Position–Time Graphs

The position–time graph is working infrastructure. Railway control rooms still schedule single-track lines on Marey-style train graphs — time across, distance along the route up the page, every train one slanted line. Where two lines would cross on a single track, dispatchers see a collision before it exists and reroute one train to a siding: this module's meeting-point reading, deployed a thousand times a day on metros and mountain railways. Your own devices draw the same picture. A fitness app's distance chart, a delivery tracker's progress line, an aircraft's flight-data recorder logging position each second — all are x–t graphs, read exactly as you have learned: flat = stopped, steep = fast, bend = braking or surging. Investigators reconstructing an incident from a recorder are doing, professionally, the board exam's "describe the motion shown".

Take one small journey and tell it three ways. In words: "Riya walks briskly from her front gate to a shop 80 m down the street in 40 s, chats at the counter for 30 s, then walks back at the same brisk pace, reaching her gate 40 s later." As data: position 0 m at time 0, position 80 m at time 40 s, still 80 m at 70 s, back to 0 m at 110 s. As a graph: a rising line, a flat shelf, a falling line. All three carry identical information — but only the graph lets you see the whole journey at once and read off any instant of it.

Notice what the graph is not. Riya's street is straight and level, yet her graph rises, flattens and falls. The vertical axis is position along the street, not altitude; the horizontal axis is time, not distance east. The falling section does not mean she walked downhill — it means her position decreased with time: she headed back. This single re-wiring of instinct — the graph is a timeline, not a photograph — is the entire foundation of graph questions, and the first thing to consciously check whenever a graph appears in an exam. Once the axes are read correctly, three visual readings unlock everything. Height answers "where?" — just trace across to the position axis. Slope answers "how fast, and which way?" — steep means fast, uphill-to-the-right means moving in the positive direction, downhill-to-the-right means moving in the negative direction. Bending answers "is the velocity changing?" — a straightening or curving line means the slope itself is drifting, which is acceleration. Every position–time question ever set is one of these three readings, or two of them chained together.

The claim "slope = velocity" is not a mnemonic; it is the definition of average velocity wearing geometric clothes. Pick any two instants $t_1$ and $t_2$ on the graph and read the positions $x_1$ and $x_2$. The average velocity over that interval divides the displacement by the elapsed time:

where: $\Delta x$ — the rise between the two points (positive if the graph climbed, negative if it fell); $\Delta t$ — the run between them, always positive because time only moves forward. But rise over run is precisely what "slope of the line joining two points" means. So the secant line through two points of an x–t graph has slope equal to the average velocity between those instants — with the sign carrying direction. In Riya's return leg the rise is $0 - 80 = -80$ m over a run of $40$ s: slope $-2$ m/s, the minus announcing "toward the gate". For a straight segment the slope is the same everywhere, so the velocity is constant: straight line = uniform motion. For a curved graph the slope is different at every point, and the velocity at one instant is the slope of the tangent — the straight line that just grazes the curve there. This is the secant-shrinking idea from the speed–velocity module, reused as pure geometry: lay a ruler along the curve at the instant you care about, and the ruler's slope is the instantaneous velocity. No formula for $x(t)$ is needed; graph questions are designed to be answered by ruler and grid squares.

A straight x–t line, however steep, is velocity locked constant. The moment the graph bends, the slope is changing — and a changing velocity is an acceleration. You can extract its sign without a single number: check whether the slope is increasing or decreasing as time passes. Read bending in two steps, always in this order. First the slope now (which way, how steep), then the trend (tangent getting steeper or flatter?). Each tangent steeper than the last means increasing velocity; flattening means decreasing. Whether that is speeding up or slowing down is a comparison of directions: velocity growing along the direction of motion is speeding up, velocity being eaten toward zero is braking. (The full sign test lives in the acceleration module; here the graph hands you both signs by eye.) One special point deserves star billing: the turning point. If an x–t curve rises, rounds off, and comes back down — a hill-shaped graph — then at the crest the tangent is horizontal: velocity zero at that instant. The object is at its farthest position, momentarily at rest, about to reverse. A ball thrown upward, a car nosing forward then reversing out of a parking spot, a runner touching the turnaround cone: all produce this crest. Exams adore it because the intuition "farthest out = moving fastest" is so tempting and so wrong — at the extreme position the object is not moving at all.

Draw two objects' graphs on the same axes and the grid becomes a chase scene. The rule that unlocks it: where the curves intersect, the objects share a position at the same instant — they meet. A crossing says nothing about their speeds; at the crossing the two lines usually have different slopes, meaning the objects pass each other at different velocities. Equal slopes — parallel tangents — mean equal velocities, and that can happen far from any crossing. Sorting "same position" from "same velocity" is the most reliably asked two-graph question on every board. Chase stories then read off the picture. A line starting higher (a head start) approached by a steeper line (a faster pursuer): the vertical gap between the curves at fixed t is the separation at that instant, shrinking until the crossing — the catch — after which the steeper line leads. Where the gap is widest, the tangents are parallel: momentarily matched velocities. These readings turn multi-step word problems into a glance.

The last row is a favourite "which graph is impossible?" MCQ. A single object's graph must also never curl back leftward — that would give it two positions at one instant. Every legal x–t graph passes the vertical-line test: one position per instant.

Board level — reading a three-part journey (CBSE/ICSE/IGCSE pattern)

A cyclist's position–time graph shows: a straight rise from $x = 0$ m at $t = 0$ to $x = 25$ m at $t = 5$ s; a horizontal shelf at 25 m until $t = 8$ s; then a straight fall from 25 m to $x = -5$ m at $t = 14$ s. Find the velocity in each phase, the total displacement, the total distance, and the average speed for the whole trip.

Step 1. Phase 1 slope: $v_1 = \dfrac{25 - 0}{5 - 0} = \boxed{+5\ \text{m/s}}$ — brisk, positive direction.

Step 2. Phase 2 slope: horizontal, so $v_2 = \boxed{0}$ — at rest for 3 s. The graph still "moves" to the right; only time passes.

Step 3. Phase 3 slope: $v_3 = \dfrac{-5 - 25}{14 - 8} = \dfrac{-30}{6} = \boxed{-5\ \text{m/s}}$ — same speed as phase 1, opposite direction, and the cyclist rides past the starting point to $-5$ m.

Step 4. Displacement is final minus initial position: $\Delta x = -5 - 0 = \boxed{-5\ \text{m}}$. Distance adds the legs as lengths: $25 + 0 + 30 = \boxed{55\ \text{m}}$.

Step 5. Average speed $= 55/14 = \boxed{3.9\ \text{m/s}}$ (and average velocity $= -5/14 \approx -0.36$ m/s — direction stated, as always). Note the marks live in the signs: the third slope must come out negative from the graph itself.

Competitive level — the catch-up read off one grid (JEE-Main/NEET pattern)

On one set of axes: line A starts at the origin with slope 8 m/s; line B starts at 48 m with slope 2 m/s. Find when and where A catches B, and the separation between them at $t = 5$ s.

Step 1. Write each line from intercept + slope: $x_A = 8t$, $\quad x_B = 48 + 2t$.

Step 2. Meeting = crossing: $8t = 48 + 2t \Rightarrow 6t = 48 \Rightarrow t = \boxed{8\ \text{s}}$, at $x = 8 \times 8 = \boxed{64\ \text{m}}$.

Step 3. Separation at 5 s is the vertical gap: $x_B - x_A = 48 - 6 \times 5 = \boxed{18\ \text{m}}$, B still ahead. Timing note: 45 seconds. The entire question is "gap closes at the slope difference, 6 m/s". $48/6 = 8$ s, done — practise until you never actually solve the equation.

Competitive level — the parabolic graph and its crest (JEE-Main pattern)

A particle's position–time graph is the parabola $x = 12t - t^2$ (metres, seconds). Read from the graph's geometry: when does the particle reverse, what is its farthest position, and what is its velocity at $t = 2$ s?

Step 1. The parabola crosses $x = 0$ at $t = 0$ and $t = 12$ s (factor: $x = t(12 - t)$). By symmetry its crest sits midway: $t = \boxed{6\ \text{s}}$ — horizontal tangent, $v = 0$, the reversal instant.

Step 2. Farthest position = height of the crest: $x(6) = 12 \times 6 - 36 = \boxed{36\ \text{m}}$.

Step 3. Velocity at $t = 2$ s = tangent slope there. For a parabola, the chord joining equal time-offsets on either side is parallel to the tangent at the middle — so use the secant from $t = 1$ to $t = 3$: $x(1) = 11$ m, $x(3) = 27$ m, slope $= \dfrac{27 - 11}{3 - 1} = \boxed{8\ \text{m/s}}$. (Equivalently: for uniformly accelerated motion, the average velocity over an interval equals the instantaneous velocity at its midpoint.)

Step 4. Sanity checks: at $t = 2$ s the particle is left of the crest, so it should still be moving out — positive velocity ✓; and the slope should be shallower than at $t = 0$ (where the secant $t=-1$ to $1$ gives 12 m/s) since the crest is approaching ✓. Timing note: 60–75 seconds. Root symmetry finds the crest with zero calculus; the midpoint-chord trick reads any tangent of a parabola. Both are pure graph geometry — no differentiation needed, though $v = 12 - 2t$ confirms every number if you know it.

Reading the graph as a picture of the road. A rising line is not uphill and a crest is not a summit; the axes are position and time. Say the axes aloud before reading anything — this trap decides more graph marks than any other.

"They cross, so they have the same speed." Crossing means same position at the same instant (a meeting). Same speed means parallel tangents. The two facts live at different points of the grid and exams swap them deliberately.

Comparing who is faster by who is higher. Height is position, not speed. The faster object at an instant is the one with the steeper tangent, even if its curve runs far below the other's.

Sign of the slope thrown away. A downhill-to-the-right segment must be reported as negative velocity (direction!), and a journey's displacement is final height minus initial height — which can be negative or zero while the distance is large.

Finding "area under the x–t graph". Area under x–t has no physical meaning (its units would be metre-seconds). Slope is the only currency here — area becomes meaningful on the velocity–time graph, the next module's business.

Distance–time vs position–time wording (IGCSE/A-Level). CIE 0625 draws distance–time graphs: distance never decreases, so those graphs can never fall — a falling section there is an impossible graph, while on a position(displacement)–time graph a fall is just return motion. Check which word the paper printed before judging a graph "impossible".

Vertical segments and doubling back. A vertical jump means teleportation (infinite velocity); a graph that curls back leftward gives one object two positions at one instant. Both are auto-eliminate options in "which graph is possible?" MCQs.

Texts. NCERT Physics XI Part 1, Ch. 3 (Motion in a Straight Line, graph sections) — the CBSE/NEET canonical treatment · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday, Resnick & Walker Ch. 2 (graphical analysis) · CIE 0625 syllabus §1.2 for the distance–time reading list. The UG deep dive with full calculus is Velocity & Acceleration (aq-mech tier).

Provenance. All readings are standard kinematics per NCERT Ch. 3. A verifier should recompute: Riya's legs $80/40 = 2$ m/s out, $-80/40 = -2$ m/s back; board example slopes $25/5 = 5$, $(-5 - 25)/6 = -5$ m/s, displacement $-5$ m, distance $55$ m, average speed $55/14 = 3.9$ m/s; catch-up $48/(8-2) = 8$ s at $64$ m, gap at 5 s $= 48 - 30 = 18$ m; parabola $x = 12t - t^2$: roots 0 and 12, crest $(6, 36)$, secant $1 \to 3$: $(27 - 11)/2 = 8$ m/s vs slope law $12 - 2t = 8$ ✓.

Prerequisites. Position, Distance & Displacement — the quantity on the vertical axis · Speed & Velocity — the quantity hiding in the slope. Leads to. Velocity–Time Graphs — the companion picture where area finally means something · Graph Reading at Speed — the technique module that drills all three graph types against the clock.