Kinematics — Displacement, Velocity and Acceleration

Spec mapping: OCR H556 Module 3.1 — Motion (kinematic definitions, scalars and vectors, graphical representation of motion). Refer to the official OCR H556 specification document for exact wording.

Kinematics is the branch of mechanics that describes how objects move, without yet asking why. It is the descriptive grammar of motion: given a clock and a way of measuring position, you can reconstruct the entire history of a moving body — where it has been, how fast it was going, how its velocity changed, and where it is heading next. The whole of A-Level mechanics, plus electric-field and gravitational-field motion later in the H556 course, depends on a watertight grasp of the three central kinematic quantities introduced here: displacement, velocity and acceleration, all of which are vectors, and their scalar partners distance, speed and the rate-of-change-of-speed.

This lesson sets up the conceptual scaffolding the rest of Module 3.1 needs. The next lesson (SUVAT) turns the graph-and-definition story into algebra; the lesson after (projectile motion) applies it in two dimensions; the dynamics block then asks the why question by introducing forces.

Key Definition: Displacement is the vector from a chosen origin to the current position of a body. It has magnitude (in metres) and direction. Distance, by contrast, is the scalar path-length actually travelled and so is always non-negative.

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Scalars and Vectors — Why It Matters From the Start

OCR mark schemes draw a hard line between scalars and vectors, and a great deal of A-Level mechanics turns on getting that line right.

• A scalar has magnitude only. Examples: mass, time, temperature, distance, speed, energy, work, charge.
• A vector has magnitude and direction. Examples: displacement, velocity, acceleration, force, momentum, impulse, electric and gravitational field strength.

Exam Tip: When OCR asks "state the difference between distance and displacement", "explain why velocity is a vector" or "give two examples each of a scalar and a vector quantity", the word direction is doing the work. A typical full-mark response is: "Distance is a scalar — magnitude only; displacement is a vector — magnitude and direction relative to a defined origin."

The distinction is not pedantic. Consider an athlete completing one full lap of a 400 m track. Their distance travelled is 400 m; their displacement (start point to finish point) is zero. Their average speed over the lap is non-zero; their average velocity is zero. A student who confuses these two pairs of quantities will lose marks on every motion question in the paper.

A vector can be specified in any of three equivalent ways: as a magnitude with a direction (e.g. "12 m due north"), as a pair of perpendicular components (e.g. $s_x = 8.5$sx​=8.5 m, $s_y = 8.5$sy​=8.5 m), or graphically as an arrow drawn to scale. You will use all three representations across A-Level, and switching fluently between them is a high-value skill.

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Distance, Displacement and the Geometry of Position

Distance — symbol $d$d — is the total path length travelled and is measured in metres (m).

Displacement — symbol $s$s or $\vec{s}$s — is the straight-line vector from origin to current position. In one dimension we encode direction with a sign convention ($+$+ or $-$−). In two dimensions we use either components $(s_x, s_y)$(sx​,sy​) or magnitude-and-bearing form.

Worked Example 1 — Triangle journey

A cyclist rides 3.0 km due east, then 4.0 km due north.

• Distance travelled $= 3.0 + 4.0 = 7.0$=3.0+4.0=7.0 km.
• Displacement magnitude $= \sqrt{3.0^2 + 4.0^2} = 5.0$=3.02+4.02​=5.0 km (Pythagoras).
• Displacement direction $= \tan^{-1}(3.0 / 4.0) = 36.9^{\circ}$=tan−1(3.0/4.0)=36.9∘ east of north (bearing $036.9^{\circ}$036.9∘).

Notice that the cyclist has travelled 7.0 km of road but is only 5.0 km from where they began. Distance and displacement are not interchangeable quantities — they happen to coincide only when the journey is a straight line in a single direction.

Worked Example 2 — Out-and-back

A runner jogs 1.6 km along a straight road, then turns around and jogs 0.9 km back.

• Distance $= 1.6 + 0.9 = 2.5$=1.6+0.9=2.5 km.
• Displacement $= 1.6 - 0.9 = 0.7$=1.6−0.9=0.7 km (in the original direction).

If they had continued back past the start by a further 0.2 km, the displacement would be $1.6 - 1.8 = -0.2$1.6−1.8=−0.2 km — i.e. 0.2 km in the opposite direction, signalled by the negative sign.

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Speed and Velocity

Speed — the rate of change of distance with time — is a scalar with SI unit metres per second (m s$^{-1}$−1).

Velocity — the rate of change of displacement with time — is the vector version, same units.

Formally:

$v_{\text{avg}} = \frac{\Delta s}{\Delta t}, \qquad v_{\text{inst}} = \frac{ds}{dt}$vavg​=ΔtΔs​,vinst​=dtds​

The average velocity over an interval is the displacement divided by the elapsed time. The instantaneous velocity at a moment is the derivative of displacement with respect to time — graphically, the gradient of the tangent to the displacement-time curve at that instant.

A body travelling in a circle at constant 10 m s$^{-1}$−1 has constant speed but changing velocity, because the direction of motion is changing continuously. This is the conceptual seed of circular-motion dynamics in Module 5.4: a changing velocity vector implies an acceleration, which (by Newton II) implies a force — and so a stone whirled on a string must be subjected to a force, the tension, despite its speed being unchanged.

Worked Example 3 — Average velocity vs average speed

A car drives 120 m due east in 8.0 s, then 40 m due west in 4.0 s.

Take east as positive.

• Total displacement $= +120 - 40 = +80$=+120−40=+80 m.
• Total time $= 8.0 + 4.0 = 12$=8.0+4.0=12 s.
• Average velocity $= 80 / 12 = +6.7$=80/12=+6.7 m s$^{-1}$−1 east.

Now the average speed:

• Total distance $= 120 + 40 = 160$=120+40=160 m.
• Average speed $= 160 / 12 = 13$=160/12=13 m s$^{-1}$−1.

Two different numerical answers, because the journey reversed direction. Always read the question carefully — velocity and speed are not synonyms.

Worked Example 4 — Instantaneous from average

A car's displacement is given by $s(t) = 4.0\,t^2 + 1.5\,t$s(t)=4.0t2+1.5t (SI units). Find (a) the average velocity over the first 3.0 s and (b) the instantaneous velocity at $t = 3.0$t=3.0 s.

(a) $s(3.0) = 4.0 \times 9 + 1.5 \times 3 = 40.5$s(3.0)=4.0×9+1.5×3=40.5 m, $s(0) = 0$s(0)=0. So $v_{\text{avg}} = 40.5 / 3.0 = 13.5$vavg​=40.5/3.0=13.5 m s$^{-1}$−1.

(b) Differentiating, $v(t) = 8.0\,t + 1.5$v(t)=8.0t+1.5. At $t = 3.0$t=3.0 s, $v_{\text{inst}} = 24 + 1.5 = 25.5$vinst​=24+1.5=25.5 m s$^{-1}$−1.

The instantaneous value is nearly double the average — because the body is accelerating, the speed at the end of the interval is much higher than the mean speed across it. Recognising this gap between instantaneous and average is a classic A-to-A* discriminator.

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Acceleration

Acceleration is the rate of change of velocity with time. It is a vector, with SI unit metres per second squared (m s$^{-2}$−2):

$a_{\text{avg}} = \frac{\Delta v}{\Delta t}, \qquad a_{\text{inst}} = \frac{dv}{dt} = \frac{d^2 s}{dt^2}$aavg​=ΔtΔv​,ainst​=dtdv​=dt2d2s​

A positive acceleration means the velocity is becoming more positive (in whichever direction you chose as positive). A negative acceleration — often loosely called deceleration — means the velocity vector is becoming more negative. Crucially, "negative acceleration" does not automatically mean "slowing down":

• If $v > 0$v>0 and $a < 0$a<0: speeding up in the negative direction is impossible while $v > 0$v>0, so the body is slowing down.
• If $v < 0$v<0 and $a < 0$a<0: the body is speeding up in the negative direction — its speed is increasing.
• If $v < 0$v<0 and $a > 0$a>0: the body is slowing down (moving toward zero from below).

Common Exam Mistake: Reading off "$a$a is negative" and writing "the object is decelerating" without checking the sign of $v$v. Slowing down requires that velocity and acceleration vectors point in opposite directions. OCR mark schemes routinely penalise the conflation of "negative" with "slowing".

Worked Example 5 — Sprinter

A sprinter accelerates from rest to 9.0 m s$^{-1}$−1 in 1.8 s.

$a = \frac{\Delta v}{\Delta t} = \frac{9.0 - 0}{1.8} = 5.0 \text{ m s}^{-2}$a=ΔtΔv​=1.89.0−0​=5.0 m s−2

For comparison, the gravitational acceleration at the Earth's surface is $g = 9.81$g=9.81 m s$^{-2}$−2, so this sprinter generates an acceleration of about $g/2$g/2 at the start of the race. Elite Olympic sprinters can briefly exceed $1\,g$1g in the first stride out of the blocks.

Worked Example 6 — Aircraft on the runway

A passenger jet accelerates uniformly from rest down a 1800 m runway and lifts off at 75 m s$^{-1}$−1.

• $u = 0$u=0, $v = 75$v=75 m s$^{-1}$−1, $s = 1800$s=1800 m, $a = ?$a=?.
• Using $v^2 = u^2 + 2as$v2=u2+2as: $a = v^2 / (2s) = 5625 / 3600 = 1.56$a=v2/(2s)=5625/3600=1.56 m s$^{-2}$−2.

Approximately $g/6$g/6 — comfortable for passengers. The take-off time is $t = v/a = 75/1.56 = 48$t=v/a=75/1.56=48 s, agreeing with the felt duration of a long take-off roll.

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Displacement–Time Graphs

A displacement–time ($s$s–$t$t) graph plots position against time. Its features encode kinematics:

• Gradient $= ds/dt = v$=ds/dt=v (velocity).
• Straight line $\Rightarrow$⇒ constant velocity.
• Curve $\Rightarrow$⇒ velocity changing (i.e. acceleration).
• Horizontal line $\Rightarrow$⇒ velocity zero (at rest).
• Negative gradient $\Rightarrow$⇒ moving back toward the origin / negative direction.

flowchart LR
  A[s-t Graph] --> B[Gradient at a point]
  B --> C[= Instantaneous velocity]
  A --> D[Shape of curve]
  D --> E[Straight line: constant v]
  D --> F[Curve concave up: increasing v]
  D --> G[Curve concave down: decreasing v]
  D --> H[Horizontal: at rest]

Reading instantaneous velocity from a curve

If the graph curves, draw a tangent at the time of interest and compute its gradient using two well-separated points on the tangent. Why two well-separated points? Because the gradient is calculated as $\Delta y / \Delta x$Δy/Δx, and the uncertainty in each reading is fixed (typically $\pm$± half a grid square). The wider the $\Delta x$Δx, the smaller the percentage uncertainty in the gradient.

Exam Tip: Always extend the tangent at least one-third the length of the graph. Mark the two points you used, label them clearly, and show your subtraction explicitly. Marks are given for the method of tangent-drawing as well as the numerical answer.

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Velocity–Time Graphs

The velocity–time ($v$v–$t$t) graph is the workhorse of A-Level kinematics, because it encodes two physical quantities in its shape:

• Gradient $= dv/dt = a$=dv/dt=a (acceleration).
• Area under the graph $= \int v\,dt = s$=∫vdt=s (displacement — signed).

Feature of $v$v–$t$t graph	Physical interpretation
Horizontal line	Constant velocity, $a = 0$a=0
Straight line, positive slope	Uniform acceleration
Straight line, negative slope	Uniform deceleration (if $v > 0$v>0)
Curve (concave up)	Acceleration increasing
Curve (concave down)	Acceleration decreasing
Area above time axis	Positive displacement (forward motion)
Area below time axis	Negative displacement (motion reversed)

Worked Example 7 — Area under a v–t graph

A train accelerates uniformly from rest to 30 m s$^{-1}$−1 in 20 s, runs at 30 m s$^{-1}$−1 for a further 40 s, then decelerates uniformly to rest in 10 s. Total distance?

Split the area into three regions:

1. Triangle 1 (acceleration phase): $\frac{1}{2} \times 20 \times 30 = 300$21​×20×30=300 m.
2. Rectangle (cruising phase): $40 \times 30 = 1200$40×30=1200 m.
3. Triangle 2 (braking phase): $\frac{1}{2} \times 10 \times 30 = 150$21​×10×30=150 m.

Total displacement $= 300 + 1200 + 150 = 1650$=300+1200+150=1650 m. Average velocity $= 1650 / 70 = 23.6$=1650/70=23.6 m s$^{-1}$−1, neatly between zero and the cruise speed.

Exam Tip: OCR's v–t graph questions almost invariably ask you to find a distance (an area) or an acceleration (a gradient). Annotate the graph with the geometric shapes you are using and write out the area formulae. Marks accrue for the explicit method.

Worked Example 8 — Non-uniform motion: the curved v–t graph

A car's velocity over the first 6 seconds of a journey is recorded as: $v(0) = 0$v(0)=0, $v(2) = 6$v(2)=6, $v(4) = 14$v(4)=14, $v(6) = 18$v(6)=18 m s$^{-1}$−1. Estimate (a) the acceleration at $t = 4$t=4 s and (b) the distance travelled in the first 6 s.

(a) Use a symmetric two-point estimate: $a(4) \approx [v(6) - v(2)] / 4 = (18 - 6) / 4 = 3.0$a(4)≈[v(6)−v(2)]/4=(18−6)/4=3.0 m s$^{-2}$−2.

(b) Use the trapezium rule with strips of width 2 s:

$s \approx \frac{2}{2}\big[v(0) + 2v(2) + 2v(4) + v(6)\big] = 1 \times (0 + 12 + 28 + 18) = 58 \text{ m}$s≈22​[v(0)+2v(2)+2v(4)+v(6)]=1×(0+12+28+18)=58 m

This is exactly the kind of question OCR poses to test whether you understand that area = displacement even when the motion is non-uniform.

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Uniform vs Non-Uniform Acceleration

A central distinction in A-Level mechanics:

• Uniform (constant) acceleration — the realm of SUVAT. The $v$v–$t$t graph is a straight line.
• Non-uniform acceleration — requires graphical or calculus methods; SUVAT is invalid.

Common scenarios of uniform acceleration:

• Free fall near Earth's surface, ignoring air resistance: $a = g \approx 9.81$a=g≈9.81 m s$^{-2}$−2.
• An object on a smooth incline under gravity.
• An object pulled by a constant resultant force on a smooth surface.

Common scenarios of non-uniform acceleration:

• A falling body with significant drag (skydiver), where $F_{\text{drag}} \propto v$Fdrag​∝v or $v^2$v2 and so the resultant force changes with speed.
• A vehicle accelerating at full throttle where engine output, drag and rolling resistance all vary with speed.
• A simple pendulum away from small-angle approximation.

The skydiver case is examined directly in Module 3.2.2: prior to terminal velocity, $a$a falls smoothly from $g$g toward zero, and the $v$v–$t$t graph is a curve asymptoting to a horizontal line.

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Worked Example 9 — Sketching a ball thrown upward

A ball is thrown vertically upward at 15 m s$^{-1}$−1 from a balcony 20 m above the ground, with air resistance neglected. Take upward positive.

• Ball decelerates at 9.81 m s$^{-2}$−2 until $v = 0$v=0 at the apex: time to apex $= 15 / 9.81 = 1.53$=15/9.81=1.53 s.
• Ball then accelerates downward at 9.81 m s$^{-2}$−2.
• It passes the balcony moving downward at $-15$−15 m s$^{-1}$−1 at $t = 3.06$t=3.06 s (symmetry).
• It strikes the ground at some later time, having travelled an additional 20 m below the balcony.

On a $v$v–$t$t graph: a straight line of slope $-9.81$−9.81 m s$^{-2}$−2, starting at $+15$+15, crossing zero at $t = 1.53$t=1.53 s, and continuing into negative territory until impact. On an $s$s–$t$t graph (measuring $s$s from the balcony, upward positive): a downward-opening parabola, peaking at the apex height of $15^2 / (2 \times 9.81) = 11.5$152/(2×9.81)=11.5 m above the balcony, then crossing back through $s = 0$s=0 at $t = 3.06$t=3.06 s and reaching $s = -20$s=−20 m at the moment of ground impact.

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Synoptic Links

• SUVAT equations (Module 3.1, next lesson) — the algebraic encoding of constant-acceleration kinematics builds directly on the definitions of $u$u, $v$v, $a$a, $s$s, $t$t established here. Every SUVAT problem is a graphical area or gradient relationship in disguise.
• Newton's second law (Module 3.2) — the cause of acceleration is force. The kinematic acceleration $a$a you compute here becomes the right-hand side of $F = ma$F=ma in the next block. Examiners pair a $v$v–$t$t gradient with a $F = ma$F=ma substitution as a standard Paper 1 synoptic question.
• Simple harmonic motion (Module 5.3) and circular motion (Module 5.4) — both rest on the recognition that changing direction of velocity is itself an acceleration. The vector-versus-scalar discipline acquired in this lesson is what makes SHM's $a = -\omega^2 x$a=−ω2x and circular motion's $a = v^2 / r$a=v2/r intelligible.
• Practical work — PAG 1 (measurements) and PAG 6 (Newtonian mechanics, free-fall determination of $g$g) both depend on graphical kinematics. A typical PAG 6 write-up computes $g$g as $2 \times$2× the gradient of an $s$s vs $t^2$t2 plot, which is precisely the manipulation of $s = ut + \frac{1}{2}at^2$s=ut+21​at2 a graphical kinematicist must master.

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Specimen question modelled on the OCR H556 paper format

Question (6 marks): A car is travelling along a straight horizontal road. The driver, on seeing a hazard ahead, takes 0.70 s to react before applying the brakes; while reacting, the car continues at a steady 26 m s$^{-1}$−1. The brakes then produce a uniform deceleration that brings the car to rest in a further 3.2 s.

(a) Sketch a labelled velocity–time graph for the motion from the moment the hazard is seen until the car stops. [2]

(b) Calculate the total displacement of the car from the moment the hazard is first seen until it stops. [3]

(c) State the meaning of the gradient of the line segment after the brakes are applied, and give its numerical value. [1]

AO breakdown

Mark	AO	Awarded for
1	AO2	$v$v–$t$t graph showing horizontal segment at $26$26 m s$^{-1}$−1 for $0.70$0.70 s
2	AO2	Straight-line negative-slope segment from $(0.70, 26)$(0.70,26) to $(3.90, 0)$(3.90,0)
3	AO2	Distance during reaction $= 26 \times 0.70 = 18.2$=26×0.70=18.2 m identified as rectangle area
4	AO2	Distance during braking $= \frac{1}{2} \times 3.2 \times 26 = 41.6$=21​×3.2×26=41.6 m identified as triangle area
5	AO2	Total $= 18.2 + 41.6 = 59.8$=18.2+41.6=59.8 m $\approx 60$≈60 m
6	AO1	Gradient = acceleration; numerical value $-26 / 3.2 = -8.1$−26/3.2=−8.1 m s$^{-2}$−2

AO split: AO1 $= 1$=1, AO2 $= 5$=5, AO3 $= 0$=0.

Mid-band response (3/6):

The car goes at 26 m s$^{-1}$−1 for the reaction time. So in 0.70 s it goes $26 \times 0.70 = 18.2$26×0.70=18.2 m. After that the brakes come on and the car slows down to zero. The braking distance is $26 \times 3.2 = 83.2$26×3.2=83.2 m. Total is $18.2 + 83.2 = 101.4$18.2+83.2=101.4 m. The gradient of the braking line is the acceleration, $-26 / 3.2 = -8.1$−26/3.2=−8.1 m s$^{-2}$−2. The graph has a horizontal line at 26 m s$^{-1}$−1 then drops down to zero.

Examiner commentary: To lift this to top-band the candidate would need to halve the braking distance — for uniform deceleration the area beneath a $v$v–$t$t line is a triangle, not a rectangle, so the correct area is $\frac{1}{2} \times 3.2 \times 26 = 41.6$21​×3.2×26=41.6 m. M1 awarded for the reaction-phase distance, M1 for identifying the gradient as acceleration, M1 for the numerical value of acceleration. The braking-distance error costs M1 (correct area formula) and M1 (correct numerical answer for total displacement). The graph description is partially worth M1 but lacks explicit axis labels and the kink at the moment the brakes are applied. A classic Grade C error — using $v \times t$v×t for a decelerating phase when only the average velocity $v/2$v/2 should multiply $t$t.

Top-band response (6/6):

(a) The graph has time on the $x$x-axis (0 to 3.90 s) and velocity on the $y$y-axis (0 to 26 m s$^{-1}$−1). From $t = 0$t=0 to $t = 0.70$t=0.70 s the line is horizontal at $v = 26$v=26 m s$^{-1}$−1 (driver reacting, constant velocity). At $t = 0.70$t=0.70 s the line kinks downward and falls as a straight line of constant negative slope, reaching $v = 0$v=0 at $t = 0.70 + 3.20 = 3.90$t=0.70+3.20=3.90 s.

(b) Total displacement is the total area under the $v$v–$t$t graph.

• Reaction phase (rectangle): $s_1 = 26 \times 0.70 = 18.2$s1​=26×0.70=18.2 m.
• Braking phase (triangle): $s_2 = \frac{1}{2} \times 3.2 \times 26 = 41.6$s2​=21​×3.2×26=41.6 m.

Total $s = s_1 + s_2 = 18.2 + 41.6 = 59.8$s=s1​+s2​=18.2+41.6=59.8 m, or about $60$60 m (2 s.f.).

(c) The gradient of the braking line is $dv/dt$dv/dt, which equals the acceleration during braking. Numerically:

$a = \frac{\Delta v}{\Delta t} = \frac{0 - 26}{3.2} = -8.1 \text{ m s}^{-2}$a=ΔtΔv​=3.20−26​=−8.1 m s−2

The negative sign indicates the acceleration vector points opposite to the velocity — i.e. the car is decelerating.

Examiner commentary: Full marks. The discriminator moves are: (i) recognising that the braking phase is a triangle, not a rectangle — many candidates lose two marks here; (ii) the explicit kink at $t = 0.70$t=0.70 s in the graph sketch; (iii) the careful interpretation of the negative gradient as "acceleration opposite to velocity" rather than the loose "deceleration". The candidate also correctly used 2 s.f. for the final answer reflecting the data precision.

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Common errors and mark-loss patterns

Pedagogical observations — no fabricated candidate percentages.

• Many candidates lose marks by using distance where displacement is required, or vice versa. Always check whether the question demands a vector (with direction) or a scalar magnitude.
• A typical pitfall is forgetting to state direction when giving a vector answer. Writing "velocity $= 12$=12 m s$^{-1}$−1" is incomplete; "velocity $= 12$=12 m s$^{-1}$−1 due east" earns the mark.
• Candidates routinely confuse gradient and area on graphs. On an $s$s–$t$t graph, area is meaningless; gradient gives velocity. On a $v$v–$t$t graph, gradient gives acceleration and area gives displacement. A useful mnemonic: gradient is steepness, area is amount.
• A widespread error is to assume "negative acceleration" implies "slowing down". It does not — see the sign-analysis above. Slowing requires $v$v and $a$a vectors pointing in opposite directions.
• Reading a tangent through just one point. Always extend the tangent right across the graph and take two well-separated readings to compute the gradient — this both reduces percentage uncertainty and demonstrates explicit method to the examiner.
• Missing or wrong units. Every numerical answer at A-Level must carry correct SI units. m s$^{-1}$−1, m s$^{-2}$−2 are non-negotiable; km h$^{-1}$−1 must be converted ($\div 3.6$÷3.6) before substitution.
• Mixing instantaneous and average values. Average velocity is total displacement over total time; it equals neither the maximum nor the final instantaneous velocity. Sketch the $v$v–$t$t graph if in doubt.
• Using $v = u + at$v=u+at on the assumption $a$a is constant when the question implies it is not — for instance ignoring drag on a "ball falling through 50 m" question. Always check whether SUVAT is justifiable.

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Going further

The kinematic ideas you have just met are, mathematically, the first steps of differential calculus applied to motion. At first-year undergraduate level, the displacement-velocity-acceleration triple becomes a single object — the position vector $\vec{r}(t)$r(t) — whose first and second time-derivatives give velocity and acceleration: $\vec{v} = d\vec{r}/dt$v=dr/dt, $\vec{a} = d^2\vec{r}/dt^2$a=d2r/dt2. The whole of Newtonian mechanics, when developed via Lagrange or Hamilton's formulation in second-year theoretical-mechanics courses, builds on this position-time framework. Lagrangian mechanics in particular re-derives every constant-acceleration formula not from forces but from a single scalar function — the Lagrangian $L = T - V$L=T−V — and an integral principle (Hamilton's principle of stationary action). It is mathematically beautiful and physically equivalent, and a standard prompt at Oxbridge physics interview is "how would you find the trajectory of a ball thrown vertically without ever mentioning forces?" — the answer is "minimise the action functional".

A second beautiful extension: calculus of variations generalises the question "what is the shortest path between two points?" (answer: a straight line) to "what is the path of fastest descent on a curved surface?" (answer: the brachistochrone, a section of a cycloid — a result that astonished Bernoulli and which combines the kinematic ideas you have here with the energy ideas to come). Galileo had partial answers in 1638; Johann Bernoulli posed the problem in 1696 and Newton produced an anonymous solution overnight, a story sometimes told as "the lion recognised by his claw".

A third frontier — relativistic kinematics — replaces the Galilean velocity-addition rule $u + v$u+v (which you use implicitly every time you add velocities here) with the Einstein composition law $\frac{u + v}{1 + uv/c^2}$1+uv/c2u+v​. At everyday speeds the relativistic correction is utterly negligible ($v / c \sim 10^{-7}$v/c∼10−7 for a jet aircraft), but it becomes dominant for particle physics. The first-year Special Relativity course in any UK physics department starts here.

Recommended reading: Physics for Scientists and Engineers (Tipler & Mosca) chapter 2; Mechanics (Kibble & Berkshire) chapter 1; for the Lagrangian re-derivation, Classical Mechanics (Goldstein) chapter 1. Interview-style prompt: "Two cars approach each other along a straight road, each at 30 m s$^{-1}$−1. From the reference frame of car A, what is the velocity of car B? Now repeat the calculation when each car's speed is half the speed of light." — the Newtonian answer is $60$60 m s$^{-1}$−1; the relativistic answer is $0.8c$0.8c, not $c$c.

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A-Level misconceptions

The subtle errors that distinguish A from A*:

• Distance $\neq$= displacement. A 400 m runner has zero displacement at the end of a lap; their displacement is zero, but their distance covered is 400 m.
• Instantaneous $\neq$= average velocity. Average velocity over an interval need not equal the velocity at any single instant within it; only when the motion is uniform does it agree with the endpoint values.
• "Constant speed" $\neq$= "constant velocity" — a body in uniform circular motion has constant speed but continuously changing velocity, hence non-zero acceleration.
• "Negative acceleration" does not equal "slowing down". It means the acceleration vector points in the negative chosen direction; whether speed increases or decreases depends on the sign of velocity.
• Sign conventions are arbitrary but must be consistent. Choosing upward as positive and then writing $g = +9.81$g=+9.81 m s$^{-2}$−2 guarantees catastrophic algebraic errors. State your axes; stick to them.
• Tangent gradients require two well-separated points to be quantitatively trustworthy; many candidates draw a tangent then use the original curve to compute the gradient, defeating the purpose.
• Mixing kinematic and dynamic language. "The force was $9.81$9.81" or "the acceleration was $9.81$9.81 N" — every year examiners flag the unit-confusion. Acceleration carries m s$^{-2}$−2; force carries N.

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Summary

• Scalars have magnitude only (distance, speed, time, mass); vectors also have direction (displacement, velocity, acceleration, force).
• Displacement is the vector from origin to current position; distance is path length. They coincide only for unidirectional straight-line motion.
• Velocity is the rate of change of displacement (vector); speed is the rate of change of distance (scalar). $v_{\text{inst}} = ds/dt$vinst​=ds/dt.
• Acceleration is the rate of change of velocity (vector). $a = dv/dt = d^2 s/dt^2$a=dv/dt=d2s/dt2. Negative $a$a means slowing only when $v > 0$v>0.
• $s$s–$t$t graph: gradient = velocity. $v$v–$t$t graph: gradient = acceleration, area = displacement.
• Uniform acceleration unlocks SUVAT (next lesson); non-uniform requires graphical or calculus methods.

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Reference: OCR A-Level Physics A (H556) specification 3.1 — Motion (refer to the official OCR H556 specification document for exact wording).