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

Sach — ek seedhi line ki slope constant hoti hai, aur $x$–$t$ ki slope velocity hai, isliye velocity nahi badlti (chahe line flat ho, jisme $v=0$ mile).

Jhooth — flat $x$–$t$ line ki slope $0$ hai, isliye $v=0$; object at rest hai, constant speed par chal nahi raha.

Sach — flat $v$–$t$ ka matlab hai $v$ ki value nahi badal rahi, yani constant velocity (aur uski slope $0$ matlab $a=0$).

Jhooth — position ko time ke upar integrate karne se units $\text{m}\cdot \text{s}$ milti hain, jo physically meaningless hai; $x$–$t$ ke liye slope lete hain, area nahi.

Jhooth — negative area ka matlab hai net displacement $-x$ direction mein hai; object zaroor move kiya, bas overall peeche ki taraf gaya.

Sach — equal positive aur negative areas cancel ho ke displacement $0$ dete hain, lekin unke magnitudes ka sum (distance) bada ho sakta hai.

Sach — agar $v\ge 0$ puri tarah, velocity kabhi reverse nahi hoti, isliye koi area subtract nahi hota aur net area total area ke barabar hota hai.

Jhooth — graph sirf position mein changes fix karta hai (areas); alag starting positions se poora $x$–$t$ curve upar ya neeche shift ho jaata hai.

Sach — speed slope ka magnitude hai, isliye ek steep line (upar ya neeche) tez motion ka matlab hai; sirf direction alag hoti hai.

Jhooth — $a=0$ ka matlab hai velocity constant hai; woh constant koi bhi nonzero value ho sakti hai, isliye object steadily move kar sakta hai.

$v$ ka sign direction hai, deceleration nahi; yeh slow ho raha hai ya nahi yeh slope ($a$) aur $v$ ke sign ke comparison pe depend karta hai, na ki $v$ ke negative hone pe.

Galat operation — velocity $x$–$t$ ki slope hoti hai; $x$–$t$ ke neeche ke area ka koi physical meaning nahi hai (units $\text{m}\cdot \text{s}$).

Height instantaneous acceleration hai; $\Delta v$ $a$–$t$ ke neeche ka area hai, yani time ke saath accumulated acceleration.

Top par $v=0$ hota hai lekin $a=-g$ puri tarah rehta hai; $v$–$t$ ki slope zero ke through unbroken rehti hai, isliye acceleration kabhi zero nahi hoti.

Upar ki taraf curvature ka matlab $a>0$ hai; yeh motion tab hi speed up karta hai jab $v$ bhi $+$ ho. Agar object $-x$ mein move kar raha hai ($v<0$), toh $a>0$ actually use slow down karta hai.

Displacement ke liye aap subtract karte ho (signed), lekin distance ke liye uska magnitude add karte ho — distance direction ignore karti hai.

Slope acceleration $a$ deta hai; us trapezium ka area displacement deta hai — slopes aur areas alag operations hain.

Uski slope $a$ deti hai aur uska area $\Delta x$ deta hai — dono operations physically meaningful quantities dete hain, unlike $x$–$t$ (sirf slope) ya $a$–$t$ (sirf area). Dekho Differentiation and Integration in Physics.

Slope lena tumhe chain $x\to v\to a$ mein neeche le jaata hai (differentiation), jabki area accumulate karna tumhe wapas $a\to v\to x$ le jaata hai (integration) — opposite operations, opposite directions.

Har thin strip $vdt$ hai, aur $v$ ek sign carry karta hai, isliye backward motion negative strips contribute karta hai; signed sum net displacement hai. Dekho Distance vs Displacement.

Constant acceleration ke saath $v$–$t$ graph ek seedhi sloped line hai, isliye do times ke beech uske neeche ka region ek trapezium hai (average height $\times$ width). Dekho Equations of Uniformly Accelerated Motion.

$v$ ka sign motion ko point karta hai; $|v|$ badhta hai ya nahi yeh depend karta hai ki $a$ motion ke saath push karta hai ($v,a$ same sign) ya against ($v,a$ opposite sign).

Curvature ka matlab hai slope (velocity) change ho rahi hai, aur changing velocity exactly wohi hai jo acceleration measure karta hai.

Velocity wahan momentarily zero hai (jaise throw ke top par), lekin agar slope nonzero hai toh object turant reverse ho jaata hai — woh rest nahi karta. Dekho Free Fall and Projectile Motion.

Zero — positive $v$–$t$ area (upar jaate hue) exactly equal negative area (wapas starting point par neeche aate hue) ko cancel karta hai.

Maximum height se do guna — aap upar aur neeche ke equal areas ke magnitudes add karte ho.

Zero area ka matlab zero $\Delta v$ hai; ek genuine velocity jump ke liye finite area chahiye, isliye infinitely thin spike kuch nahi badalta.

Us instant par undefined — slope jump karta hai, matlab velocity discontinuously change hoti hai, jiske liye infinite (idealised) acceleration chahiye. Dekho Instantaneous vs Average Velocity.

Yeh infinite slope imply karta hai, yani infinite velocity — physically impossible, isliye real $x$–$t$ graphs kabhi vertical nahi hote.

Woh magnitude mein equal ho jaate hain, kyunki koi area subtract nahi hota; $v$ mein sign changes hi woh cheez hai jo unhe alag karti hai. Dekho Vectors — Components and Signs.