Equations of motion (SUVAT) — derivations from calculus: Worked examples

1.1.16 · D3 · Physics › Measurement, Vectors & Kinematics › Equations of motion (SUVAT) — derivations from calculus

Scenario matrix

Kuch bhi work karne se pehle, yeh poora map hai. Hum har SUVAT problem ko case classes mein split karte hain — ek case class ka matlab sirf itna hai ki "problem ka genuinely alag flavour", ek jo alag trick ya watch-out maangta hai. Degenerate input ka matlab hai ki paanch quantities mein se ek kisi special value par set hai (jaise $u = 0$ ya $a = 0$ ) jisse kuch terms gayab ho jaate hain. Last column us example ka naam deta hai jo har row cover karta hai.

#	Case class	Kya alag banaata hai	Sign / edge trap	Example
A	Speeding up, sab positive	$u > 0, a > 0$	kuch nahi — "easy" base case	Ex 1
B	Rukne tak slow karna	$a < 0$ , $v = 0$ par khatam	$a$ opposes $v$ , toh $a$ negative hai	Ex 2
C	Rest se start karna	$u = 0$ (degenerate input: $u$ -term gayab)	$u$ -terms gayab ho jaate hain	Ex 3
D	Reversal: upar phir neeche	$u > 0, a < 0$ , $v = 0$ se guzarta hai	displacement ≠ distance; mid-flight sign flip	Ex 4
E	$t$ ke liye do roots	quadratic $s (t)$ ke do solutions	kaun sa root physical hai?	Ex 5
F	Limiting value / $a \to 0$	acceleration zero ki taraf shrink hoti hai (degenerate input)	formulas ko constant-speed motion par reduce karna chahiye	Ex 6
G	Real-world word problem	prose se SUVAT letters extract karne padte hain	pehle positive direction choose karo	Ex 7
H	Exam twist: do phases	accelerate, phir brake	SUVAT har phase par apply hota hai, across nahi	Ex 8

Hum neeche har row work karte hain.

Case A — base case (sab positive)

Worked example Ex 1 · Speeding up

Ek train station se $u = 8 m/s$ par chhootati hai aur $a = 1.5 m/s^{2}$ se $t = 12 s$ tak accelerate karti hai. Iski final speed $v$ aur tayi distance $s$ nikalo.

Forecast: speed $8$ se kaafi aage chadh jaani chahiye; distance $8 \times 12 = 96$ se zyada hogi (kyunki speed badh rahi hai). Numbers guess karo pehle padhne se.

$v$ ke liye equation choose karo. Missing variable $s$ hai → $v = u + a t$ use karo. Yeh step kyun? Hume $u, a, t$ pata hai aur $v$ chahiye; timeless aur average equations dono ko pehle kuch compute karna padega. $v = 8 + 1.5 \times 12 = 26 m/s$
$s$ ke liye equation choose karo. Ab hume $u, v, t$ pata hai → average-velocity form sabse saaf hai. Yeh step kyun? $s = \frac{1}{2} (u + v) t$ mein kuch re-square nahi karna padta. $s = \frac{1}{2} (8 + 26) \times 12 = \frac{1}{2} \times 34 \times 12 = 204 m$

Verify: $s$ ko $s = u t + \frac{1}{2} a t^{2} = 8 (12) + \frac{1}{2} (1.5) (144) = 96 + 108 = 204 m$ ✓ se cross-check karo. Aur $204 > 96$ forecast ke according hai. Units: $m/s \times s = m$ ✓.

Case B — rukne tak brake karna

Worked example Ex 2 · Slow karna (

$a$ ka sign) Ek cyclist $u = 12 m/s$ par $3 m/s^{2}$ se brake karta hai aur ruk jaata hai. Kitna time lagta hai aur kitna roll karte hain?

Forecast: rukne ka time kuch seconds hona chahiye; distance kuch dozen metres.

Positive direction = motion ki direction choose karo. Tab brake peeche push karta hai, toh $a = - 3 m/s^{2}$ . Yeh step kyun? SUVAT "brake" nahi jaanta — sirf signs jaanta hai. Deceleration chosen positive axis ke relative negative acceleration hai.
$t$ nikalo $v = u + a t$ se aur $v = 0$ (ruk gaya). Yeh step kyun? Hume time chahiye aur $u, v, a$ pata hai; yeh equation $s$ ko omit karti hai jo abhi chahiye nahi. $0 = 12 + (- 3) t \Rightarrow t = \frac{12}{3} = 4 s$
$s$ nikalo $v^{2} = u^{2} + 2 a s$ se, $v = 0$ . Yeh step kyun? Timeless form seedha speeds se distance tak jaata hai. $0 = 1 2^{2} + 2 (- 3) s \Rightarrow s = \frac{144}{6} = 24 m$

Verify: $s = \frac{1}{2} (u + v) t = \frac{1}{2} (12 + 0) (4) = 24 m$ ✓. Positive $s$ — achha hai, rukne ke dauran bhi aage hi move kar rahe hain.

Case C — rest se start karna (degenerate input $u = 0$ )

Worked example Ex 3 · Rest se drop karna

Ek ball drop ki jaati hai ( $u = 0$ ) aur $t = 2 s$ tak freely girti hai. Yahan $g$ gravitational acceleration ka magnitude hai, $g = 9.8 m/s^{2}$ (dekho Free fall and g). Is problem ke liye hum neeche ko positive choose karte hain, toh acceleration simply $a = + g = + 9.8 m/s^{2}$ hai. Iski speed aur kitni door girti hai nikalo.

Forecast: har second speed mein ~ $9.8$ add hota hai, toh ~ $20$ m/s; distance kahin $20$ m ke paas.

$u = 0$ set karo. $u$ waale har term gayab ho jaate hain. Yeh step kyun? "From rest" woh special case hai jo $u t$ term khatam kar deta hai — yeh wahi degenerate input hai jiske baare mein matrix warn karta hai.
Speed: $v = u + a t = 0 + 9.8 (2) = 19.6 m/s$ (positive = neeche, jaisa choose kiya).
Distance: $s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} (9.8) (2^{2}) = 19.6 m$ . Yeh step kyun? $u = 0$ ke saath yeh sirf $\frac{1}{2} a t^{2}$ tak collapse ho jaata hai.

Verify: timeless check $v^{2} = u^{2} + 2 a s \Rightarrow 19. 6^{2} = 0 + 2 (9.8) (19.6)$ . Left $= 384.16$ , right $= 384.16$ ✓.

Case D — reversal (displacement ≠ distance)

Yeh woh trap hai jo sabse zyada logon ko pakadti hai, isliye isko ek figure milta hai.

Worked example Ex 4 · Seedha upar throw karna

Ek patthar upar $u = 20 m/s$ se throw kiya jaata hai. Yahan hum upar ko positive choose karte hain, toh gravity (jo neeche kheenchti hai) $a = - g = - 9.8 m/s^{2}$ deti hai. Nikalo (a) top tak time, (b) max height, (c) $t = 3 s$ par velocity, (d) $t = 3 s$ par displacement.

Forecast: ~2 s chadha, ~20 m par peak, aur 3 s tak already neeche aa raha hai.

Top tak time: peak par patthar momentarily ruka hota hai, $v = 0$ . $v = u + a t$ use karo. Yeh step kyun? Top $v = 0$ se define hota hai; yahi hamaari target condition hai. $0 = 20 + (- 9.8) t \Rightarrow t = \frac{20}{9.8} = 2.0408 s$
Max height: $v^{2} = u^{2} + 2 a s$ ke saath $v = 0$ . $0 = 2 0^{2} + 2 (- 9.8) s \Rightarrow s = \frac{400}{19.6} = 20.408 m$
$t = 3$ s par velocity: $v = 20 + (- 9.8) (3) = - 9.4 m/s$ . Yeh step kyun? Negative sign batata hai ki ab yeh neeche move kar raha hai — formula automatically direction track karta hai. Figure mein red arrow dekho: woh flip ho gaya hai.
$t = 3$ s par displacement: $s = u t + \frac{1}{2} a t^{2} = 20 (3) + \frac{1}{2} (- 9.8) (9) = 60 - 44.1 = 15.9 m$ . Yeh step kyun? $s$ launch point se measure hota hai. Yeh peak height se kam hai kyunki patthar wapas aana shuru ho gaya hai — displacement total path length nahi hai.

Verify: $t = 3$ s par patthar $+ 15.9$ m par hai (haath ke upar abhi bhi) phir bhi velocity $- 9.4$ hai (neeche ja raha hai) — "top ke baad, gir raha hai" ke saath consistent hai. Distance travelled $= 20.408$ upar $+ (20.408 - 15.9) = 4.508$ neeche $= 24.9$ m, jo $15.9$ m displacement se zyada hai ✓ — exactly distance-vs-displacement point.

Case E — do roots: kaun sa time physical hai?

Worked example Ex 5 · Kisi dii hui height par kab pahunchta hai?

Wahi throw use karte hue ( $u = 20$ , $a = - 9.8$ , up positive), patthar height $s = 15 m$ par kab hai?

Forecast: do answers — ek baar jaate hue, ek baar aate hue.

Quadratic set up karo: $s = u t + \frac{1}{2} a t^{2}$ . $15 = 20 t + \frac{1}{2} (- 9.8) t^{2} \Rightarrow 4.9 t^{2} - 20 t + 15 = 0$ Yeh step kyun? Height as a function of time ek parabola hai, toh " $s = 15$ kab hai?" poochna hai ki parabola ek horizontal line ko kahan hit karta hai — generally do crossings hoti hain.
Quadratic formula se solve karo $t = \frac{20 \pm 2 0 ^{2} - 4 ( 4.9 ) ( 15 )}{2 ( 4.9 )}$ . Yeh step kyun? Equation $t$ mein quadratic hai (isme $t^{2}$ term hai), aur quadratic formula general tool hai jo dono roots ek saath return karta hai — exactly wahi chahiye jab do crossing times suspect ho. Kuch aur (linear rearrangement) dono nahi dhundh paata. $t = \frac{20 \pm 400 - 294}{9.8} = \frac{20 \pm 106}{9.8}$ $t_{1} = \frac{20 - 10.2956}{9.8} = 0.9902 s, t_{2} = \frac{20 + 10.2956}{9.8} = 3.0914 s$
Roots turant screen karo. $t_{1}$ aur $t_{2}$ dono positive aur real hain, toh dono launch ke baad hote hain — yahan koi discard nahi. (Agar koi root negative aata, toh hum use turant throw kar dete: negative time matlab "throw se pehle", jo hua hi nahi.) Yeh step kyun? Root selection ek physics decision hai, algebra nahi: sirf wahi times rakho jo us motion ke andar fall karte hain jise hum describe kar rahe hain.
Survivors interpret karo. $t_{1} \approx 0.99$ s = $15$ m se upar jaate hue guzarna; $t_{2} \approx 3.09$ s = same height se neeche aate hue guzarna.

Verify: inka sum $2 \times$ (time to top) ke barabar hona chahiye parabola ki symmetry se: $0.9902 + 3.0914 = 4.0816 = 2 \times 2.0408$ ✓.

Case F — limiting behaviour ( $a \to 0$ )

Worked example Ex 6 · Agar (almost) koi acceleration na ho?

Ek puck $u = 6 m/s$ par slide karta hai jisme thodi si deceleration $a = - 0.01 m/s^{2}$ hai $t = 5 s$ ke liye. $s$ nikalo, aur dikhao ki $a \to 0$ hone par result plain constant-speed motion se match karta hai.

Forecast: itne chhote $a$ ke saath, distance ≈ $6 \times 5 = 30$ m, thodi si kam.

Full formula: $s = u t + \frac{1}{2} a t^{2} = 6 (5) + \frac{1}{2} (- 0.01) (25) = 30 - 0.125 = 29.875 m$ .
Limit lo $a \to 0$ : $\frac{1}{2} a t^{2}$ term gayab ho jaata hai, $s \to u t = 30 m$ bachta hai. Yeh step kyun? Constant acceleration mein zero acceleration bhi ek boundary case hai. Yeh check karna ki formula gracefully $s = u t$ tak degrade hoti hai ek sanity check hai ki hamaari physics consistent hai — dekho Velocity-time graphs, jahan $a = 0$ ek flat horizontal line hai aur area sirf rectangle $u \times t$ hai.

Verify: correction term $\frac{1}{2} ∣ a ∣ t^{2} = 0.125$ m, $30$ m ka $0.4%$ hai — chhota, jaisa "nearly-frictionless" puck hona chahiye ✓.

Case G — word problem se SUVAT extract karna

Worked example Ex 7 · Overtaking lorry

Ek car steadily $25 m/s$ par chal rahi hai aur ek stationary police bike ke paas se guzarti hai. Jis instant car guzarti hai, bike rest se $a = 4 m/s^{2}$ se chal padti hai. Bike car ko pakadne mein kitna time lagega, aur kitni door tak gaye?

Forecast: bike zyada tezi se accelerate karti hai lekin zero se start karti hai, toh pakadne mein kai seconds lagte hain.

Positive direction = road aage choose karo. Har vehicle ka SUVAT list karo. Yeh step kyun? Do objects matlab do motions; ek shared positive direction positions directly compare karne deta hai.
- Car: $u = 25$ , $a = 0$ → position $s_{car} = 25 t$ .
- Bike: $u = 0$ , $a = 4$ → position $s_{bike} = \frac{1}{2} (4) t^{2} = 2 t^{2}$ .
Catch-up matlab equal positions: $s_{bike} = s_{car}$ . Yeh step kyun? "Pakad lena" ek position condition hai, speed nahi. $2 t^{2} = 25 t \Rightarrow t (2 t - 25) = 0 \Rightarrow t = 0 or t = 12.5 s$
$t = 0$ discard karo (woh guzarne ka instant hai) → catch-up $t = 12.5 s$ par.
Distance: $s = 25 \times 12.5 = 312.5 m$ .

Verify: bike ki distance $= 2 (12.5)^{2} = 2 \times 156.25 = 312.5 m$ — match karta hai ✓. Catch-up par bike ki speed $= a t = 4 (12.5) = 50 m/s$ , car ki double — plausible ki woh pakad payi.

Case H — exam twist: do phases

Worked example Ex 8 · Accelerate phir brake

Ek drone rest se start karta hai, $2 m/s^{2}$ se $6 s$ tak accelerate karta hai, phir turant $3 m/s^{2}$ se rukne tak brake karta hai. Total distance nikalo.

Forecast: ek decent speed tak pahunchta hai, phir rukne mein kam distance lagti hai (hard braking).

Phase 1 (accelerate). $u_{1} = 0, a_{1} = 2, t_{1} = 6$ . Split kyun? SUVAT ko ek time par ek constant $a$ chahiye. Switch par acceleration change ho jaata hai, toh har leg alag treat karo — yahi is cell ka poora point hai. $v_{1} = 0 + 2 (6) = 12 m/s, s_{1} = \frac{1}{2} (0 + 12) (6) = 36 m$
Phase 2 (brake). Phase 1 ki end speed hi phase 2 ki start speed hai: $u_{2} = 12$ , $a_{2} = - 3$ , $v_{2} = 0$ par khatam. Yeh step kyun? Motion continuous hai — velocity carry over hoti hai bhale hi acceleration jump kare. $0 = 1 2^{2} + 2 (- 3) s_{2} \Rightarrow s_{2} = \frac{144}{6} = 24 m$
Total distance: $s = s_{1} + s_{2} = 36 + 24 = 60 m$ .

Verify: braking time $t_{2} = \frac{12}{3} = 4$ s; har phase ki average speed $6$ m/s hai, toh $s_{1} = 6 \times 6 = 36$ , $s_{2} = 6 \times 4 = 24$ — dono match ✓. Total time $= 10$ s.

Recall Kaun sa cell kaun sa hai?

Speeding up all-positive kaun sa cell hai ::: A (Ex 1). Woh cell jahan displacement distance se differ karta hai ::: D (Ex 4, upar throw kiya). Woh cell jo ek hi height ke liye do valid times deta hai ::: E (Ex 5). Jab $a \to 0$ , $s = u t + \frac{1}{2} a t^{2}$ reduce ho jaata hai ::: $s = u t$ (cell F) par. "Accelerate then brake" mein SUVAT apply karna padhta hai ::: ek baar har phase pe (cell H).

Connections

Equations of motion (SUVAT) — derivations from calculus — parent note; yahan use kiya gaya har formula wahan derive hota hai.
Velocity-time graphs — cells A–F sirf $v$ – $t$ plot par areas aur slopes hain.
Vectors — cells B aur D mein signs positive axis direction choose karne se aate hain.
Projectile motion — cell D ek projectile ka vertical half hai.
Free fall and g — cell C mein $a = g$ use hota hai.
Simple Harmonic Motion — ek reminder ki agar $a$ constant nahi hoti, toh in cells mein se koi bhi apply nahi hota.