9.1 Units, Measurements & Kinematics

9.1 Units, Measurements & Kinematics

Quick Answer: SI fluency, dimensional checks, and clean use of constant-acceleration equations are strong foundations for technical physics. This guide does not claim an official PAF physics weight, level, or item format.

Every later topic in this physics review—Newton’s laws, circular motion, fluids, and thermodynamics—rests on measurable quantities and consistent units. The material is useful for general engineering preparation; it is not evidence that PAF publishes a physics paper or any particular count or timing for this induction.

Why Units Come First

An aeronautical engineer cannot confuse mass with weight, or speed with velocity, and still reason safely about aircraft loads. Many tricky exercises rely on unit traps or dimensional mismatches rather than hard algebra.

SI Base Quantities Used in Mechanics

Derived mechanical units you must recognise instantly:

Worked unit check: Convert 72 km/h to m/s.

$72\,\text{km/h} = 72 \times \frac{1000\,\text{m}}{3600\,\text{s}} = 20\,\text{m/s}$

Memorise the shortcut: divide km/h by 3.6 to get m/s (or multiply m/s by 3.6 for km/h).

Significant Figures, Precision & Error Sense

A reported result cannot be more precise than the least precise measurement feeding it. If lengths are given to two significant figures, quoting acceleration to five decimals is incorrect reporting. Prefer answers that match the same order of precision as the data.

Percentage error (concept): if a measured length is $L \pm \Delta L$, then

$\text{Percentage error} = \frac{\Delta L}{L} \times 100\%$

When quantities multiply, relative errors add (to first order)—useful if an MCQ asks which measurement dominates uncertainty.

Dimensional Analysis

Every physical equation must be dimensionally homogeneous: both sides share the same combination of mass (M), length (L), and time (T).

Common dimensions:

Worked example: Is $v = u + at$ dimensionally consistent?

• $[u] = LT^{-1}$
• $[at] = (LT^{-2})(T) = LT^{-1}$

Yes—both terms are velocities. Dimensional analysis cannot prove an equation is true (numerical coefficients can still be wrong), but it can eliminate distractors that mix energy with force or acceleration with velocity.

Another check: kinetic energy $\tfrac{1}{2}mv^{2}$ has dimensions $[M][LT^{-1}]^{2} = [ML^{2}T^{-2}]$, matching work. Potential energy $mgh$ has $[M][LT^{-2}][L] = [ML^{2}T^{-2}]$—same family.

Scalars vs Vectors

Distance is the path length travelled (scalar). Displacement is the straight-line change in position from start to finish (vector). A drone that flies 3 km east then 4 km north travels 7 km (distance) but has displacement magnitude 5 km (by Pythagoras).

Average vs Instantaneous Motion

$\text{Average speed} = \frac{\text{total distance}}{\text{total time}}$

$\vec{v}_{\text{avg}} = \frac{\Delta \vec{s}}{\Delta t}$

Instantaneous velocity is the limit of average velocity as the time interval shrinks toward zero—the slope of the position–time graph at a point. Instantaneous acceleration is the slope of the velocity–time graph.

Graph reading tips that appear on MCQs:

• Position–time: slope = velocity; curved graph ⇒ changing velocity ⇒ acceleration present.
• Velocity–time: slope = acceleration; area under the curve = displacement.
• Acceleration–time: area under the curve = change in velocity.

Equations of Motion (Constant Acceleration)

When acceleration $a$ is constant along a straight line:

$v = u + at$

$s = ut + \tfrac{1}{2}at^{2}$

$v^{2} = u^{2} + 2as$

Also useful: $s = \dfrac{(u+v)}{2}t$ (average velocity × time).

Sign convention: pick a positive direction and stick to it. If upward is positive, free-fall acceleration is $a = -g$.

Worked Example 1 — Runway Acceleration

An aircraft starts from rest and accelerates uniformly at $3.0,\text{m/s}^{2}$ for $12,\text{s}$. Find final speed and distance travelled.

$v = 0 + (3.0)(12) = 36\,\text{m/s}$

$s = 0 + \tfrac{1}{2}(3.0)(12)^{2} = 216\,\text{m}$

Worked Example 2 — Using $v^{2} = u^{2} + 2as$

A vehicle slows from $25,\text{m/s}$ to rest in $50,\text{m}$. Find acceleration.

$0 = (25)^{2} + 2a(50)$

$0 = 625 + 100a \implies a = -6.25\,\text{m/s}^{2}$

The negative sign means deceleration opposite the chosen positive direction of motion.

Free Fall Near Earth’s Surface

Neglecting air resistance, all objects near Earth’s surface accelerate at $g \approx 9.8,\text{m/s}^{2}$ downward (often take $g = 10,\text{m/s}^{2}$ in quick MCQs if the stem allows).

Dropped from rest: $u = 0$, $a = +g$ if downward is positive.

$v = gt, \quad s = \tfrac{1}{2}gt^{2}, \quad v^{2} = 2gs$

Thrown upward with speed $u$: time to apex $t = u/g$; maximum height $H = u^{2}/(2g)$; total flight time for return to launch height $T = 2u/g$.

Relative Velocity (1-D and Simple 2-D)

If A and B move along a line,

$\vec{v}_{AB} = \vec{v}_{A} - \vec{v}_{B}$

Two cars travelling the same direction at $30,\text{m/s}$ and $20,\text{m/s}$ have relative speed $10,\text{m/s}$. Opposite directions add magnitudes for closing speed.

Crosswind / riverboat intuition (preview for later vectors): the ground velocity is the vector sum of velocity relative to the medium and the medium’s velocity. Engineering papers love asking which heading cancels a cross component—resolve perpendicular components.

Practice Strategy for This Section

1. Convert all quantities to SI before plugging into formulas.
2. Check dimensions if two options look numerically close.
3. Draw a quick axis and mark $u$, $a$, and the unknown—most mistakes are sign errors.
4. Build speed gradually with self-timed drills after the method is accurate; do not treat a practice clock as an official PAF limit.

Master units and kinematics thoroughly; Section 9.2 builds Newton’s laws and energy directly on these definitions.