (exponential approach to a limit under linear drag, hyperbolic-tangent approach under quadratic drag) can be anticipated from this single observation, which is why experienced candidates always identify the terminal velocity first and integrate second.

One more subtlety deserves attention. After integrating, the constant of integration must always be fixed by an initial condition. With an indefinite integral this is explicit; with a definite integral the condition becomes the lower limit. Forgetting the constant — or, just as bad, evaluating a definite integral with the wrong limits — is the single most frequent way to lose accuracy marks in this topic. Treat the initial condition as part of the method, not an afterthought, and write it down at the moment you integrate.

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3 Worked examples with mark scheme

Example 1 — force as a function of time

A particle of mass $2\,\text{kg}$2kg starts from rest; a force $F = 6t\,\text{N}$F=6tN acts. Find $v$v and the displacement at $t = 3\,\text{s}$t=3s.

$$2\frac{\mathrm{d}v}{\mathrm{d}t} = 6t \ \Rightarrow\ \frac{\mathrm{d}v}{\mathrm{d}t} = 3t \ \Rightarrow\ v = \tfrac{3}{2}t^2\ (\,v=0\text{ at }t=0\,). \quad (\textbf{M1 A1})$$2dtdv​=6t ⇒ dtdv​=3t ⇒ v=23​t2 (v=0 at t=0).(M1 A1) $$\text{At }t=3:\ v = \tfrac32(9) = 13.5\ \text{m s}^{-1}. \qquad x = \int_0^3 \tfrac32 t^2\,\mathrm{d}t = \left[\tfrac12 t^3\right]_0^3 = \tfrac{27}{2} = 13.5\ \text{m}. \quad (\textbf{M1 A1})$$At t=3: v=23​(9)=13.5 m s−1.x=∫03​23​t2dt=[21​t3]03​=227​=13.5 m.(M1 A1)

(M1 $F = m\,\mathrm{d}v/\mathrm{d}t$F=mdv/dt and integrate; A1 $v = \tfrac32 t^2$v=23​t2; M1 integrate $v$v for displacement; A1 both values. The constant is fixed by $v(0)=0$v(0)=0 — forgetting it is the standard error.)

Example 2 — force as a function of position

A particle of mass $3\,\text{kg}$3kg starts from rest at $x = 0$x=0; a force $F = 12 - 2x\,\text{N}$F=12−2xN acts. Find the speed at $x = 4\,\text{m}$x=4m, and the position where it first comes momentarily to rest.

Use $a = v\,\mathrm{d}v/\mathrm{d}x$a=vdv/dx: (M1)

$$\tfrac12(3)v^2 = \int_0^4 (12 - 2x)\,\mathrm{d}x = \left[12x - x^2\right]_0^4 = 48 - 16 = 32 \ \Rightarrow\ v^2 = \tfrac{64}{3} \ \Rightarrow\ v = \tfrac{8}{\sqrt3} \approx 4.62\ \text{m s}^{-1}. \quad (\textbf{M1 A1})$$21​(3)v2=∫04​(12−2x)dx=[12x−x2]04​=48−16=32 ⇒ v2=364​ ⇒ v=3​8​≈4.62 m s−1.(M1 A1)

The particle is momentarily at rest where the cumulative work returns to zero: $\int_0^X (12-2x)\,\mathrm{d}x = 12X - X^2 = 0 \Rightarrow X(12 - X) = 0 \Rightarrow X = 12\,\text{m}$∫0X​(12−2x)dx=12X−X2=0⇒X(12−X)=0⇒X=12m (the force does positive work out to $x = 6$x=6, then negative work, bringing $v$v back to $0$0 at $x = 12$x=12). (M1 A1)

(M1 choose $v\,\mathrm{d}v/\mathrm{d}x$vdv/dx; M1 integrate the force; A1 $v \approx 4.62$v≈4.62; M1 set net work $=0$=0; A1 $X = 12$X=12. Note the speed is greatest at $x = 6$x=6, where the force changes sign.)

Example 3 — inverse-square attraction (conservative force)

A particle of mass $m$m is attracted to a fixed point $O$O by $F = k/x^2$F=k/x2 towards $O$O. It starts from rest at $x = a$x=a. Find its speed at distance $x$x $(x < a)$(x<a).

The force is conservative with potential $U(x) = -k/x$U(x)=−k/x (since $F = -\mathrm{d}U/\mathrm{d}x = -k/x^2$F=−dU/dx=−k/x2, i.e. towards $O$O). Energy conservation from rest at $x = a$x=a: (M1)

$$\tfrac12 mv^2 = U(a) - U(x) = \left(-\frac{k}{a}\right) - \left(-\frac{k}{x}\right) = k\!\left(\frac{1}{x} - \frac{1}{a}\right) > 0. \quad (\textbf{M1})$$21​mv2=U(a)−U(x)=(−ak​)−(−xk​)=k(x1​−a1​)>0.(M1) $$v = \sqrt{\frac{2k}{m}\!\left(\frac{1}{x} - \frac{1}{a}\right)}. \quad (\textbf{A1})$$v=m2k​(x1​−a1​)​.(A1)

(M1 identify the conservative potential or set up $mv\,\mathrm{d}v/\mathrm{d}x = -k/x^2$mvdv/dx=−k/x2; M1 integrate/apply energy with the limits; A1 the speed. Equivalently $\int mv\,\mathrm{d}v = \int_a^x -k/x^2\,\mathrm{d}x$∫mvdv=∫ax​−k/x2dx gives the same result directly.)

Example 4 — velocity-dependent resistance, time form

A particle of mass $2\,\text{kg}$2kg moves horizontally on a smooth surface, subject only to a resistance $4v\,\text{N}$4vN. It has speed $10\,\text{m s}^{-1}$10m s−1 at $t = 0$t=0. Find $v$v as a function of time, and the time to slow to $2\,\text{m s}^{-1}$2m s−1.

The force opposes motion, so $m\dfrac{\mathrm{d}v}{\mathrm{d}t} = -4v$mdtdv​=−4v, i.e. $\dfrac{\mathrm{d}v}{\mathrm{d}t} = -2v$dtdv​=−2v. (M1) Separating,

$$\int \frac{\mathrm{d}v}{v} = -\int 2\,\mathrm{d}t \ \Rightarrow\ \ln v = -2t + \ln 10 \ \Rightarrow\ v = 10\,e^{-2t}\ \text{m s}^{-1}. \quad (\textbf{M1 A1})$$∫vdv​=−∫2dt ⇒ lnv=−2t+ln10 ⇒ v=10e−2t m s−1.(M1 A1)

For $v = 2$v=2: $2 = 10e^{-2t} \Rightarrow e^{-2t} = 0.2 \Rightarrow t = \dfrac{\ln 5}{2} \approx 0.805\,\text{s}$2=10e−2t⇒e−2t=0.2⇒t=2ln5​≈0.805s. (A1)

(M1 $\dfrac{\mathrm{d}v}{\mathrm{d}t} = -2v$dtdv​=−2v; M1 separate and integrate; A1 $v = 10e^{-2t}$v=10e−2t; A1 the time. With linear resistance the speed decays exponentially but never reaches zero in finite time — yet the distance travelled is finite, $\int_0^\infty 10e^{-2t}\,\mathrm{d}t = 5\,\text{m}$∫0∞​10e−2tdt=5m.)

Example 5 — quadratic resistance via partial fractions

A particle of mass $m$m falls from rest under gravity with resistance $mkv^2$mkv2. Find $v$v as a function of time, using partial fractions.

Down-positive: $m\dfrac{\mathrm{d}v}{\mathrm{d}t} = mg - mkv^2 \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} = g - kv^2 = k\!\left(\dfrac{g}{k} - v^2\right)$mdtdv​=mg−mkv2⇒dtdv​=g−kv2=k(kg​−v2). Write $V^2 = g/k$V2=g/k (the terminal speed squared). (M1) Then

$$\frac{\mathrm{d}v}{V^2 - v^2} = k\,\mathrm{d}t, \qquad \frac{1}{V^2 - v^2} = \frac{1}{2V}\!\left(\frac{1}{V-v} + \frac{1}{V+v}\right). \quad (\textbf{M1 partial fractions})$$V2−v2dv​=kdt,V2−v21​=2V1​(V−v1​+V+v1​).(M1 partial fractions)

Integrating with $v(0) = 0$v(0)=0:

$$\frac{1}{2V}\ln\!\left(\frac{V+v}{V-v}\right) = kt \ \Rightarrow\ \frac{V+v}{V-v} = e^{2Vkt}. \quad (\textbf{M1})$$2V1​ln(V−vV+v​)=kt ⇒ V−vV+v​=e2Vkt.(M1)

Solving for $v$v and using $Vk = g/V$Vk=g/V (since $V^2 = g/k$V2=g/k) gives

$$v = V\,\frac{e^{2Vkt} - 1}{e^{2Vkt} + 1} = V\tanh(Vkt) = \sqrt{\frac{g}{k}}\,\tanh\!\left(\sqrt{gk}\;t\right). \quad (\textbf{A1})$$v=Ve2Vkt+1e2Vkt−1​=Vtanh(Vkt)=kg​​tanh(gk​t).(A1)

(M1 form $g - kv^2$g−kv2 and identify $V^2 = g/k$V2=g/k; M1 partial fractions; M1 integrate and apply the condition; A1 the $\tanh$tanh solution. As $t\to\infty$t→∞, $\tanh\to1$tanh→1 so $v\to V = \sqrt{g/k}$v→V=g/k​, the terminal velocity — exactly the $a=0$a=0 balance. The hyperbolic-tangent form links neatly to the hyperbolic-functions material in the pure papers.)

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4 Specimen-style exam question — resisted vertical motion

(specimen-style — not from any past paper)

A particle of mass $5\,\text{kg}$5kg is released from rest and falls vertically. The air resistance is $2v\,\text{N}$2vN when the speed is $v\,\text{m s}^{-1}$vm s−1. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2. (a) Show that $5\dfrac{\mathrm{d}v}{\mathrm{d}t} = 49 - 2v$5dtdv​=49−2v. (b) Find $v$v as a function of $t$t. (c) State the terminal velocity and the time to reach $99\%$99% of it.

Model solution. (a) Taking downwards positive, the forces are weight $mg = 5(9.8) = 49\,\text{N}$mg=5(9.8)=49N down and resistance $2v$2v up, so Newton's second law $m\dfrac{\mathrm{d}v}{\mathrm{d}t} = mg - 2v$mdtdv​=mg−2v gives $5\dfrac{\mathrm{d}v}{\mathrm{d}t} = 49 - 2v$5dtdv​=49−2v.

(b) Separate variables:

$$\frac{5\,\mathrm{d}v}{49 - 2v} = \mathrm{d}t \ \Rightarrow\ -\tfrac{5}{2}\ln|49 - 2v| = t + C.$$49−2v5dv​=dt ⇒ −25​ln∣49−2v∣=t+C.

At $t = 0,\ v = 0$t=0, v=0: $C = -\tfrac52\ln 49$C=−25​ln49. Hence $-\tfrac52\ln\!\left(\dfrac{49 - 2v}{49}\right) = t$−25​ln(4949−2v​)=t, so $49 - 2v = 49e^{-2t/5}$49−2v=49e−2t/5 and