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

1.1.18 · D3 · Physics › Measurement, Vectors & Kinematics › Graphs — x-t, v-t, a-t; areas and slopes meaning

Yeh page Graphs — x-t, v-t, a-t; areas and slopes meaning ka drill hall hai. Hum yahan ideas dobara nahi padhate — hum har tarah ki situation ko dhundte hain jo ek graph problem mein aa sakti hai, aur har ek ke liye ek clean example karte hain. Agar koi symbol ya rule unfamiliar lage, parent note pe wapas jaao; neeche sab kuch sirf do golden rules assume karta hai:

Scenario matrix

Har graph question asal mein neeche diye kisi ek cell jaisa hota hai. Hum har cell ko kam se kam ek worked example se cover karenge, aur batayenge ki kaunsa cell hai.

#	Case class	Tricky kyon hai	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	seedha displacement	Ex 3
D	v–t, sign change (area below axis)	displacement vs distance differ karte hain	Ex 4
E	a–t, area gives Δv	ladder upar chadna	Ex 5
F	Degenerate: zero slope / zero area	flat lines, rest ka instant	Ex 6
G	Limiting behaviour	kya hota hai jab ek phase zero ho jaata hai	Ex 7
H	Real-world word problem	words → graph → numbers mein translate karna	Ex 8
I	Exam twist: same graph, do questions	ek v–t, dono displacement & distance	Ex 9

Geometric cells (A, C, D, F, H, I) ke saath figures hain.

Cell A — x–t graph se velocity padhna

Worked example Ex 1 · x–t slope, teeno signs

Ek chinti ki position yeh hai: $t = 0$ pe woh $x = 2$ m pe hai; woh is tarah chalti hai ki $t = 2$ s pe $x = 8$ m pe hai; $t = 5$ s tak $x = 8$ m pe ruk jaati hai; phir $t = 7$ s pe $x = 2$ m pe wapas aa jaati hai. Har phase mein velocity nikalo.

Forecast: Pehle signs guess karo — kaun sa phase $+$ hai, kaun sa $0$ , kaun sa $-$ ?

Phase 1 slope $= \frac{Δ x}{Δ t} = \frac{8 - 2}{2 - 0} = + 3$ m/s. Yeh step kyon? x–t ka slope hi velocity hai; red segment pe rise-over-run.
Phase 2 slope $= \frac{8 - 8}{5 - 2} = 0$ m/s (flat line ⇒ rest mein). Yeh step kyon? Horizontal x–t matlab position change nahi ho raha — chinti ruki hui hai, nahin "constant speed se chal rahi."
Phase 3 slope $= \frac{2 - 8}{7 - 5} = \frac{- 6}{2} = - 3$ m/s. Yeh step kyon? x–t pe downhill ⇒ $- x$ direction mein chal raha hai; sign direction hai, "slow ho raha hai" nahi.

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

Dekho Instantaneous vs Average Velocity — yahan har phase seedha hai, toh chord slope barabar hai tangent slope ke.

Cell B — curvature acceleration batati hai

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

Do balls mein se har ek x–t curve follow karta hai. Ball P: $x = t^{2}$ (metres, seconds). Ball Q: $x = 10 t - t^{2}$ . $t = 1$ s aur $t = 4$ s pe har ball ki velocity aur acceleration describe karo.

Forecast: Kaun sa ball speed up kar raha hai, kaun sa slow down? Compute karne se pehle guess karo.

Velocity = slope = derivative. P ke liye: $v_{P} = 2 t$ . Q ke liye: $v_{Q} = 10 - 2 t$ . Yeh step kyon? Slope ladder se neeche jaata hai; $x \to v$ matlab differentiate karo. (Dekho Differentiation and Integration in Physics.)
Acceleration = $v$ ka slope = second derivative. $a_{P} = 2$ m/s² (constant, curve upar bend karti hai). $a_{Q} = - 2$ m/s² (curve neeche bend karti hai). Yeh step kyon? Curving-up graph = slope badh raha hai = $a > 0$ ; curving-down = slope ghatt raha hai = $a < 0$ .
Evaluate karo. P: $t = 1$ pe, $v_{P} = 2$ m/s; $t = 4$ pe, $v_{P} = 8$ m/s — $v$ aur $a$ dono $+$ ⇒ speed up. Q: $t = 1$ pe, $v_{Q} = 8$ m/s; $t = 4$ pe, $v_{Q} = 2$ m/s — $v > 0$ lekin $a < 0$ ⇒ slow down. Yeh step kyon? Slow down ⇔ $v$ aur $a$ ke opposite signs hain (parent ka rule).

Verify: Q apne turning point (position hill ki top) pe pahunchta hai jab $v_{Q} = 0$ : $10 - 2 t = 0 \Rightarrow t = 5$ s, jo dono sample times ke baad hai, toh Q abhi bhi $+ x$ mein chal raha hai lekin poore time decelerate kar raha hai — consistent. ✓

Cell C — positive-only v–t area se displacement

Worked example Ex 3 · v–t, saari area axis ke upar

Ek train ki velocity: $6$ s mein $0$ se $15$ m/s tak badhti hai, $4$ s tak $15$ m/s hold karti hai, phir $2$ s mein $0$ tak girti hai. Total displacement nikalo.

Forecast: Shape hai triangle + rectangle + triangle. Guess karo: $150$ m se zyada ya kam?

Ramp-up area (triangle) $= \frac{1}{2} (6) (15) = 45$ m. Yeh step kyon? v–t ke neeche area = displacement; triangle $\frac{1}{2} base \times height$ hota hai.
Cruise area (rectangle) $= (4) (15) = 60$ m. Yeh step kyon? 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: $12$ s mein average velocity same dena chahiye: displacement $/$ time $= 120/12 = 10$ m/s, jo $0$ aur $15$ ke beech hai — is trapezoid-ish trip ke liye plausible. ✓ Saari area axis ke upar hai, toh distance $=$ displacement $= 120$ m.

Cell D — sign change: axis ke neeche area

Worked example Ex 4 · v–t zero cross karta hai — displacement ≠ distance

Ek drone ki velocity: $3$ s tak $v = + 8$ m/s, phir $6$ s tak $v = - 4$ m/s. (a) displacement aur (b) distance nikalo.

Forecast: Net displacement ka sign guess karo — kya drone apne start se aage, peeche, ya wahi pe end hua?

Area above axis $= (+ 8) (3) = + 24$ m. Yeh step kyon? Positive velocity ⇒ positive (forward) displacement.
Area below axis $= (- 4) (6) = - 24$ m. Yeh step kyon? Time-axis ke neeche, height negative hai, toh yeh rectangle negative displacement count hoti hai ( $- x$ mein motion).
Displacement $= + 24 + (- 24) = 0$ m. Yeh step kyon? Net (signed) area = displacement.
Distance $= ∣ + 24 ∣ + ∣ - 24 ∣ = 48$ m. Yeh step kyon? Distance direction ignore karta hai — areas ke magnitudes add karo. (Dekho Distance vs Displacement.)

Verify: Drone $24$ m bahar gaya aur $24$ m wapas aaya ⇒ start pe end (displacement $0$ ) lekin $48$ m road travel ki. Signs consistent. ✓

Cell E — upar chadna: a–t ke neeche area se Δv milta hai

Worked example Ex 5 · a–t, non-constant (two-step) acceleration

Ek cart $u = 2$ m/s se shuru hota hai. Uska acceleration pehle $4$ s ke liye $a = 3$ m/s² hai, phir agle $2$ s ke liye $a = - 1$ m/s². Final velocity nikalo.

Forecast: Guess karo ki final velocity $14$ m/s se upar hai ya neeche.

First-phase Δv $= (3) (4) = + 12$ m/s. Yeh step kyon? a–t ke neeche area $Δ v$ hai — hum ladder $a \to v$ pe area accumulate karke upar chadhte hain.
Second-phase Δv $= (- 1) (2) = - 2$ m/s. Yeh step kyon? Negative acceleration ⇒ axis ke neeche area ⇒ velocity kam hoti hai.
Final velocity $= u + 12 + (- 2) = 2 + 10 = 12$ m/s. Yeh step kyon? Total Δv starting velocity pe add hota hai.

Verify: $v = u + a t$ se har phase ke liye cross-check karo. Phase 1 ke baad: $v = 2 + 3 (4) = 14$ m/s. Phase 2 ke baad: $v = 14 + (- 1) (2) = 12$ m/s. ✓ (Dekho Equations of Uniformly Accelerated Motion.)

Cell F — degenerate inputs: zero slope, zero area, rest ka ek instant

Worked example Ex 6 · flat pieces aur rest ka ek point

Seedha upar pheka gaya ek ball $v = 20 - 10 t$ (m/s, upar positive), $g = 10$ m/s². (a) Ek pal ke liye woh rest mein kab hota hai? (b) Us instant pe acceleration kya hai? (c) $0 \leq t \leq 4$ s pe a–t graph mein area kya hai, aur us graph ka slope kya batata hai?

Forecast: Bilkul top pe, kya acceleration zero hai? (Dhyan raho — yeh classic trap hai.)

Rest ka instant: $v = 0$ set karo: $20 - 10 t = 0 \Rightarrow t = 2$ s. Yeh step kyon? "Momentarily at rest" matlab v–t line time-axis ko cross karti hai — ek single point, flat stretch nahi.
Wahan acceleration: v–t ka slope $= - 10$ m/s² har jagah, $t = 2$ s pe bhi. Yeh step kyon? v–t line seedhi hai, toh uska slope (acceleration) same constant hai chahe us moment $v = 0$ ho. Velocity zero ka matlab acceleration zero nahi hota.
a–t graph: yeh $- 10$ pe horizontal line hai. $0$ se $4$ s tak area $= (- 10) (4) = - 40$ m/s (yeh $Δ v$ hai: $+ 20$ se $- 20$ tak). Uska slope $= 0$ (jerk zero hai — acceleration change nahi ho raha). Yeh step kyon? Flat a–t ⇒ zero slope ⇒ constant acceleration; area abhi bhi meaning rakhta hai (Δv), slope degenerate/zero case hai.

Verify: $Δ v = v (4) - v (0) = (20 - 40) - (20) = - 40$ m/s, area se match karta hai. ✓ (Dekho Free Fall and Projectile Motion.)

Cell G — limiting behaviour: ek phase ko zero tak shrink karo

Worked example Ex 7 · kya hota hai jab middle phase gayab ho jaata hai

Ex 3 ki train dobara sochte hain, lekin cruise phase shrink karo. Ramp-up (triangle, $6$ s, peak $15$ ) aur ramp-down ( $2$ s) wahi rakho, lekin flat cruise $τ$ seconds tak chalay. Total displacement ko $τ$ ke function ke roop mein likho, aur $τ \to 0$ pe iska limit nikalo.

Forecast: Jab flat top gayab hota hai, kya area zero jaata hai, ya koi positive constant pe?

Displacement $= ramp-up 45 + cruise rectangle 15 τ + ramp-down 15 = 60 + 15 τ$ m. Yeh step kyon? Dono triangles $τ$ pe depend nahi karte; sirf rectangle ki width shrink hoti hai.
Limit lo $τ \to 0$ : displacement $\to 60$ m. Yeh step kyon? Zero width ke rectangle ka area zero hota hai — cruise kuch contribute nahi karta, sirf do triangles bachte hain. Graph smoothly ek single tent shape mein degenerate ho jaata hai.

Verify: $τ = 0$ pe v–t ek triangle hai jo $6$ s mein utha phir $2$ s mein gira, dono ka peak $15$ : area $= \frac{1}{2} (6) (15) + \frac{1}{2} (2) (15) = 45 + 15 = 60$ m. ✓ Koi jump nahi — limit degenerate shape se match karta hai, bilkul jaisa areas ko behave karna chahiye.

Cell H — real-world word problem

Worked example Ex 8 · lift (elevator) trip

Ek lift rest se shuru hoti hai, $2$ s ke liye $1.5$ m/s² se upar accelerate karti hai, $5$ s tak constant speed se chalti hai, phir $3$ m/s² se rest tak decelerate karti hai. Woh kul kitna upar uthti hai?

Forecast: Cruise speed v–t tent ki peak hai. Guess karo: total $15$ m se zyada hai ya kam?

Cruise speed: $v = 0 + (1.5) (2) = 3$ m/s. Yeh step kyon? Ramp-up ke end ki velocity = a–t ke neeche area = $a t$ ; yeh flat top ki height ban jaati hai.
Decel time: $3$ m/s se $0$ tak $3$ m/s² pe: $t = 3/3 = 1$ s. Yeh step kyon? $Δ v = a t \Rightarrow t = Δ v / a$ ; aakhri triangle size karne ke liye zaroori.
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. Yeh step kyon? v–t ke neeche area = displacement, har phase mein.
Total height $= 3 + 15 + 1.5 = 19.5$ m.

Verify: Saari velocity positive hai (lift sirf upar jaati hai), toh distance = displacement = $19.5$ m. Units: $m/s \times s = m$ har jagah. ✓

Cell I — exam twist: ek graph, do questions

Worked example Ex 9 · wahi v–t, do tarike se answer

Ek particle ki velocity: $2$ s ke liye $v = + 10$ m/s, phir agle $4$ s mein linearly $- 10$ m/s tak ramp down, phir $1$ s ke liye $- 10$ m/s hold. (a) displacement aur (b) total distance nikalo.

Forecast: Ramp beech mein kahin zero cross karta hai. Guess karo kab $v = 0$ hota hai, aur kya displacement positive hai ya negative.

$v$ zero kahan cross karta hai? Ramp ke dauran, $v$ $4$ s mein $+ 10 \to - 10$ jaata hai ( $t = 2$ se $t = 6$ tak). Yeh linear hai, toh zero beech mein hai: $t = 4$ s pe. Yeh step kyon? Distance ke liye humein sign change pe split karna hoga — upar aur neeche ki area alag-alag handle karni hogi.
Signed areas (displacement):
- Flat top $t = 0..2$ : $(+ 10) (2) = + 20$ m.
- Axis ke upar ramp $t = 2..4$ : triangle $\frac{1}{2} (2) (10) = + 10$ m.
- Axis ke neeche ramp $t = 4..6$ : triangle $\frac{1}{2} (2) (10) = - 10$ m.
- Flat bottom $t = 6..7$ : $(- 10) (1) = - 10$ m.
- Displacement $= 20 + 10 - 10 - 10 = + 10$ m. Yeh step kyon? Net signed area = displacement; axis ke neeche ke pieces negative hote hain.
Distance = magnitudes ka sum: $∣20∣ + ∣10∣ + ∣ - 10 ∣ + ∣ - 10 ∣ = 50$ m. Yeh step kyon? Distance direction ignore karta hai — absolute areas add karo.

Verify: Displacement ( $+ 10$ m) $<$ distance ( $50$ m), aur yeh exactly is liye differ karte hain kyunki $v$ ka sign badla — rule ke saath consistent. Axis ke upar area $= 20 + 10 = 30$ , axis ke neeche $= 10 + 10 = 20$ ; net $= 30 - 20 = + 10$ ✓, total $= 30 + 20 = 50$ ✓.

Connections

Instantaneous vs Average Velocity — chord vs tangent slope Ex 1–2 mein dikhta hai.
Equations of Uniformly Accelerated Motion — Ex 5, Ex 8 ke liye cross-check.
Differentiation and Integration in Physics — slope ≡ derivative (Ex 2), area ≡ integral (Ex 3–5).
Free Fall and Projectile Motion — Ex 6 ka up-then-down v–t.
Distance vs Displacement — Ex 4 aur Ex 9 ka core.
Vectors — Components and Signs — kyun $v$ ka sign direction hai, size nahi.