Velocity–Time Graphs

After a serious road accident, investigators pull one object from the wreckage before anything else: the tachograph — the sealed recorder that has been drawing the vehicle's velocity–time graph, minute by minute, for the whole journey. From that single wiggling line they recover everything a court needs. How hard did the driver brake? Read the steepness of the line. How far did the truck travel after the brakes went on? Measure the area trapped under the line. No odometer, no witnesses — one graph, two readings, the full story. That is what makes the velocity–time graph the physicist's favourite picture of motion. A position–time graph hands you velocity through its slope, and that is essentially all. A v–t graph pays out twice: its slope at any moment is the acceleration, and the area between the line and the time axis is the displacement. Every uniform-acceleration formula you will meet in the next module is really just this graph's rectangles, triangles and trapeziums wearing algebraic clothes. You already know how to read an x–t graph and what acceleration means, including its sign. Here those two skills meet on one set of axes. The prize is real exam currency: multi-phase journey graphs, braking problems, and the single most-tested subtlety in the chapter — what happens to the area when the line dives below the time axis and the graph quietly starts subtracting.

Key takeaways

On a velocity–time graph, slope = acceleration: $a = \Delta v / \Delta t$ , read as rise over run. Steeper means larger $|a|$ ; a downhill slope means acceleration in the negative direction; a horizontal line means constant velocity — not rest.
Area between the line and the time axis = displacement. The units work: height (m/s) × base (s) = metres.
Area is signed: regions below the time axis count negative (motion in the negative direction). Displacement = areas above − areas below; distance = all areas added as positives.
A ball thrown straight up has a v–t graph that is one straight line crossing the axis: the crossing instant is the top of the flight, and the up-triangle and down-triangle record the rise and the fall.
The trapezium under a uniform-acceleration segment gives $s = \tfrac{1}{2}(u+v)t$ — the bridge to the kinematic equations of the next module.
Key idea: one graph, two readings — the slope of the v–t line is the acceleration, and the signed area under it is the displacement, so a single curve carries the object's entire motion history.

The core insight

The velocity–time graph encodes motion twice over: its slope at any instant equals the acceleration, and the signed area between the curve and the time axis over any interval equals the displacement — area above the axis counting forward, area below counting backward.

Where this lives today

Every commercial truck and coach in the EU and India carries a tachograph that legally must record the vehicle's speed–time graph; accident investigators compute braking distances as areas under the recorded trace, and courts accept the geometry as evidence. The same two readings run motorsport: an F1 telemetry screen is a wall of v–t traces, where engineers compare braking points by slope and corner-exit performance by the area recovered between two markers. Electric vehicles put the signed-area idea to commercial work. A trip computer integrates the v–t record continuously — that is how "distance to empty" updates — while regenerative braking turns every steep negative-slope segment into recovered energy, logged against the deceleration profile. And the accelerometer chip in your phone runs the whole logic in reverse twice over: it measures acceleration directly, then reconstructs velocity and displacement by accumulating areas — step counting, screen rotation and fitness tracking are v–t graph readings performed a hundred times a second in silicon.

The physical picture

Watch a car brake for a red light: it is doing 24 m/s when the driver sees the signal, and it stops 6 seconds later, slowing steadily. The v–t graph of those six seconds is a single straight line falling from 24 to 0. Now read it twice. Steepness: the line drops 24 m/s over 6 s, so the velocity is bleeding away at $24/6 = 4$ m/s every second — the acceleration is $-4\ \mathrm{m/s^2}$ , the minus sign recording that it points against the motion. Area: the line and the axes enclose a right triangle, base 6 s and height 24 m/s, area $\tfrac{1}{2} \times 6 \times 24 = 72$ . Seventy-two metres — the braking distance, conjured out of geometry.

Figure

One braking car, two readings — slope gives the acceleration, area gives the stopping distance

Why should area know about distance? Strip the graph back to the simplest case: a metro train holding a steady 10 m/s for 5 s. Displacement is velocity × time $= 50$ m. On the graph that same multiplication is a rectangle: height 10, base 5, area 50. The area is the physics formula, drawn instead of written — and the units multiply along with the numbers, (m/s) × s = m, which is why an "area" on this graph comes out in metres, not square metres. Everything else in this module is that one rectangle, generalised. The second big idea is what the graph does when the line crosses the time axis. Below the axis, velocity is negative — the object is moving backwards — and every second spent there removes displacement instead of adding it. The graph subtracts. A thrown ball, a bouncing cricket ball, a car reversing out of a parking spot: their v–t lines dive under the axis, and reading them correctly is where exams separate students who look at graphs from students who read them.

Slope: the acceleration reading

Take any two points on a straight v–t segment and form rise over run. The rise is a velocity change $\Delta v$ , the run a time interval $\Delta t$ , and their ratio is precisely the definition of acceleration:

a = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1} \qquad (1)

where: $v_1, v_2$ — velocities read off the graph at times $t_1, t_2$ ; subtract both in the same order or the sign of $a$ flips. A horizontal segment has $\Delta v = 0$ : zero acceleration, constant velocity. A downhill segment has negative slope: the acceleration points in the negative direction — which, by the $a$ – $v$ sign test from the acceleration module, means slowing down only while $v$ is still positive. For a curved graph the ratio between two points gives only an average; the instantaneous acceleration is the slope of the tangent at that instant, exactly as the tangent on an x–t graph gave instantaneous velocity. Two habits make the reading reliable under exam pressure. Pick points far apart on the straight section (a long run divides away reading errors), and take the values from the axes, not from estimated positions between gridlines. And keep the units honest: a graph with velocity in km/h and time in seconds must be converted (×5/18) before the slope is quoted in m/s².

Area: the displacement reading

The rectangle argument generalises to any motion. Chop the time interval into thin slices of width $\Delta t$ . During one slice the velocity is nearly constant at some value $v$ , so that slice contributes displacement $v\,\Delta t$ — the area of the thin rectangle standing on it. Total displacement is the sum of all the strips:

\Delta x = \text{signed area between the } v\text{–}t \text{ curve and the time axis} \qquad (2)

where: "signed" is the load-bearing word — strips above the axis have positive height ( $v > 0$ , forward motion) and add; strips below have negative height and subtract. This works for any curve shape, which is why examiners love it: no formula for the motion is needed, only geometry. In practice every school problem decomposes into three shapes — rectangle ( $v \times t$ ), triangle ( $\tfrac{1}{2} \times$ base $\times$ height), trapezium ( $\tfrac{1}{2}(v_1 + v_2)\,t$ ) — plus square-counting for genuine curves.

Derivation AWhy the area under a velocity time graph equals the displacement

We want to justify equation (2) — that a geometric area equals a physical displacement — without calculus, using only the definition of velocity and a staircase of rectangles. The strategy: prove it exactly for constant velocity, then approximate any changing motion by short constant-velocity steps and watch the approximation become perfect as the steps shrink. Watch the units as much as the numbers — the argument also explains why this "area" is measured in metres.

Step 1. Constant velocity first. An object holding velocity $v$ for a time $t$ has displacement $\Delta x = v\,t$ , straight from the definition of velocity. Its graph is a horizontal line at height $v$ ; the region under it is a rectangle of height $v$ and base $t$ ; the rectangle's area is $v \times t$ . Formula and area are the same multiplication — and the units multiply too: (m/s) × s = m.

Step 2. Now let velocity change. Freeze the motion into $n$ short intervals of width $\Delta t$ , and during each interval pretend the velocity holds its value at the start of that interval. Each pretend-interval is Step 1 again: it contributes displacement $v_i\,\Delta t$ , the area of one thin rectangle. The pretend journey's total displacement is the area of the whole staircase of rectangles. Why the pretence is harmless: within a short enough interval, a smoothly changing velocity simply has no room to differ from its starting value by more than a whisker — the staircase hugs the true curve more tightly every time the steps are halved.

Step 3. Shrink the steps. As $\Delta t \to 0$ the staircase's jagged top melts onto the true v–t curve, and the staircase's total area settles onto the exact area under the curve. The displacement — which the staircase was computing all along — is therefore exactly that area.

Step 4. Keep the sign. If the object moves backwards during some interval, $v_i$ is negative, the strip's contribution $v_i \, \Delta t$ is negative, and geometrically the strip hangs below the time axis. Backward motion subtracts displacement; below-axis area subtracts area. The bookkeeping matches perfectly, which is why equation (2) says signed area. Verified: for the metro rectangle, $10 \times 5 = 50$ m either way; for the braking triangle, the staircase limit gives $\tfrac{1}{2} \times 6 \times 24 = 72$ m, matching $\bar{v}\,t = 12 \times 6 = 72$ m computed from the average of 24 and 0.

Go deeper · Derivation B · 1 minThe trapezium that becomes the second kinematic equation

Derivation BThe trapezium that becomes the second kinematic equation

We want the displacement of an object that accelerates uniformly from velocity $u$ to velocity $v$ in time $t$ , by reading its v–t graph — a straight ramp from height $u$ to height $v$ . The strategy: cut the trapezium under the ramp into a rectangle plus a triangle, compute each, and recognise the result as the average-velocity formula that the next module will adopt as a kinematic equation. This is the single most exam-useful area on any graph.

Step 1. Decompose. Under the ramp sits a rectangle of height $u$ and base $t$ (the displacement the object would have had with no acceleration at all), area $u\,t$ . On top of it sits a right triangle: base $t$ , height $v - u$ (the extra velocity gained), area $\tfrac{1}{2}(v-u)\,t$ — the bonus displacement contributed by the acceleration.

Step 2. Add the pieces:

s = u\,t + \tfrac{1}{2}(v - u)\,t

Step 3. Combine over a common factor of $t$ :

s = \left(\frac{2u + v - u}{2}\right) t = \frac{(u + v)}{2}\,t \qquad (3)

Why this shape is worth memorising: $\tfrac{1}{2}(u+v)$ is just the average of the starting and finishing velocities — so the trapezium says that under uniform acceleration, the object travels exactly as far as if it had cruised at its average velocity the whole time. That is the medieval mean-speed theorem, proved with one rectangle and one triangle.

Step 4. Sanity-check with numbers: $u = 10$ m/s, $v = 20$ m/s, $t = 4$ s. Rectangle $10 \times 4 = 40$ m, triangle $\tfrac{1}{2} \times 4 \times 10 = 20$ m, total 60 m; formula: $\tfrac{1}{2}(10+20) \times 4 = 60$ m. ✓ The next module takes equation (3), eliminates variables from it, and mass-produces the remaining kinematic equations — every one of them is secretly this trapezium. Verified: with $u = 0$ the trapezium degenerates to a triangle and (3) gives $s = \tfrac{1}{2}v t$ — for the braking car run backwards (rest to 24 m/s in 6 s), $\tfrac{1}{2} \times 24 \times 6 = 72$ m, the same triangle as the hero figure.

When the line crosses the axis: the graph that subtracts

Throw a ball straight up at 20 m/s and take $g = 10\ \mathrm{m/s^2}$ , up as positive. Its velocity obeys $v = 20 - 10t$ : one straight line, slope $-10$ the whole flight, sliding from $+20$ through zero (at $t = 2$ s) down to $-20$ (at $t = 4$ s, back at your hand). The single most important reading: the instant the line crosses the time axis is the top of the flight — the ball's highest point, not its return. Velocity passes through zero there while the slope (the acceleration) never changes.

Figure

A thrown ball's v–t line — the axis crossing is the top of the flight, and the two triangles cancel

Now read the areas. The up-phase triangle above the axis: $\tfrac{1}{2} \times 2 \times 20 = +20$ m — the height of the flight. The down-phase triangle below: $\tfrac{1}{2} \times 2 \times 20 = 20$ m of area, counted negative. Add them with signs and the displacement is zero — correct, the ball is back in your hand. Add their magnitudes and the distance is 40 m — also correct, 20 up plus 20 down. One graph delivers both answers, provided you respect the sign:

\Delta x = A_{\text{above}} - A_{\text{below}}, \qquad d = A_{\text{above}} + A_{\text{below}} \qquad (4)

where: $A_{\text{above}}, A_{\text{below}}$ — the geometric (positive) areas of the regions above and below the time axis. Displacement can be positive, negative or zero; distance only grows. When a question asks for average velocity, divide the signed sum by total time; for average speed, divide the unsigned sum. For the ball: average velocity $0/4 = 0$ , average speed $40/4 = 10$ m/s.

you are asked for…	read the graph as…	anchor (thrown ball)
acceleration	slope of the segment (tangent if curved)	$-10$ m/s² everywhere
displacement	area above − area below	$20 - 20 = 0$
distance	all areas as positives	$20 + 20 = 40$ m
average velocity	signed area ÷ total time	$0 \div 4 = 0$
average speed	unsigned area ÷ total time	$40 \div 4 = 10$ m/s
moment of direction reversal	where the line crosses the t-axis	$t = 2$ s, the top

Limiting cases

case	graph feature	meaning
constant velocity	horizontal line	$a = 0$ ; area is a plain rectangle $vt$
starts from rest	line leaves the origin	triangle area $\tfrac{1}{2}vt$ ; equation (3) with $u=0$
line touches the axis and rises again	$v = 0$ for an instant	momentary stop or turnaround — not parked
line runs along the axis	$v = 0$ over an interval	genuinely at rest for that interval
line crosses the axis	sign of $v$ flips	direction reversal; area bookkeeping switches sign
two segments, same height difference, different widths	steeper vs gentler	same $\Delta v$ ; the narrow one is the harder acceleration
symmetric triangles above and below	equal + and − areas	round trip: displacement 0, distance $2A$

Common trapReading the v–t graph as a picture of the journey

The single costliest graph mistake in exams: treating the v–t curve as if it showed where the object is. On this graph, height is speed, not position. A line sloping down does not mean "moving down" or "coming back" — it means slowing (or accelerating backwards). The point where the line meets the time axis does not mean "returned to the start" — it means velocity zero, which for a thrown ball is the highest point of the whole motion. And a tall flat section does not mean "far away"; it means "moving fast". Whenever you catch yourself narrating a v–t graph as a landscape the object travels over, stop and re-read every feature through the two legal readings: slope → acceleration, area → displacement. Nothing else on this graph has any meaning.

Worked examples

Board level — a three-phase journey graph (CBSE/ICSE/IGCSE pattern)

A car starts from rest and its velocity–time graph shows three straight phases: it accelerates uniformly to 20 m/s in 8 s, holds 20 m/s for the next 5 s, then decelerates uniformly to rest in a further 4 s. Find (a) the acceleration in each phase, (b) the total displacement, and (c) the average velocity for the whole journey.

Step 1. Sketch and label the phases: a rising ramp (0 → 8 s), a plateau (8 → 13 s), a falling ramp (13 → 17 s). Total time $= 17$ s.

Step 2. Accelerations from slopes, equation (1). Phase 1: $(20-0)/8 = \boxed{2.5\ \mathrm{m/s^2}}$ . Phase 2: horizontal line, $\boxed{0}$ . Phase 3: $(0-20)/4 = \boxed{-5\ \mathrm{m/s^2}}$ — the sign records deceleration.

Step 3. Displacement from areas, equation (2). Triangle: $\tfrac{1}{2} \times 8 \times 20 = 160/2 = 80$ m. Rectangle: $20 \times 5 = 100$ m. Triangle: $\tfrac{1}{2} \times 4 \times 20 = 40$ m. Total $80 + 100 + 40 = \boxed{220\ \text{m}}$ .

Step 4. Average velocity $= 220/17 = \boxed{12.9\ \text{m/s}}$ — and not some average of 0 and 20; multi-phase journeys must go through total-area-over-total-time.

Competitive level — displacement vs distance across an axis crossing (JEE-Main/NEET pattern)

A particle moves along a straight line with initial velocity $+20$ m/s and constant acceleration $-5\ \mathrm{m/s^2}$ for 6 s. Using its v–t graph, find the displacement, the distance travelled, and the average speed.

Step 1. Build the line: $v = 20 - 5t$ . It crosses the axis at $t = 20/5 = 4$ s and reaches $v = 20 - 30 = -10$ m/s at $t = 6$ s. One crossing → split the area there. Never integrate blindly across a crossing when distance is asked.

Step 2. Area above (0 → 4 s): triangle $\tfrac{1}{2} \times 4 \times 20 = +40$ m. Area below (4 → 6 s): triangle $\tfrac{1}{2} \times 2 \times 10 = 10$ m, negative.

Step 3. Displacement $= 40 - 10 = \boxed{+30\ \text{m}}$ . Distance $= 40 + 10 = \boxed{50\ \text{m}}$ .

Step 4. Average speed $= 50/6 = \boxed{8.3\ \text{m/s}}$ (average velocity would be $30/6 = 5$ m/s — the options will contain both; read which is asked). Timing note: 60–90 seconds. The whole question is Step 1 — find the crossing time first; the two triangles then read off instantly. Students who plug $t=6$ into a displacement formula get +30 correctly but then report 30 for distance too, which is the planted wrong option.

Competitive level — when does it return to the start? (JEE-Main pattern)

A body is projected along a straight line with velocity $+10$ m/s under constant acceleration $-5\ \mathrm{m/s^2}$ . From the v–t graph, find when it returns to its starting point.

Step 1. Returning to the start means displacement $= 0$ : the negative area must grow until it exactly cancels the positive area.

Step 2. Positive lobe: the line $v = 10 - 5t$ crosses zero at $t = 2$ s, enclosing $\tfrac{1}{2} \times 2 \times 10 = 10$ m.

Step 3. Negative lobe: from $t = 2$ to time $t$ , the triangle has base $(t-2)$ and height $5(t-2)$ , area $\tfrac{1}{2} \times 5(t-2)^2$ . Set it equal to 10: $(t-2)^2 = 4$ , so $t - 2 = 2$ and $t = \boxed{4\ \text{s}}$ .

Step 4. Notice the symmetry worth banking: under constant acceleration, the return to start takes exactly twice the time to the turnaround ( $t_{\text{return}} = 2u/|a|$ ). The two lobes are congruent triangles — read it straight off the picture next time. Timing note: 45 seconds with the symmetry; 2 minutes by algebra. In an MCQ, sketch, mark the crossing at $u/|a|$ , double it, done.

Exam traps

Height read as position. The graph's height is velocity. "The graph comes back down to the axis" means slows to zero, not comes back home. The thrown ball's highest point is the axis crossing, where the graph looks most unremarkable.

Distance computed as if it were displacement. Any time the line crosses the time axis, split the area at the crossing and keep two ledgers: signed sum for displacement, unsigned sum for distance. The planted MCQ options are always the two answers you get by keeping or dropping the sign.

Accelerations compared by height. Which graph shows the greater acceleration? The steeper one — not the one that reaches higher. A line to 40 m/s in 20 s ( $a = 2$ ) loses to a line to 10 m/s in 2 s ( $a = 5$ ).

Touching the axis read as "stops and stays stopped". A vertex touching $v = 0$ and rising again is a momentary halt (or a turnaround if it crosses). Only a segment lying along the axis is genuine rest.

Mixed units on the axes. Velocity in km/h against time in seconds is a favourite ambush: convert with $\times \tfrac{5}{18}$ before taking slopes or areas — or your acceleration is wrong by a factor of 3.6.

Board wording: "speed–time" vs "velocity–time". IGCSE (0625) works with speed–time graphs, which never go below the axis — deceleration appears as a downhill run to zero, and the area is always the plain distance. CBSE/JEE/NEET/AP/IB use full velocity–time graphs where the negative branch and signed areas are live ammunition. Know which graph your paper draws.

Using $s = \tfrac{1}{2}(u+v)t$ across phases with different accelerations. The trapezium formula belongs to one uniform-acceleration segment. For multi-phase journeys, area each phase separately and add — as the board example above does.

Origins · 1 minIsaac Barrow and how this was discovered

Isaac Barrow

1630–1677 · Cambridge

Isaac Barrow held Cambridge's Lucasian chair of mathematics before his student Isaac Newton, and in his 1670 Lectiones Geometricae he proved — geometrically, a generation before calculus had a name — that the tangent problem and the area problem are inverses of each other: the two operations undo one another. The velocity–time graph is that theorem wearing physics: the slope (tangent) of the curve gives acceleration, the area under it gives displacement, and the two readings are inverse routes between the same three quantities. Barrow eventually resigned the chair in Newton's favour, having handed him both the job and the mathematical machinery that would become, in Newton's hands, the calculus of motion.

History

1335

— William Heytesbury and the Merton School state the mean-speed theorem: uniform acceleration covers the same distance as steady motion at the average speed.

1350

— Nicole Oresme draws the theorem: velocity graphed against time, with the area under the graph standing for distance — the first v–t area argument.

1638

— Galileo's Two New Sciences derives the odd-number rule (distances 1:3:5:7 in successive seconds) from uniformly accelerated motion.

1670

— Barrow's Lectiones Geometricae proves tangents and areas are inverse problems, the geometric seed of the fundamental theorem of calculus.

1687

— Newton's Principia turns slopes and areas into fluxions: the modern kinematics of continuously varying motion.

Going further

Texts. NCERT Physics XI Part 1, Ch. 2 (Motion in a Straight Line, graphical analysis) — the CBSE/NEET canonical treatment · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday, Resnick & Walker Ch. 2 · Galileo, Two New Sciences, Third Day (the odd-number rule from uniform acceleration). The calculus-language version of both readings is built in Calculus in Kinematics; the full UG treatment is Velocity & Acceleration (aq-mech tier deep dive).

Provenance. All content is standard graphical kinematics per NCERT Ch. 2. A verifier should recompute: braking hero $24/6 = 4$ m/s² and $\tfrac{1}{2}\times 6 \times 24 = 72$ m; metro rectangle $10 \times 5 = 50$ m; thrown ball crossing at $20/10 = 2$ s, lobes $\pm 20$ m, distance 40 m, average speed $40/4 = 10$ m/s; trapezium check $40 + 20 = 60 = \tfrac{1}{2}(30)(4)$ ; board journey $80 + 100 + 40 = 220$ m, accelerations $2.5, 0, -5$ m/s², average $220/17 = 12.9$ m/s; competitive crossing at 4 s, $40 - 10 = 30$ m vs $40 + 10 = 50$ m, $50/6 = 8.3$ m/s; return-to-start $(t-2)^2 = 4 \Rightarrow t = 4$ s $= 2u/|a|$ .

Prerequisites. Position–Time Graphs — the slope-reading habit this module extends · Acceleration — the quantity our slopes now measure. Leads to. a–t Graphs & Graph Interconversion — the third graph and the full translation table · Kinematic Equations — the trapezium of Derivation B, industrialised.

Why does the area come out in metres when area should be in square units?

Because the "lengths" being multiplied are not lengths — they carry the units of the axes. The base of a strip is a time (seconds), its height a velocity (metres per second), so the area multiplies to (m/s) × s = m. Graph areas always inherit the product of the axis units: under an acceleration–time graph the area comes out in m/s (a velocity change), and in later physics the area under a force–extension graph comes out in joules. The square-centimetre instinct comes from geometry class, where both axes happen to be lengths. On physics graphs, always multiply the units of the axes to discover what the area means — it is also a quick way to catch your own mistakes.

Does the v–t graph tell me where the object actually is?

Not by itself. Areas give the change in position over an interval, not the position. If the graph shows +30 m of signed area between $t = 0$ and $t = 6$ s, the object is 30 m ahead of wherever it started — but the graph carries no record of where that was. Questions that want a final position must supply the initial one ("a particle starts at $x = 5$ m…"), and you add: $x = x_0 + \Delta x$ . This is also why two objects with identical v–t graphs can be in completely different places: same motion, different starting posts.

What is the difference between the line touching the time axis and crossing it?

Touching (a vertex that meets $v = 0$ and lifts off again on the same side) is a momentary stop: the object pauses for an instant and resumes in the same direction. Crossing (the line passes through the axis to the other side) is a reversal: velocity changes sign, the object turns around, and from that instant its area contributions flip sign. A parked object is neither — its graph lies along the axis for a whole interval. The three cases look similar in a thumbnail sketch and mean three different motions; exams test the distinction directly.

How do I find acceleration and displacement when the v–t graph is curved?

The readings survive; only the tools change. Acceleration at an instant = slope of the tangent drawn at that point — lay a ruler along the curve's direction there and compute rise over run of the ruler line, exactly as tangents worked on the x–t graph. Displacement = area under the curve, which you estimate by counting grid squares (work out how many metres one square is worth from the axis scales, then count, taking half-squares sensibly) or by slicing the region into approximate trapezium strips. Boards that set curved graphs (CBSE, A-Level data questions) award full marks for a clean tangent construction and a square-count within a stated tolerance.

Is a negative slope the same thing as slowing down?

No — negative slope means the acceleration is negative, and whether that slows or speeds the object depends on which side of the axis the line is. Above the axis ( $v > 0$ ) a falling line is genuine slowing. Below the axis ( $v < 0$ ) that same downhill slope is making the velocity more negative — the object is speeding up in the reverse direction. The thrown ball shows both on one line: the same $-10$ m/s² slope slows the rise, then quickens the fall. Apply the sign test from the acceleration module: opposite signs of $a$ and $v$ mean slowing, same signs mean gaining speed.

My board's paper says "speed–time graph" — is that the same thing?

Almost, and the difference is exactly where marks hide. A speed–time graph plots the magnitude of velocity, so it can never dip below the axis; a thrown ball's speed–time graph is a V shape (down to zero at the top, back up during the fall), while its velocity–time graph is one straight line crossing the axis. On a speed–time graph the area is always the plain distance. IGCSE 0625 uses speed–time graphs throughout; CBSE, ISC, JEE, NEET, AP and IB expect the full velocity version with signed areas. If a question hands you a V-shaped "velocity" graph of a bouncing or thrown object, suspect it is actually plotting speed.

Why is average velocity not just (u + v)/2 for the whole journey?

The formula $\tfrac{1}{2}(u+v)$ is the average velocity only across one segment of uniform acceleration — it is the trapezium result, equation (3). A journey with several phases (accelerate, cruise, brake) has a different average: total signed area over total time, no shortcuts. In the board example above, the endpoint velocities are 0 and 0, but the average velocity is $220/17 \approx 12.9$ m/s, not zero. Use $\tfrac{1}{2}(u+v)$ within a phase; use area ÷ time for anything composite. Options built from the wrong choice appear on nearly every graph MCQ.