5.13 KinematicsIB Maths: Applications and Interpretation HL: Revision notes

For motion in a straight line, displacement $s$ is the position relative to a fixed origin (it can be negative), velocity $v$ is the rate of change of displacement and acceleration $a$ is the rate of change of velocity: $v=\frac{ds}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.$ Dots are shorthand for derivatives with respect to time: $\dot{x}=\frac{dx}{dt}$ and $\ddot{x}=\frac{d^2x}{dt^2}$. Going the other way, integrate: $v=\int a\,dt$ and $s=\int v\,dt$, with a constant fixed by given values such as $s=0$ when $t=0$. Units: $s$ in m, $v$ in m s$^{-1}$, $a$ in m s$^{-2}$.

Speed is the magnitude of velocity, $|v|$, so it is never negative. A negative velocity means motion in the negative direction. The particle is instantaneously at rest when $v=0$. Solve $v(t)=0$; the velocity usually changes sign there, so the particle turns round. The particle is speeding up when $v$ and $a$ have the same sign, and slowing down when they have opposite signs. Example: $v=t^2-7t+10$ is $0$ at $t=2$ and $t=5$. At $t=3$, $v=-2$ and $a=2t-7=-1$, so it speeds up and the speed is $2$ m s$^{-1}$.

When velocity is given as a function of displacement, use the chain rule: $a=\frac{dv}{dt}=\frac{dv}{ds}\cdot\frac{ds}{dt}=v\frac{dv}{ds}.$ Example: $v=s^2-4$. Then $\frac{dv}{ds}=2s$ and $a=(s^2-4)(2s)$. At $s=3$, $a=5\times6=30$ m s$^{-2}$. Note that $v\frac{dv}{ds}=\frac12\frac{d}{ds}(v^2)$, so be careful not to quote $\frac{d}{ds}(v^2)$ as the acceleration.

The displacement (change in position) between $t_1$ and $t_2$ is $\int_{t_1}^{t_2}v(t)\,dt.$ It is signed: movement in the negative direction counts as negative. The position at time $t_2$ is the initial position plus the displacement. Example: $v=3t^2-12t+9$ gives displacement $\int_0^2v\,dt=\left[t^3-6t^2+9t\right]_0^2=2$ m. With a GDC, evaluate the definite integral directly, or use the integral function on the graph of $v$.

The total distance travelled counts all movement as positive: $\int_{t_1}^{t_2}|v(t)|\,dt.$ Find the times when $v=0$ in the interval, split the integral there, and add the magnitudes of the parts. A GDC can integrate $|v|$ directly. Example: $v=4\cos(0.5t)$ on $0\le t\le2\pi$. Displacement is $\int_0^{2\pi}v\,dt=0$, but the robot moves $8$ m forwards and $8$ m back, so the distance is $16$ m. Distance is always at least the size of the displacement. They are equal only if the particle never changes direction.

A particle has $\dot{x}=t^2-7t+10$ for $0\le t\le6$ and starts at $x=3$ m. Displacement: $\int_0^6\dot{x}\,dt=6$ m, so $x(6)=3+6=9$ m. At rest: $(t-2)(t-5)=0$, so $t=2$ and $t=5$. Pieces: $\frac{26}{3}$ forwards, $\frac{9}{2}$ backwards, $\frac{11}{6}$ forwards. Distance: $\frac{26}{3}+\frac92+\frac{11}{6}=15$ m, confirmed by the GDC with $\int_0^6|t^2-7t+10|\,dt=15$. State units, and give 3 s.f. unless exact.