Differential equation

Once the equation of motion is written with the correct form of acceleration, it is a separable first-order differential equation. The whole method is three moves: separate, integrate, apply conditions.

Step 1 — Separate. Rearrange so every $v$ (and $dv$) is on one side and every $t$ or $x$ (with its differential) is on the other. For a force in time, $m\frac{dv}{dt} = f(t)$ gives: $dv = \frac{1}{m}f(t)\,dt$ For a force in displacement, $mv\frac{dv}{dx} = f(x)$ gives: $v\,dv = \frac{1}{m}f(x)\,dx$

Step 2 — Integrate both sides. The integration techniques are restricted to those in 9709 Pure Mathematics 3: standard polynomial integrals, $\int\frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b|$, partial fractions, and the standard exponential/trig results. You will not need anything beyond P3.

Step 3 — Don't forget $+C$. Every indefinite integral carries a constant; it is fixed in the next section by the initial condition. A common version of the same idea is to integrate with limits instead, e.g. $\int_{u}^{v} dv = \frac{1}{m}\int_{0}^{t} f(t)\,dt$, which skips the constant altogether.

Worked sketch. $\frac{dv}{dt} = 6 - v$ separates as $\frac{dv}{6 - v} = dt$. Integrating: $-\ln|6 - v| = t + C$.

Cambridge tip. If an integral looks awkward, check whether partial fractions (e.g. for $\frac{1}{v(a - v)}$) or a simple $\frac{1}{ax+b}$ log will do it — those are the only tools the examiner expects.

The constant of integration $C$ is fixed by the initial (or boundary) condition stated in the question — almost always the speed at $t = 0$ or at a given displacement.

The routine:

1. Integrate to get a general relation, e.g. $v = 3t^2 + C$ or $\frac{1}{2}v^2 = 10x - x^2 + C$.
2. Substitute the given values (e.g. "starts from rest" means $v = 0$ when $t = 0$; "passes through $O$ with speed $3$" means $v = 3$ when $x = 0$) to find $C$.
3. Write the particular solution and use it to answer what is asked.

Reading the wording:

• "from rest" $\Rightarrow v = 0$ at $t = 0$;
• "with speed $u$" $\Rightarrow v = u$ at the stated point;
• "comes to rest" / "stops" $\Rightarrow v = 0$ at the unknown time/place you are finding.

Worked sketch. From $-\ln|6 - v| = t + C$ with $v = 0$ at $t = 0$: $-\ln 6 = C$, so $-\ln|6 - v| = t - \ln 6$, i.e. $\ln\frac{6}{6 - v} = t$, giving $v = 6(1 - e^{-t})$.

Cambridge tip. Substitute the initial condition immediately after integrating, before any further algebra. Carrying an un-evaluated $C$ through several lines is where sign errors breed.

A body that is driven forward against a resistance speeds up while the driving force exceeds the resistance, and stops speeding up when the two balance. At that instant the acceleration is zero, so: $\frac{dv}{dt} = 0 \quad\Longleftrightarrow\quad F_{\text{resultant}} = 0$

Maximum speed. Set the resultant force to zero and solve for $v$. For example, with $m\frac{dv}{dt} = F_0 - kv$, the maximum speed satisfies $F_0 - kv = 0$, i.e. $v_{\max} = \frac{F_0}{k}$.

Terminal (limiting) velocity. For a body falling under gravity against resistance $kv$ (per unit mass), $\frac{dv}{dt} = g - kv$. Setting $\frac{dv}{dt} = 0$ gives the terminal speed $v_T = \frac{g}{k}$. The body never quite reaches it: the velocity rises and levels off asymptotically, as shown below.

Cambridge tip. "Find the maximum speed" almost never means differentiating $v(t)$ — it means setting the acceleration (the resultant force) to zero. That is the fast, intended route.

A very common Paper 3 set-up: a body of mass $m$ moving with no driving force, slowed only by a resistance. The form you choose depends on what is asked.

Resistance $R = kv$ (linear).

• Distance (use $v\frac{dv}{dx}$): $mv\frac{dv}{dx} = -kv \Rightarrow m\frac{dv}{dx} = -k$, so $v$ falls linearly with $x$ — a tidy result.
• Time (use $\frac{dv}{dt}$): $m\frac{dv}{dt} = -kv \Rightarrow \frac{dv}{v} = -\frac{k}{m}dt$, integrating to $v = u\,e^{-kt/m}$ — exponential decay, so the body never quite stops in finite time.

Resistance $R = kv^2$ (quadratic).

• Distance: $mv\frac{dv}{dx} = -kv^2 \Rightarrow m\frac{dv}{v} = -k\,dx$, giving a logarithmic $v$–$x$ relation.
• Time: $m\frac{dv}{dt} = -kv^2 \Rightarrow m\frac{dv}{v^2} = -k\,dt$, giving a relation in $\frac{1}{v}$.

Driving force and resistance together (e.g. $m\frac{dv}{dt} = F_0 - kv^2$) often needs partial fractions to integrate $\frac{1}{F_0 - kv^2}$ — squarely within P3.

Worked sketch (distance, linear resistance). With $2v\frac{dv}{dx} = -4v$ from $v = 10$: $\frac{dv}{dx} = -2$, so $v = 10 - 2x$. The body stops ($v = 0$) at $x = 5\,\text{m}$.

Cambridge tip. Decide distance vs time first — it fixes whether you use $v\frac{dv}{dx}$ or $\frac{dv}{dt}$, and therefore whether the integral becomes a clean log/linear result or an exponential.