Graphs — x-t, v-t, a-t; areas and slopes meaning: Worked examples

This page is the drill hall for Graphs — x-t, v-t, a-t; areas and slopes meaning. We do not re-teach the ideas — we hunt down every kind of situation a graph problem can throw at you and work one clean example for each. If a symbol or rule feels unfamiliar, jump back to the parent note; everything below assumes only the two golden rules:

The scenario matrix

Every graph question is really one of the cells below. We will hit each cell with at least one worked example, and label which cell it belongs to.

#	Case class	What makes it tricky	Example
A	x–t, slope reading	slope = velocity; sign & steepness	Ex 1
B	x–t, curvature	curving ⇒ acceleration; concave up/down	Ex 2
C	v–t, positive area only	straightforward displacement	Ex 3
D	v–t, sign change (area below axis)	displacement vs distance differ	Ex 4
E	a–t, area gives Δv	stepping up the ladder	Ex 5
F	Degenerate: zero slope / zero area	flat lines, instant of rest	Ex 6
G	Limiting behaviour	what happens as a phase shrinks to zero	Ex 7
H	Real-world word problem	translate words → graph → numbers	Ex 8
I	Exam twist: same graph, two questions	one v–t, both displacement & distance	Ex 9

Figures accompany the geometric cells (A, C, D, F, H, I).

Cell A — reading velocity off an x–t graph

Worked example Ex 1 · x–t slope, all three signs

An ant's position is: at $t = 0$ it is at $x = 2$ m; it moves so that at $t = 2$ s it is at $x = 8$ m; it stays at $x = 8$ m until $t = 5$ s; then returns to $x = 2$ m at $t = 7$ s. Find the velocity in each phase.

Forecast: Guess the signs first — which phase is $+$ , which is $0$ , which is $-$ ?

Phase 1 slope $= \frac{Δ x}{Δ t} = \frac{8 - 2}{2 - 0} = + 3$ m/s. Why this step? Slope of x–t is velocity; rise-over-run on the red segment.
Phase 2 slope $= \frac{8 - 8}{5 - 2} = 0$ m/s (flat line ⇒ at rest). Why this step? A horizontal x–t means position isn't changing — the ant is parked, not "moving at constant speed."
Phase 3 slope $= \frac{2 - 8}{7 - 5} = \frac{- 6}{2} = - 3$ m/s. Why this step? Downhill on x–t ⇒ moving in the $- x$ direction; the sign is direction, not "slowing."

Verify: Net displacement $= 2 - 2 = 0$ m; check via velocities $\times$ times: $(+ 3) (2) + (0) (3) + (- 3) (2) = 6 + 0 - 6 = 0$ m. ✓ Units: $m/s \times s = m$ . ✓

See Instantaneous vs Average Velocity — here each phase is straight, so the chord slope equals the tangent slope.

Cell B — curvature tells you acceleration

Worked example Ex 2 · x–t curving, concave up vs down

Two balls each follow an x–t curve. Ball P: $x = t^{2}$ (metres, seconds). Ball Q: $x = 10 t - t^{2}$ . Describe each ball's velocity and acceleration at $t = 1$ s and $t = 4$ s.

Forecast: Which ball is speeding up, which is slowing down? Guess before computing.

Velocity is the slope = derivative. For P: $v_{P} = 2 t$ . For Q: $v_{Q} = 10 - 2 t$ . Why this step? Slope slides down the ladder; $x \to v$ means differentiate. (See Differentiation and Integration in Physics.)
Acceleration is the slope of $v$ = second derivative. $a_{P} = 2$ m/s² (constant, curve bends up). $a_{Q} = - 2$ m/s² (curve bends down). Why this step? Curving-up graph = slope increasing = $a > 0$ ; curving-down = slope decreasing = $a < 0$ .
Evaluate. P: at $t = 1$ , $v_{P} = 2$ m/s; at $t = 4$ , $v_{P} = 8$ m/s — $v$ and $a$ both $+$ ⇒ speeding up. Q: at $t = 1$ , $v_{Q} = 8$ m/s; at $t = 4$ , $v_{Q} = 2$ m/s — $v > 0$ but $a < 0$ ⇒ slowing down. Why this step? Slowing down ⇔ $v$ and $a$ have opposite signs (rule from parent).

Verify: Q reaches its turning point (top of the position hill) when $v_{Q} = 0$ : $10 - 2 t = 0 \Rightarrow t = 5$ s, which is after both sample times, so Q is still moving $+ x$ but decelerating throughout — consistent. ✓

Cell C — displacement from a positive-only v–t area

Worked example Ex 3 · v–t, all area above the axis

A train's velocity: rises from $0$ to $15$ m/s over $6$ s, holds $15$ m/s for $4$ s, then drops to $0$ over $2$ s. Find the total displacement.

Forecast: The shape is triangle + rectangle + triangle. Guess: more or less than $150$ m?

Ramp-up area (triangle) $= \frac{1}{2} (6) (15) = 45$ m. Why this step? Area under v–t = displacement; a triangle is $\frac{1}{2} base \times height$ .
Cruise area (rectangle) $= (4) (15) = 60$ m. Why this step? Constant $v$ ⇒ rectangular strip, height $v$ , width $Δ t$ .
Ramp-down area (triangle) $= \frac{1}{2} (2) (15) = 15$ m.
Total $= 45 + 60 + 15 = 120$ m.

Verify: Average velocity over $12$ s should give the same: displacement $/$ time $= 120/12 = 10$ m/s, which lies between $0$ and $15$ — plausible for this trapezoid-ish trip. ✓ All area is above the axis, so distance $=$ displacement $= 120$ m.

Cell D — sign change: area below the axis

Worked example Ex 4 · v–t crosses zero — displacement ≠ distance

A drone's velocity: $v = + 8$ m/s for $3$ s, then $v = - 4$ m/s for $6$ s. Find (a) displacement and (b) distance.

Forecast: Guess the sign of the net displacement — did the drone end ahead of, behind, or at its start?

Area above axis $= (+ 8) (3) = + 24$ m. Why this step? Positive velocity ⇒ positive (forward) displacement.
Area below axis $= (- 4) (6) = - 24$ m. Why this step? Below the time-axis, height is negative, so this rectangle counts as negative displacement (motion in $- x$ ).
Displacement $= + 24 + (- 24) = 0$ m. Why this step? Net (signed) area = displacement.
Distance $= ∣ + 24 ∣ + ∣ - 24 ∣ = 48$ m. Why this step? Distance ignores direction — add the magnitudes of the areas. (See Distance vs Displacement.)

Verify: The drone went $24$ m out and $24$ m back ⇒ ends at start (displacement $0$ ) but travelled $48$ m of road. Signs consistent. ✓

Cell E — stepping up: area under a–t gives Δv

Cell F — degenerate inputs: zero slope, zero area, an instant of rest

Worked example Ex 6 · flat pieces and a single point of rest

A ball thrown straight up has $v = 20 - 10 t$ (m/s, up positive), $g = 10$ m/s². (a) At what instant is it momentarily at rest? (b) What is the acceleration at that instant? (c) On the a–t graph over $0 \leq t \leq 4$ s, what is the area, and what does the slope of that graph tell you?

Forecast: At the very top, is the acceleration zero? (Careful — this is the classic trap.)

Instant of rest: set $v = 0$ : $20 - 10 t = 0 \Rightarrow t = 2$ s. Why this step? "Momentarily at rest" means the v–t line crosses the time-axis — a single point, not a flat stretch.
Acceleration there: slope of v–t $= - 10$ m/s² everywhere, including $t = 2$ s. Why this step? The v–t line is straight, so its slope (acceleration) is the same constant even at the moment $v = 0$ . Velocity zero does not mean acceleration zero.
a–t graph: it's a horizontal line at $- 10$ . Area over $0$ to $4$ s $= (- 10) (4) = - 40$ m/s (that is $Δ v$ : from $+ 20$ to $- 20$ ). Its slope $= 0$ (jerk is zero — acceleration isn't changing). Why this step? Flat a–t ⇒ zero slope ⇒ constant acceleration; the area still carries meaning (Δv), the slope is the degenerate/zero case.

Verify: $Δ v = v (4) - v (0) = (20 - 40) - (20) = - 40$ m/s, matching the area. ✓ (See Free Fall and Projectile Motion.)

Cell G — limiting behaviour: shrink a phase to zero

Worked example Ex 7 · what happens as the middle phase vanishes

Reconsider Ex 3's train, but let the cruise phase shrink. Keep ramp-up (triangle, $6$ s, peak $15$ ) and ramp-down ( $2$ s), but let the flat cruise last $τ$ seconds. Write the total displacement as a function of $τ$ , and find its limit as $τ \to 0$ .

Forecast: As the flat top disappears, does the area go to zero, or to a positive constant?

Displacement $= ramp-up 45 + cruise rectangle 15 τ + ramp-down 15 = 60 + 15 τ$ m. Why this step? The two triangles don't depend on $τ$ ; only the rectangle's width shrinks.
Take the limit $τ \to 0$ : displacement $\to 60$ m. Why this step? A rectangle of zero width has zero area — the cruise contributes nothing, leaving just the two triangles. The graph degenerates smoothly into a single tent shape.

Verify: At $τ = 0$ the v–t is a triangle rising over $6$ s then falling over $2$ s, both to peak $15$ : area $= \frac{1}{2} (6) (15) + \frac{1}{2} (2) (15) = 45 + 15 = 60$ m. ✓ No jump — the limit matches the degenerate shape, exactly as areas should behave.

Cell H — real-world word problem

Worked example Ex 8 · lift (elevator) trip

A lift starts at rest, accelerates upward at $1.5$ m/s² for $2$ s, travels at constant speed for $5$ s, then decelerates to rest at $3$ m/s². How high does it rise in total?

Forecast: The cruise speed is the peak of the v–t tent. Guess: is the total more or less than $15$ m?

Cruise speed: $v = 0 + (1.5) (2) = 3$ m/s. Why this step? End of ramp-up velocity = area under a–t = $a t$ ; this becomes the height of the flat top.
Decel time: from $3$ m/s to $0$ at $3$ m/s²: $t = 3/3 = 1$ s. Why this step? $Δ v = a t \Rightarrow t = Δ v / a$ ; needed to size the last triangle.
Ramp-up area $= \frac{1}{2} (2) (3) = 3$ m; cruise area $= (5) (3) = 15$ m; ramp-down area $= \frac{1}{2} (1) (3) = 1.5$ m. Why this step? Area under v–t = displacement, phase by phase.
Total height $= 3 + 15 + 1.5 = 19.5$ m.

Verify: All velocity is positive (lift only goes up), so distance = displacement = $19.5$ m. Units: $m/s \times s = m$ throughout. ✓

Cell I — exam twist: one graph, two questions

Worked example Ex 9 · the same v–t answered two ways

A particle's velocity: $v = + 10$ m/s for $2$ s, ramps linearly down to $- 10$ m/s over the next $4$ s, then holds $- 10$ m/s for $1$ s. Find (a) displacement and (b) total distance.

Forecast: The ramp crosses zero somewhere in the middle. Guess when $v = 0$ , and whether displacement is positive or negative.

Where does $v$ cross zero? During the ramp, $v$ goes $+ 10 \to - 10$ over $4$ s (from $t = 2$ to $t = 6$ ). It's linear, so zero is halfway: at $t = 4$ s. Why this step? For distance we must split at the sign change — area above and below must be handled separately.
Signed areas (displacement):
- Flat top $t = 0..2$ : $(+ 10) (2) = + 20$ m.
- Ramp above axis $t = 2..4$ : triangle $\frac{1}{2} (2) (10) = + 10$ m.
- Ramp below axis $t = 4..6$ : triangle $\frac{1}{2} (2) (10) = - 10$ m.
- Flat bottom $t = 6..7$ : $(- 10) (1) = - 10$ m.
- Displacement $= 20 + 10 - 10 - 10 = + 10$ m. Why this step? Net signed area = displacement; below-axis pieces are negative.
Distance = sum of magnitudes: $∣20∣ + ∣10∣ + ∣ - 10 ∣ + ∣ - 10 ∣ = 50$ m. Why this step? Distance ignores direction — add absolute areas.

Verify: Displacement ( $+ 10$ m) $<$ distance ( $50$ m), and they differ exactly because $v$ changed sign — consistent with the rule. Above-axis area $= 20 + 10 = 30$ , below-axis $= 10 + 10 = 20$ ; net $= 30 - 20 = + 10$ ✓, total $= 30 + 20 = 50$ ✓.

Connections

Instantaneous vs Average Velocity — chord vs tangent slope shows up in Ex 1–2.
Equations of Uniformly Accelerated Motion — cross-check for Ex 5, Ex 8.
Differentiation and Integration in Physics — slope ≡ derivative (Ex 2), area ≡ integral (Ex 3–5).
Free Fall and Projectile Motion — the up-then-down v–t of Ex 6.
Distance vs Displacement — the heart of Ex 4 and Ex 9.
Vectors — Components and Signs — why sign of $v$ is a direction, not a size.