Variable acceleration in a straight lineEdexcel International A Level Further Maths: Revision notes

For motion in a straight line, with displacement $x$, velocity $v$ and acceleration $a$: $v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}=v\frac{dv}{dx}.$ The last form follows from the chain rule: $\frac{dv}{dt}=\frac{dv}{dx}\times\frac{dx}{dt}$. The formulae $v=u+at$ and $s=ut+\frac12at^2$ hold only when $a$ is constant. When $a$ depends on $t$ or on $x$, you must integrate (or separate variables) instead. Remember that velocity and acceleration are vectors: the sign shows direction.

If $a=f(t)$, integrate to find $v$, then integrate again to find $x$: $v=\int f(t)\,dt,\qquad x=\int v\,dt.$ Each integration brings a constant of integration which you find from the initial conditions (for example $v=9$ at $t=0$, or $x=0$ at $t=0$). Example: $a=6t-12$, $v=9$ and $x=0$ at $t=0$: $v=3t^2-12t+9=3(t-1)(t-3)$ and $x=t^3-6t^2+9t$. To find when the particle is at rest, solve $v=0$. To find the distance travelled (not displacement), split the motion at the times when $v=0$ and add the magnitudes.

If $a=f(x)$, use $a=v\frac{dv}{dx}$ and separate the variables: $\int v\,dv=\int f(x)\,dx\quad\Rightarrow\quad\tfrac12v^2=\int f(x)\,dx+c.$ This gives $v^2$ as a function of $x$ without involving time. Example: $a=3-x$ and $v=4$ at $x=0$: $\frac12v^2=3x-\frac12x^2+8$, so $v^2=6x-x^2+16$. At $x=3$, $v=5$, which is the greatest speed because $a=0$ there. The particle is at rest where $v^2=0$: $x=8$ (or $x=-2$ for motion in the other direction). Taking the square root, choose the sign of $v$ from the direction of motion.

If $\frac{dx}{dt}=f(t)$, integrate directly: $x=\int f(t)\,dt$. If $\frac{dx}{dt}=f(x)$, separate the variables: $\int\frac{1}{f(x)}\,dx=\int dt$. For $\frac{dx}{dt}=2x$ this gives $\ln x=2t+c$, so $x=Ae^{2t}$, and the initial value fixes $A$. To find the acceleration when $v$ is given as a function of $x$, use $a=v\frac{dv}{dx}$; if $v=2x$ then $a=2x\times2=4x$. If $v=\frac4x$ then $a=\frac4x\times\left(-\frac4{x^2}\right)=-\frac{16}{x^3}$. The calculus needed is no more than in P1 to P4: polynomials, trigonometric, exponential and logarithmic integrals.

$P$: $\frac{dv}{dt}=-8\sin2t$, $v=4$ and $x=0$ at $t=0$. Then $v=4\cos2t$ and $x=2\sin2t$. $v=0$ when $t=\frac{\pi}{4}$, where $x=2$, the greatest distance from $O$. Method: (1) identify which of $a=f(t)$, $a=f(x)$, $v=f(x)$ or $v=f(t)$ you have; (2) choose the integral that matches (integrate in $t$, or separate variables in $x$); (3) use the initial conditions for each constant; (4) interpret: $v=0$ for rest, the sign of $v$ for direction; (5) check units and give exact values (such as $\ln2$) where convenient.