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Motion in One Dimension (Kinematics)

Source lecture(s): SC133 Lec 2

Intuition

Kinematics is the description of motion — where something is, how fast it moves, how that speed changes — without yet asking why (that's Newton's laws). Three quantities do all the work: position, its rate of change velocity, and its rate of change acceleration. Each is the slope of the graph of the one before.

The three quantities

Displacement \(\Delta x = x_f - x_i\) — the change in position (a vector; can be negative). Distinct from distance travelled, the path length, which never decreases.

Velocity — rate of change of position:

\[v_\text{avg} = \frac{\Delta x}{\Delta t}, \qquad v = \frac{dx}{dt}\]

Average velocity is the slope of the secant on an \(x\)–\(t\) graph; instantaneous velocity is the slope of the tangent.

Acceleration — rate of change of velocity:

\[a_\text{avg} = \frac{\Delta v}{\Delta t}, \qquad a = \frac{dv}{dt} = \frac{d^2x}{dt^2}\]

The kinematic equations (constant acceleration)

When \(a\) is constant, integrating \(a = dv/dt\) and \(v = dx/dt\) gives the four workhorse equations (full derivation on the equation page):

\[v = v_0 + at \qquad x = x_0 + v_0 t + \tfrac12 a t^2\]
\[v^2 = v_0^2 + 2a(x - x_0) \qquad x = x_0 + \tfrac12(v_0 + v)t\]

The most useful trick: pick the equation that omits the variable you neither know nor want.

Reading motion graphs

Graph Slope gives Area under gives
\(x\) vs \(t\) velocity —
\(v\) vs \(t\) acceleration displacement
\(a\) vs \(t\) jerk change in velocity

Fluency at moving between these three graphs is worth more on the exam than memorizing the four equations.

Worked example: free fall

A ball is thrown straight up at \(v_0 = 20\,\text{m/s}\) (\(g = 9.8\,\text{m/s}^2\) downward). How high does it go, and when does it return?

At the top \(v = 0\): from \(v^2 = v_0^2 - 2gh\), \(h = v_0^2/2g = 400/19.6 \approx 20.4\,\text{m}\). Time up: \(t = v_0/g \approx 2.04\,\text{s}\); total flight \(\approx 4.08\,\text{s}\) (symmetry: up-time equals down-time in free fall).

Common mistakes

  • Confusing displacement with distance (and velocity with speed). Displacement and velocity are vectors; a round trip has zero displacement but nonzero distance.
  • Using the constant-\(a\) equations when \(a\) isn't constant (e.g. with drag). Then you must integrate the actual \(a(t)\).
  • Sign errors. Choose a positive direction once and stick to it — \(g\) is \(-9.8\) if up is positive.

Knowledge graph position

Prerequisites: Measurement & units. Leads to: Vectors, projectile motion, circular motion, Newton's laws.

Quiz

Q1 (computational). A car brakes from \(30\,\text{m/s}\) to rest over \(75\,\text{m}\). What is its (constant) acceleration?

Answer

\(v^2 = v_0^2 + 2a\Delta x \Rightarrow 0 = 900 + 2a(75) \Rightarrow a = -6\,\text{m/s}^2\).

Q2 (conceptual). On a \(v\)–\(t\) graph, what does the area between the curve and the time axis represent?

Answer

Displacement: \(\Delta x = \int v\,dt\). Area above the axis is positive displacement, area below is negative.

Q3 (conceptual). Two balls are dropped one second apart. Does the gap between them stay constant, grow, or shrink?

Answer

Grows. Both accelerate identically, but the first always leads in velocity by \(g\tau\), so the separation increases linearly: \(\Delta = g\tau t\) with \(\tau=1\) s.

Q4 (multiple choice). At the very top of its flight, an object thrown upward has: (a) zero velocity and zero acceleration (b) zero velocity, acceleration \(g\) downward (c) maximum velocity

Answer

(b). Velocity passes through zero, but gravity never switches off — \(a=g\) downward the entire time.