Graphs — x-t, v-t, a-t; areas and slopes meaning: Visual walkthrough

1.1.18 · D2 · Physics › Measurement, Vectors & Kinematics › Graphs — x-t, v-t, a-t; areas and slopes meaning

Step 0 — Teen words jo hum use karenge (pehle draw karo)

Kisi bhi graph se pehle, teen plain-English quantities fix karte hain aur unhe chhote names dete hain.

Hum sab kuch ek assumption se build karte hain: acceleration $a$ poore time constant rehta hai. Yahi uniformly accelerated motion ka case hai — ek ball seedhi ramp pe roll karti hai, ya ek stone free fall mein.

Step 1 — Acceleration graph banao (sabse flat possible picture)

Figure — Graphs — x-t, v-t, a-t; areas and slopes meaning

KYU hum yahan se shuru karte hain: acceleration teeno mein sabse simple hai jab wo constant ho — ek flat line. Flat line se hum $a \to v \to x$ ladder ko areas use karke upar chadh sakte hain, exactly jaisa parent ke mnemonic ne promise kiya tha.

PICTURE: burnt-orange line bilkul level hai. Uske neeche, time $0$ se time $t$ tak, ek shaded rectangle baitha hai. Uske do sides dekho: width $t$ (across), height $a$ (upar). Us rectangle ko apne dimag mein rakho — yahi Step 2 ka poora content hai.

Step 2 — Rectangle ka area hi velocity gained hai

KYA kiya humne: flat rectangle ka area compute kiya (rectangle ka area = height × width).

YEH velocity gained ke barabar kyun hai: parent note ne dikhaya tha ki $a$ – $t$ ke neeche area $v$ mein change ko accumulate karta hai. Yahan intuition yeh hai — har second tum apni speed mein $a$ metres/second add karte ho; $t$ seconds baad tumne use $t$ baar add kiya hai, jo $a t$ deta hai. ( $Δ$ bas "change in" ka shorthand hai.)

Yeh tool kyun, koi aur kyun nahi? Humne area choose kiya (slope nahi) kyunki hum ladder $a \to v$ pe upar ja rahe hain. Slopes hume sirf neeche le jaate hain. Flat rectangle ka area kisi calculus ki zaroorat nahi — bas length × width.

Ab, "velocity gained" plus "velocity jo shuru mein thi" se milti hai abhi ki velocity:

$v = start speed u + ab tak gained a t ⟹ v = u + a t$

PICTURE: Step 1 ka rectangle ab starting level $u$ ke upar stack ho gaya hai. Time $t$ par shaded region ki total height exactly $v$ hai.

Step 3 — "Velocity abhi" ko velocity graph mein badlo

Straight line kyun, curve kyun nahi: $v$ har second same amount ( $a$ ) se badh rahi hai — equal steps ek straight ramp banate hain. Is line ka slope $a$ hai (rise $Δ v$ over run $Δ t$ equals $a$ ), jo ek accha self-check hai: ladder $v \to a$ pe neeche slope lekar hamara constant $a$ wapas milta hai. Sab kuch consistent hai.

PICTURE: teal line $u$ se left pe shuru hokar $v = u + a t$ tak right pe jaati hai. Line aur time axis ke beech ka gap woh region hai jiska area hum aage attack karte hain — woh area distance travelled hai.

Step 4 — Distance, velocity line ke neeche ka area hai

Trapezium ka area nikalne ka clean tarika hai average height × width:

$Δ x = dono ends ka average (\frac{u + v}{2}) \times width t$

Term by term: $u$ left edge par height hai, $v$ right edge par height hai; unka average sloped roof ki typical height hai; width $t$ se multiply karna poora area sweep karta hai.

$x = \frac{u + v}{2} t$

PICTURE: trapezium shaded hai, uski left side $u$ label hai, right side $v$ label hai, aur ek dashed line average height ko unke beech mein mark karti hai.

Step 5 — Trapezium ko rectangle + triangle mein kaato (punchline)

Yeh cut kyun: yeh "distance jo tum apni starting speed par coast karte cover karte" aur "extra distance kyunki tum speed up hue" ko alag karta hai. Har piece ek aisi shape hai jo hum primary school geometry se measure kar sakte hain.

Rectangle — bottom slab: $area_{rect} = height u \times width t = u t$ Yahan tum hote agar acceleration zero hoti — pure coasting.

Triangle — top wedge. Uska base $t$ hai; uski height woh velocity gained hai jo humne Step 2 mein $a t$ find ki thi: $area_{tri} = \frac{1}{2} \times base t \times height = Δ v a t = \frac{1}{2} a t^{2}$

Dono pieces add karo: $x = coasting part u t + speeding-up part \frac{1}{2} a t^{2} ⟹ x = u t + \frac{1}{2} a t^{2}$

PICTURE: wahi trapezium, ab split — orange rectangle neeche, plum triangle upar, har ek apne area ke saath labelled. Yahi ek cut derivation hai. (Yeh Step 4 ke answer ke barabar hai agar $v = u + a t$ substitute karo; calculus route ke liye integration viewpoint dekho.)

Step 6 — Edge case A: rest se shuru karna ( $u = 0$ )

$x = u t^{0} + \frac{1}{2} a t^{2} = \frac{1}{2} a t^{2}$

KYU yeh matter karta hai: yeh free-fall-from-rest formula hai (with $a = g$ ). Picture se yeh obvious hai — koi bottom slab nahi, pure triangle. PICTURE: ek single plum triangle origin se rise karta hai; missing rectangle ko faint dotted outline se draw kiya gaya hai jo dikhata hai kya gayab ho gaya.

Step 7 — Edge case B: deceleration, phir reversing ( $a < 0$ )

KYU sign bookkeeping: axis ke upar area positive displacement hai (upar jaana); neeche area negative displacement hai (neeche aana). Wahi formula $x = u t + \frac{1}{2} a t^{2}$ yeh automatically handle karta hai, kyunki $a$ minus sign carry karta hai aur eventually negative $\frac{1}{2} a t^{2}$ , positive $u t$ ko overwhelm kar deta hai.

Sign rule (sab cases): object tab slow down ho raha hai jab $v$ aur $a$ ke opposite signs hoon, aur tab speed up ho raha hai jab unka sign same ho — dekho signs of vector components. Isliye "negative velocity" ≠ "slowing down."

PICTURE: teal line positive se shuru hoti hai, zero cross karti hai, negative ho jaati hai; upar triangle orange shaded (positive area) aur neeche triangle plum shaded (negative area). Agar dono triangles equal hain, net displacement zero hai — ball wahin wapas hai jahan se shuru hua tha — lekin total distance (dono areas ke magnitudes ka sum) zero nahi hai. Yeh distance-vs-displacement split Distance vs Displacement ka dil hai.

Step 8 — Pictures ka numeric check

Ek-picture summary

Recall Feynman retelling — walkthrough plain words mein

Ek car ka speedometer imagine karo jiska driver har second pedal ko thoda sa same amount press karta hai. Acceleration graph par woh steady press ek flat ceiling hai; uske neeche ka box, itne seconds chauda aur itna lamba, woh extra speed hai jo tumne pick up ki — isliye $v = u + a t$ . Ab speed ko khud draw karo: kyunki yeh equal steps mein badhti hai, yeh ek straight ramp draw karta hai jo tumhari opening speed $u$ se shuru hoti hai. Us ramp ke neeche ki zameen woh ground hai jo tum cover karte ho. Ramp ke across height $u$ par ek knife slide karo: neeche ek plain rectangle baitha hai — woh distance jo tum apni starting speed par coast karte cover karte, $u t$ — aur upar ek triangle ride karta hai — speed up hone se bonus distance, $\frac{1}{2} a t^{2}$ . Dono ko stack karo, $x = u t + \frac{1}{2} a t^{2}$ . Rest se shuru karo aur rectangle gayab ho jaata hai, sirf triangle $\frac{1}{2} a t^{2}$ bachta hai. Ball ko upar phenko aur ramp line ke neeche dip ho jaata hai: upar ki zameen upar jaana count hoti hai, neeche ki zameen neeche aana, aur agar dono balance karein tum wahin land karte ho jahan se shuru kiya tha — bahut kuch travel karke, displaced kuch bhi nahi.

Connections

Equations of Uniformly Accelerated Motion — yahan derive kiye gaye dono boxed results.
Differentiation and Integration in Physics — "area = integral" ke peeche calculus.
Instantaneous vs Average Velocity — average height $\frac{u + v}{2}$ average velocity hai.
Free Fall and Projectile Motion — Step 7 ka upward-throw picture with $a = - g$ .
Distance vs Displacement — kyun signed area ≠ total area.
Vectors — Components and Signs — $v$ aur $a$ ka $+ / -$ padhna.