Variable acceleration in one dimensionEdexcel A-Level Further Maths: Revision notes

When acceleration is not constant, the suvat equations no longer apply. Use calculus instead: $v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$ Going from acceleration to velocity to displacement means integrating; going the other way means differentiating. Every indefinite integral needs a constant of integration, found from the initial conditions (for example $v=9$ and $x=0$ at $t=0$).

Integrate $a=f(t)$ to get $v$, then integrate $v$ to get $x$. Example: $a=6t-12$, $v=9$ and $x=0$ at $t=0$. $v=3t^2-12t+9,\qquad x=t^3-6t^2+9t.$ At $t=2$: $v=-3$ m s⁻¹ and $x=2$ m. The particle is instantaneously at rest when $v=0$: $3(t-1)(t-3)=0$, so $t=1$ and $t=3$. A negative velocity means the particle moves in the negative direction.

If the velocity is given, integrate for displacement and differentiate for acceleration. Trigonometric and exponential functions appear: $\int\cos kt\,dt=\frac1k\sin kt$ and $\int e^{kt}\,dt=\frac1ke^{kt}$. Use radians. Example: $v=6\cos2t$, $x=1$ at $t=0$. Then $x=3\sin2t+1$, $a=-12\sin2t$, and the greatest displacement is 4 (when $\sin2t=1$). Distance travelled is not the same as displacement if the particle reverses.

When the acceleration depends on $v$, separate the variables: $\int\frac{dv}{f(v)}=\int dt$. Then use the initial condition to find the constant. Example: $\frac{dv}{dt}=\frac{60-v}{20}$ from rest. $\int\frac{dv}{60-v}=\int\frac{dt}{20}$ gives $-\ln(60-v)=\frac{t}{20}+c$, and $v=0$ at $t=0$ gives $v=60\left(1-e^{-0.05t}\right)$. The speed tends to the limiting speed 60 as $t\to\infty$. Example: deceleration $0.02v^2$ from 10 m s⁻¹: $\int\frac{dv}{v^2}=-0.02\int dt$, so $\frac1v=0.02t+0.1$ and $v=\frac{10}{1+0.2t}$.

The full specification also allows acceleration to depend on displacement: $v\frac{dv}{dx}=f(x)$ or $f(v)$, and $\frac{dx}{dt}=f(x)$. These use the chain-rule form $a=v\frac{dv}{dx}$ and are examined in the A2 content. At AS, questions use only $\frac{dv}{dt}=f(t)$, $\frac{dv}{dt}=f(v)$ and $\frac{dx}{dt}=f(t)$.

Decide what you are given (a function of $t$ or of $v$) and what is asked. Write the differential equation first, with correct signs. Separate or integrate, include the constant, and use initial conditions straight away. Check your final expression with the initial values. Give answers to 3 significant figures and state units.