Equations of motion (SUVAT) — derivations from calculus: Visual walkthrough

1.1.16 · D2 · Physics › Measurement, Vectors & Kinematics › Equations of motion (SUVAT) — derivations from calculus

Step 1 — Velocity–time graph hota kya hai

KYA HAI. Do number lines draw karo jo ek corner par milti hain. Horizontal wali time $t$ measure karti hai (seconds, left se right). Vertical wali velocity $v$ measure karti hai (metres per second, upar ki taraf). Is grid par ek dot kehta hai "is waqt par, object is speed se move kar raha tha".

KYUN. Pehle "graph ke neeche area" ki baat karne se pehle, humein graph chahiye. Velocity woh cheez hai jo hum curve ki tarah draw kar sakte hain kyunki yeh time ke saath badlti hai — aur us change ki picture exactly ek velocity–time graph hai.

PICTURE. Figure mein, blue dot time $t_{0}$ ke upar height $v_{0}$ par baitha hai. Ise map ki tarah padho: time tak right jao, speed tak upar jao.

Figure — Equations of motion (SUVAT) — derivations from calculus

Step 2 — Constant acceleration straight line kyun banata hai

KYA HAI. Acceleration $a$ hai "har second mein kitni velocity gain hoti hai". Constant $a$ ka matlab har single second mein velocity ka same chunk gain hona. Plot karo: height $u$ se shuru karo (initial velocity, $t = 0$ par humari speed), phir har second dot $a$ se upar uthta hai. Dots ko connect karo — ek straight, tilted line milti hai.

KYUN. Straight line sabse simple possible graph hai, aur iska steepness hi acceleration hai. Hum $a$ = constant isliye use karte hain kyunki yeh graph ko ek line banata hai, aur line ke neeche ka area ek shape hai jo hum primary school se jaante hain (rectangles aur triangles). Line nahi, toh easy area bhi nahi.

PICTURE. Line vertical axis par yellow dot $(0, u)$ se shuru hoti hai aur steadily climb karti hai. Iska steepness — har ek step right mein kitna rise hota hai — $a$ hai. Zyada steep line ka matlab bada acceleration.

Step 3 — Claim: line ke neeche area = displacement

KYA HAI. Displacement $s$ hai object ne kitna move kiya (direction ke saath). Jo claim hum build karenge, assume nahi, woh yeh hai: line aur time axis ke beech ka shaded region $s$ ke barabar hai.

KYUN. Distance = speed × time. Agar speed kabhi nahi badlti, toh yeh sirf ek rectangle hoga (speed tall, time wide) — aur rectangle ka area height × width = speed × time = distance. Toh area already hi distance hai jab speed constant ho. Neeche poori trick woh case handle karne ke liye hai jahan speed badhti rehti hai.

PICTURE. Sloped line ke neeche ka region green shade hai. Hum iska area do alag tareekon se measure karenge aur dikhayenge dono se same $s$ milta hai.

Step 4 — "Area = distance" changing speed ke liye bhi kyun (thin-strip idea)

KYA HAI. Time axis ko bahut pateeli vertical strips mein kato, har ek ki width $d t$ ho (time ka ek chhota sa sliver — itna chhota ki speed us par barely badlti ho). Ek sliver mein speed basically height $v$ par constant hai, toh us sliver ki distance height × width $= v d t$ hai. Har sliver add karo total milta hai.

KYUN. Yeh woh single idea hai jiske peeche integration hai: curvy area ko ek saath grab karna impossible hai, lekin rectangle trivial hai. Toh hum impossible shape ko laakhon trivial rectangles mein kaatate hain aur sum karte hain. Infinitely many infinitely thin strips add karna exactly woh hai jo integral sign $\int$ ka matlab hai.

PICTURE. Width $d t$ ki ek highlighted red strip line ke neeche height $v$ par baitha hai. Iska tiny area $v d t$ total distance ka ek crumb hai; saare crumbs ka sum Step 3 ke green region ko bhar deta hai.

Step 5 — Region ko ek rectangle + ek triangle mein split karo

KYA HAI. Strips ek ek karke sum karne ki jagah, notice karo ki shaded region do simple pieces se bani hai jo stack hain:

ek rectangle neeche baitha, height $u$ , width $t$ ;
ek triangle upar rakha, rectangle ke flat top aur sloped line ke beech ka gap fill karta hua.

KYUN. Hum in do shapes ke areas dil se jaante hain. Yeh ek integral ko schoolroom geometry mein badal deta hai — aur mysterious $\frac{1}{2} a t^{2}$ term literally ek triangle ke area ki tarah appear hota hai, jo is poore page ka payoff hai.

PICTURE. Rectangle blue shaded hai ("agar speed $u$ hamesha rehti" wali distance). Triangle yellow shaded hai ("extra distance jo acceleration ne add ki"). Saath mein yeh green region ko exactly tile karte hain.

Step 6 — Har piece measure karo aur add karo

KYA HAI. Ab picture se dimensions plug in karo.

Rectangle: height $u$ , width $t$ → area $= u \times t = u t$ .
Triangle: iska base time axis ke along width $t$ hai; iska height hai kitni velocity us time mein rise hui, jo $a t$ hai ( $v = u + a t$ se, line $a t$ se climb hui). Triangle area $= \frac{1}{2} \times base \times height = \frac{1}{2} t (a t) = \frac{1}{2} a t^{2}$ .

KYUN. Dono areas add karne se total shaded region milta hai, jo Step 3 se hi displacement $s$ hai. Ek baar shapes naam ho jayein toh koi calculus symbols nahi chahiye — Step 4 ka integral aur Step 5 ki geometry identical answers dete hain.

PICTURE. Har shape ki har side par uski length label ki gayi hai exactly wahan: rectangle ka $u$ aur $t$ , triangle ka base $t$ aur height $a t$ .

Step 7 — Har case, picture par check kiya

Picture ko saare inputs mein survive karna chahiye, toh har knob ghuma ke dekhte hain.

KYUN karo yeh. Ek derivation jis par aap sirf ek friendly example ke liye trust kar sako, woh worthless hai. Hum degenerate aur sign-flipped cases directly shapes par test karte hain.

Case $a = 0$ (koi acceleration nahi). Line flat ho jaati hai, triangle zero height par collapse ho jaata hai. Sirf blue rectangle bachta hai: $s = u t$ . Correct — $u$ par coasting.

Case $u = 0$ (rest se gira). Rectangle gayab ho jaata hai (zero height). Sirf yellow triangle bachta hai: $s = \frac{1}{2} a t^{2}$ . Yeh $a = g$ ke saath free fall hai.

Case $a < 0$ (braking). Line neeche slope karti hai. Triangle ab flat top ke neeche baitha hai aur iska area subtract hota hai — yeh rectangle mein se khaata hai. Toh $s$ , $u t$ se kam hai: aap utna nahi gaye jitna speed hold karte toh jaate. Agar kaafi brake karo toh line axis cross karti hai; axis ke neeche area negative displacement hai (backwards move karna).

Case bahut bada $t$ . $\frac{1}{2} a t^{2}$ triangle $t^{2}$ ki tarah grow karta hai jabki rectangle sirf $t$ ki tarah grow karta hai. Eventually triangle dominate karta hai — acceleration hamesha enough time milne par head-start ko beat kar deta hai.

Ek-picture summary

Ek figure poori kahani rakhti hai: tilted line $v = u + a t$ , blue rectangle $u t$ , aur yellow triangle $\frac{1}{2} a t^{2}$ stacked jo us area ko fill karte hain jo hi displacement hai.

Recall Feynman retelling — plain words mein walkthrough

Draw karo koi cheez kitni fast ja rahi hai, second by second, page par upar ki taraf jaate hue jab time right ko run kare. Agar ise evenly push kiya jaaye, woh drawing ek straight slanted line hai. Ab yeh magic hai: us line ke neeche floor space exactly hai kitni door woh cheez gayi. Kyun? Floor ko pateeli vertical splinters mein kato — har splinter hai "kitni fast × time ka ek blink" = ek tiny distance — aur sab pile up karo.

Woh floor space sirf ek box hai jis par ek triangle baitha hai. Box hai "wahan aap hote agar aapne kabhi speed up nahi ki" = starting-speed times time = $u t$ . Upar ka triangle speed up karne ka bonus hai; ek triangle half rectangle hota hai, iska width time $t$ hai aur iska height extra speed gained $a t$ hai, toh iska area $\frac{1}{2} t \cdot a t = \frac{1}{2} a t^{2}$ hai. Box plus triangle add karo: $s = u t + \frac{1}{2} a t^{2}$ . Agar koi push nahi, triangle gayab. Agar rest se start karo, box gayab. Agar brake kar rahe ho, triangle box mein se ek bite ban jaata hai. Har baar same picture.

Recall

$v$ – $t$ graph par, area kiske barabar hai? ::: Displacement $s$ . Extra-distance term mein $t^{2}$ kyun hai? ::: Yeh ek triangle hai jiska base aur height dono $t$ ke saath badhte hain, isliye iska area $t^{2}$ ke saath badhta hai. Blue rectangle physically kya represent karta hai? ::: Constant starting speed $u$ par cover ki gayi distance, yaani $u t$ . $u = 0$ ke saath, $s = u t + \frac{1}{2} a t^{2}$ kya ban jaata hai aur kyun? ::: $s = \frac{1}{2} a t^{2}$ — rectangle ki height zero hai, sirf triangle bachta hai. $a = 0$ ke saath, graph par kya bachta hai? ::: Line flat hai, triangle gayab, $s = u t$ bachta hai.

Connections

Equations of motion (SUVAT) — derivations from calculus — parent; yeh iska picture-proof hai.
Velocity-time graphs — poora page ek par hi jeeta hai.
Differentiation and Integration — thin strips ka area mein sum hona integration hai.
Free fall and g — $u = 0$ case, $s = \frac{1}{2} g t^{2}$ .
Projectile motion — yeh per-axis apply karo.
Vectors — yaad raho $s, u, v, a$ sign/direction carry karte hain.