Kinematics (Displacement, Velocity, Acceleration)

A particle moves on a straight line; its displacement $s(t)$, velocity $v(t)$ and acceleration $a(t)$ form a ladder connected by calculus:

$v = \frac{ds}{dt}$, $a = \frac{dv}{dt}$, differentiate to go down the ladder integrate to come back up, with constants fixed by initial conditions

State the relationship you’re using ("$v = \frac{ds}{dt}$") before computing; the stated ladder step is a recurring method mark.

The translation dictionary

Kinematics questions are written in motion-words; each translates to an equation, and each written translation is typically a mark:

The words	The equation
”at rest” / “comes to rest"	$v = 0$
"at O” / “at the origin” / “returns to O"	$s = 0$
"initially"	$t = 0$
"maximum velocity”	$\frac{dv}{dt} = 0$ (i.e. $a = 0$)
“constant velocity"	$a = 0$
"changes direction”	$v = 0$ and $v$ changes sign

$v = 3t^2 - 12t + 9$. Find when the particle is at rest, and its acceleration then. $v = 0$: $3(t - 1)(t - 3) = 0 \to t = 1, 3$ $a = \frac{dv}{dt} = 6t - 12 \to a(1) = -6$, $a(3) = +6$ (units: $\text{m/s}^2$, state them)

Integrating back: the constant is the question

Given $a = 6t - 4$ with $v = 2$ when $t = 0$: $v = \int a \,dt = 3t^2 - 4t + c$; initial condition $\to c = 2$. The +c discipline isn’t optional here, the constant carries the physics, and “initially at rest” exists precisely to fix it. Same again from $v$ to $s$.

Distance vs displacement, the discriminating mark

Displacement is net position change: $s(t_2) - s(t_1)$. Total distance counts backtracking: check whether $v = 0$ inside the interval; if the particle turned around, integrate (or evaluate $s$) piecewise and add magnitudes, exactly the below-axis area logic applied to the $v$–$t$ graph. “Find the total distance travelled in the first 4 seconds” with a turn at $t = 3$ means $|s(3) - s(0)| + |s(4) - s(3)|$. Answering with $s(4) - s(0)$ is the classic final-part error.

Common mistakes

• Differentiating when the ladder goes the other way (and vice versa)
• Constants of integration dropped, initial conditions unused
• Distance answered as displacement across a direction change
• Translation lines ($v = 0$ etc.) computed but never written
• Units omitted

Full topic context: Calculus notes, contrast with constant-velocity vector kinematics.