Kinematics graphs

What you are looking at

Velocity is the rate of change of displacement, $v = \dfrac{ds}{dt}$, and acceleration is the rate of change of velocity, $a = \dfrac{dv}{dt}$. The three graphs share one time axis, so at any instant the gradient of the $s$–$t$ graph is the height of the $v$–$t$ graph, and the gradient of the $v$–$t$ graph is the height of the $a$–$t$ graph. On the track above them, the green arrow shows the velocity and the yellow arrow the acceleration.

The shaded region on the $v$–$t$ graph is the area from $t = 0$ up to the current time. Displacement is the signed area, $s(t) - s(0) = \int_0^t v\,dt$, so area below the axis counts as negative. Distance travelled is $\int_0^t |v|\,dt$, which counts every part as positive. The two differ once the particle turns round, where $v = 0$ and changes sign; with the formulas shown, those instants are marked on the time slider and ringed on the track.

Try this

1. 1With the Cubic motion, $s(t) = t^3 - 6t^2 + 9t$, drag to $t = 5$. The displacement is $20$ m but the distance travelled is $28$ m, because the particle turns back at $t = 1$ and again at $t = 3$.
2. 2Stop at $t = 1.5$ and then at $t = 2.5$. The velocity is $-2.25\text{ m s}^{-1}$ both times, yet the panel says speeding up at the first and slowing down at the second. Compare the signs of $v$ and $a$.
3. 3Switch to Oscillation, $s = 3\sin t$, and press Play. After one full period, at $t = 2\pi$, the displacement is $0$ but the distance travelled is $12$ m.
4. 4Choose Braking and find where the velocity arrow shrinks to nothing, at $t = 2\ln 2.5 \approx 1.83$ s. After that the particle moves backwards and speeds up, because $v$ and $a$ are now both negative.

Mistakes students make

• Treating distance and displacement as the same thing. Distance is $\int |v|\,dt$: when the velocity changes sign, split the integral at the times where $v = 0$, or integrate $|v|$ directly on a calculator.
• Assuming negative acceleration means slowing down. A particle slows down when $v$ and $a$ have opposite signs and speeds up when they have the same sign, whichever sign that is.
• Reading 'at rest' as $a = 0$. At rest means $v = 0$; the acceleration at that instant is usually not zero, which is exactly why the particle turns round (in the Cubic motion, $a = -6$ at $t = 1$).

Where it is taught

• IB Maths AA SL/HL 5.9 · Kinematics
• AP Calculus AB/BC Unit 4 · Straight-line motion: position, velocity and acceleration (4.2)
• AP Calculus AB/BC Unit 8 · Position, velocity and acceleration using integrals (8.2)
• Edexcel A-Level Maths (9MA0) · Mechanics · Kinematics with variable acceleration (calculus)