Variable forces and motion in one dimensionEdexcel A-Level Further Maths: Revision notes

For motion in a straight line, $F=ma$ still holds when $F$ changes. The force may depend on time $t$, displacement $x$ or velocity $v$, and the acceleration must be written in the form that matches: $a=\frac{dv}{dt}=v\frac{dv}{dx}=\frac{d^2x}{dt^2}.$ Use $m\frac{dv}{dt}=F(t)$ or $F(v)$; use $mv\frac{dv}{dx}=F(x)$ or $F(v)$. Take care with signs: choose a positive direction and write forces opposing it as negative.

Use $m\frac{dv}{dt}=F(t)$, integrate to get $v$, then integrate again for $x$. Include the constant each time, using the initial conditions. Example: $m=2$, $F=6t$, start from rest at $O$. $\frac{dv}{dt}=3t$, so $v=\frac32t^2$ (since $v=0$ at $t=0$), and $x=\frac12t^3$. At $t=2$: $v=6$ m s⁻¹ and $x=4$ m. Do not use the constant-acceleration (suvat) formulae, because they only apply when $a$ is constant.

When $F$ depends on $x$, use $mv\frac{dv}{dx}=F(x)$ and separate variables: $\int mv\,dv=\int F(x)\,dx$. This gives $\frac12mv^2$ in terms of $x$ directly (it is the work-energy principle). Example: $2v\frac{dv}{dx}=\frac{16}{(x+2)^2}$. Then $\frac{v^2}{2}=-\frac{8}{x+2}+c$. With $v=1$ at $x=0$: $c=\frac92$, so $v^2=9-\frac{16}{x+2}$. As $x\to\infty$ the speed tends to the limit 3 m s⁻¹.

A resistance depending on velocity gives equations such as $m\frac{dv}{dt}=-kv$. Separate: $\int\frac{dv}{v}=-\frac km\int dt$, so $v=ue^{-kt/m}$. Example: $m=0.5$, resistance $2v$, $u=8$: $\frac{dv}{dt}=-4v$, so $v=8e^{-4t}$. The speed never reaches zero in finite time. To find the distance to rest, use $v\frac{dv}{dx}=-4v$ instead: $\frac{dv}{dx}=-4$, $v=8-4x$, which gives $v=0$ at $x=2$ m. Choose the form of $a$ that gives the quantity you are asked for: time or velocity from $\frac{dv}{dt}$; distance from $v\frac{dv}{dx}$.

Newton's law of gravitation gives a force $\frac{GMm}{x^2}$ at distance $x$ from the centre of a body. At the surface (radius $R$) this equals $mg$, so $GM=gR^2$ and the force is $\frac{mgR^2}{x^2}$, directed towards the centre. A particle projected upwards: $mv\frac{dv}{dx}=-\frac{mgR^2}{x^2}$, so $v^2=u^2-2gR+\frac{2gR^2}{x}$. Setting $v=0$ gives the greatest distance from the centre. As $x\to\infty$, the last term tends to zero, so the particle escapes if $u^2\ge2gR$: the escape speed is $\sqrt{2gR}$, about $1.12\times10^4$ m s⁻¹ for the Earth.

1. Choose a positive direction and draw all forces. 2. Write $F=ma$ with the matching form of $a$. 3. Separate variables and integrate, adding a constant, and use the initial conditions to find it. 4. Answer the question set: speed, time, distance or limit. Explain limits in words, for example the term tends to zero as $x\to\infty$. Give answers to 3 significant figures unless exact values are asked for.