M3.1 - Further kinematics

For motion along a straight line, displacement $x$, signed velocity $v$ and signed acceleration $a$ are linked by derivatives. Choose the form whose independent variable matches the information given; this avoids introducing an unnecessary unknown function.

The displacement form of acceleration follows from the chain rule:$$\frac{\mathrm dv}{\mathrm dt}=\frac{\mathrm dv}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}=v\frac{\mathrm dv}{\mathrm dx}.$$Use it when acceleration or velocity is expressed in terms of $x$. If the required quantity is time and $v=f(x)$, separate variables instead: $\mathrm dt=\mathrm dx/v(x)$.

Every indefinite integration needs a constant. Apply a stated condition such as $x=x_0$ and $v=v_0$ at $t=t_0$ only after integrating, or use definite integrals with those values as limits. Keep $v$ signed: a negative velocity represents motion in the negative $x$-direction, while speed is $|v|$.

Suppose $v=12/(x+2)$ metres per second and $x=1$ when $t=0$. First,$$a=v\frac{\mathrm dv}{\mathrm dx}=\frac{12}{x+2}\left(-\frac{12}{(x+2)^2}\right)=-\frac{144}{(x+2)^3}.$$At $x=4$, $a=-2/3\,\mathrm{m\,s^{-2}}$. For the elapsed time to reach $x=4$,$$t=\int_1^4\frac{\mathrm dx}{v}=\int_1^4\frac{x+2}{12}\,\mathrm dx=\left[\frac{x^2}{24}+\frac{x}{6}\right]_1^4=\frac98\text{ s}.$$The negative acceleration means velocity is decreasing here; the deceleration magnitude is $2/3\,\mathrm{m\,s^{-2}}$.

Check that the final variable matches the question, substitute the initial condition back into the integrated relation, and verify units: $v$ has units $\mathrm{m\,s^{-1}}$ and $a$ has units $\mathrm{m\,s^{-2}}$. If a derivation divides by $v$ or by another expression that can be zero, solve on intervals where that division is valid and inspect the zero case separately.