Acceleration as a function of position

What you’ll learn

• Choose a useful differential form for acceleration.
• Use initial conditions to solve non-uniform motion.

Understand the idea

Variable acceleration needs calculus rather than constant-acceleration shortcuts. If acceleration is given as a function of time, integrate $dv/dt=a(t)$. If it is given as a function of displacement, use the chain rule to write $a=v\,dv/dx$, then integrate with respect to position.

The integration constant represents the initial state and must be found before evaluating another point. A squared-velocity result gives speed, but the sign of velocity still depends on the direction of motion. If velocity is specified directly as a function of position, separate $dx/dt=f(x)$ to relate position to time.

Key formulae

$$a=\frac{dv}{dt}=v\frac{dv}{dx}$$

$$\frac12(v^2-u^2)=\int_{x_0}^{x}a(s)\,ds$$

$$v=\frac{dx}{dt}$$

$$\int\frac{dx}{f(x)}=t+C\quad\text{when }\frac{dx}{dt}=f(x)$$

Worked example

A particle has acceleration $a=2x\,\mathrm{m\,s^{-2}}$ when its displacement is $x\,\mathrm m$. At $x=0$ its velocity is $3\,\mathrm{m\,s^{-1}}$. Find its speed at $x=2$.

1. Use $v\,dv/dx=2x$.
2. Integrate to obtain $v^2/2=x^2+C$; the initial condition gives $C=9/2$.
3. Hence $v^2=2x^2+9$, which is $17$ at $x=2$.

Answer: $\sqrt{17}\,\mathrm{m\,s^{-1}}$.