Variable forcesEdexcel International A Level Further Maths: Revision notes

When the force on a particle moving in a straight line is not constant, the constant-acceleration (suvat) equations do not apply. Instead use Newton's second law, $F=ma$, with the acceleration written as a derivative, and solve by integration. The force is positive in the direction of increasing $x$, so a force towards the origin, or a resistance, is negative. Initial conditions give the constants of integration.

The acceleration can be written in three ways, and you choose by what the force depends on: $a=\frac{dv}{dt}=\frac{d^2x}{dt^2}=v\frac{dv}{dx}.$ If the force is a function of time, use $a=\frac{dv}{dt}$ and integrate with respect to $t$. If it is a function of displacement, use $a=v\frac{dv}{dx}$ and separate the variables $v$ and $x$. If it depends on velocity, use $a=\frac{dv}{dt}$ (to find $v$ in terms of $t$) or $a=v\frac{dv}{dx}$ (to find $v$ in terms of $x$), whichever is asked for.

Time: $F=(6t+4)$ N on a $2$ kg particle starting from rest gives $2\frac{dv}{dt}=6t+4$, so $v=\frac32t^2+2t$ and $x=\frac12t^3+t^2$ (constants are zero). Displacement: a resistance $2x$ on a $0.5$ kg particle gives $0.5\,v\frac{dv}{dx}=-2x$, so $v\,dv=-4x\,dx$ and $v^2=v_0^2-4x^2$. The particle stops where $v=0$.

For a resistance $kv$ on a particle of mass $m$: $m\frac{dv}{dt}=-kv$. Separating variables, $\int\frac1v\,dv=-\frac km\int dt$, so $v=v_0e^{-kt/m}$. To find how far the particle travels, either integrate $v$ with respect to $t$, or use $mv\frac{dv}{dx}=-kv$, which simplifies to $\frac{dv}{dx}=-\frac km$, a linear relation between $v$ and $x$. For a resistance proportional to $v^2$ use $v\frac{dv}{dx}$ to get $\frac{dv}{v}=-\frac kmdx$.

Newton's law of gravitation gives a force of magnitude $\frac{GMm}{x^2}$ at distance $x$ from the centre of the Earth. At the surface $x=R$ the force is $mg$, so $GM=gR^2$ and the force is $\frac{mgR^2}{x^2}$. This is an inverse square law. For a body moving radially away from the Earth, $v\frac{dv}{dx}=-\frac{gR^2}{x^2}$. Integrating, $\frac12v^2=\frac{gR^2}{x}+C$. Using $v=u$ at $x=R$: $v^2=u^2-2gR+\frac{2gR^2}{x}$.

A body fired from the surface never returns if $v^2>0$ for all $x$. As $x\to\infty$, $v^2\to u^2-2gR$, so the escape speed is $u=\sqrt{2gR}\approx 11.2$ km s$^{-1}$ with $R=6.4\times10^6$ m. If $u$ is smaller, $v=0$ at a finite maximum distance. Example: at $x=2R$ with speed $u$, $v^2=u^2-gR+\frac{2gR^2}{x}$, so the least $u$ to escape from $x=2R$ is $\sqrt{gR}$.