Question bank — Equations of motion (SUVAT) — derivations from calculus

False ek equation ke roop mein — cut-out par acceleration change ho jaata hai, isliye poore trip mein $a$ constant nahi hai. Tumhe do constant-$a$ phases mein split karna hoga aur har ek par alag se SUVAT apply karna hoga.

False. $s$ displacement hai; seedha upar phénka gaya ball tumhare haath mein wapas aata hai $s=0$ ke saath, chahe usne upar aur wapas ek badi distance travel ki ho.

False. Negative $a$ ka matlab hai acceleration negative direction mein point kar raha hai; kya yeh speed up karta hai ya slow down, yeh velocity ke sign par depend karta hai. Ek ball neeche girte waqt, jab neeche negative chosen hai, negative $a$ hai lekin speed up ho rahi hai.

True — solve karne par milta hai $v=\pm \sqrt{u^2+2as}$. Dono roots physical hain agar object us displacement par kisi bhi direction mein move kar sakta ho (jaise ek ball us height par jaate waqt aur wapas aate waqt).

False. Yeh bas do definitions $v=\frac{ds}{dt}$, $a=\frac{dv}{dt}$ ke consequences hain jo constant $a$ ke under integrate kiye gaye hain — new physics nahi, sirf calculus bookkeeping.

False. Woh midpoint rule tab hi kaam karta hai jab $v$ linearly badhti hai, yaani constant $a$. Varying $a$ ke liye velocity–time graph curve karta hai aur average true integral hai, na ki endpoint midpoint.

False. Omitted variable woh hai jo us particular relation mein appear nahi karta; woh phir bhi exist karta hai aur kisi doosri equation se recover kiya ja sakta hai. "Missing $t$" ka matlab hai tum $t$ jaane bina solve kar sakte ho, na ki time matter karna band ho gaya.

False — yahi toh is ka poora point hai. Yeh $a$ ko omit karta hai, isliye tum do velocities aur time se displacement find kar sakte ho jab bhi acceleration ki value unknown ho (jab tak woh constant rahi ho).

Formula theek hai lekin direction galat hai: agar neeche positive hai toh $s$ ek neeche (positive) fall hai. "Upar" chosen sign convention se contradict karta hai. Signs padhne se pehle hamesha ek positive direction fix karo.

Error: peak par velocity momentarily zero hoti hai lekin acceleration poore time $g$ downward rehta hai. $a$ hai $v$ ka rate of change, aur $v$ zero se guzarte waqt sabse tezi se change ho rahi hai, isliye $a\ne 0$.

Error: braking motion ko oppose karti hai, isliye motion positive ke saath acceleration $a=-5$ hona chahiye. $+5$ use karna speed up describe karta hai, yahi wajah hai ki algebra ne ek nonsensical negative $s$ diya.

Error: SUVAT requires constant $a$, mean value nahi. $s=ut+\frac{1}{2}a{t}^2$ mein mean acceleration use karna silently linear-$v$ profile assume kar leta hai, jo air resistance violate karta hai — answer sirf ek rough estimate hai, exact nahi. Dekho Free fall and g ke liye jab approximation safe hai; wahan ki figure dikhati hai kaise real $v$–$t$ curve straight SUVAT line se bend away karta hai.

Error: SUVAT per-axis apply hota hai (Vectors). Horizontally $a=0$ hai isliye tum sirf horizontal velocity component use karte ho; vertical component alag equation mein jaata hai $a=-g$ ke saath. Dekho Projectile motion, jahan velocity arrow ko uske do component sketches mein split kiya gaya hai.

Error: Simple Harmonic Motion mein acceleration hai $a=-{\omega }^{2}x$, jo position ke saath change karta hai. Constant-$a$ formulas apply nahi hote; tumhe actual $a$ ko integrate karna hoga — iska $v$–$t$ graph ek sine wave hai, straight line nahi.

Kyunki yeh acceleration ko time ki bajaye position ke respect mein derivative ke roop mein rewrite karta hai, isliye $s$ ke over integrate karne par $t$ kabhi introduce nahi hota — time construction se eliminate ho jaata hai.

Agar $a$ constant hai toh yeh $\int adt$ se bahar factor out ho jaata hai, giving $at$; agar $a=a(t)$ hai toh yeh andar rehta hai aur integral woh antiderivative ban jaata hai jo $a(t)$ ke paas hai — generally ek neat SUVAT term nahi.

$ut$ woh displacement hai jo tum fixed initial speed $u$ par travel karte hue paate; $\frac{1}{2}a{t}^2$ woh extra displacement hai jo acceleration pile on karta hai. Inhe split karna constant-speed motion plus ek acceleration correction dikhata hai.

Limits initial condition ($v=u$ at $t=0$) ko directly encode karti hain, isliye constant already bake in hai — definite integration aur "constant + solve for C" dono same result dete hain. Dekho Differentiation and Integration.

Kyunki $ds=vdt$, isliye height $v$ aur width $dt$ ke tiny strips ke areas ko sum karna (unke areas) total displacement accumulate karta hai — woh sum hai integral $\int vdt$. Dekho Velocity-time graphs aur neeche figure mein shaded strip.

$v$–$t$ line seedhi height $u$ se height $v$ tak jaati hai, ek trapezium enclose karti hai jiske parallel sides $u,v$ hain aur width $t$ hai, jiska area $\frac{1}{2}(u+v)t$ exactly $s$ hai. Figure mein yeh trapezium shaded dikhaya gaya hai.

$a=0$ ke saath: $v=u$ (constant velocity), $s=ut$, aur $v^2=u^2$. Motion steady speed tak reduce ho jaati hai — yeh ek projectile ki horizontal axis hai.

Sab starting state dete hain: $v=u$, $s=0$, aur $v^2=u^2$ trivially. Ek sanity check ki $t=0$ ka matlab hai "abhi motion start nahi hua".

$v=at$, $s=\frac{1}{2}a{t}^2$, $v^2=2as$, aur $s=\frac{1}{2}vt$. Yeh classic dropped-stone / free-fall setup hai Free fall and g mein.

Peak par Velocity $v=0$ hoti hai, lekin acceleration $a=-g$ unchanged rehta hai aur displacement $s$ apne maximum (non-zero) par hota hai. Sirf $v$ vanish hoti hai; "instant of rest" acceleration ke na hone ka instant nahi hai.

Displacement $s=0$, isliye $v^2=u^2+0=u^2$, giving $v=-u$: same speed, wapsi par opposite direction. Equation cleanly symmetric landing speed predict karti hai.

Object us displacement se do baar guzarta hai — jaise ek ball ek height par jaate waqt aur wapas aate waqt. Dono times real events hain; woh choose karo jo us phase se match kare jisme tumhari interest hai.

Iska matlab hai object us displacement tak kabhi nahi pahunchta — jaise ek ball apni energy se allow kiye gaye se upar nahi uth sakta, isliye $u^2+2as<0$ "yeh position unreachable hai" signal karta hai na ki ek computational slip.