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

List: $u=3$, $a=2$, $t=5$, chahiye $v$, $s$ ki parwah nahi. Missing letter: $s$ → use $v=u+at$. Yeh kyun? Yeh woh akela equation hai jisme $s$ nahi hai, toh $s$ kabhi guess nahi karna padta. $v=3+(2)(5)=13\text{m/s}$

List: $s=60$, $u=4$, $v=16$, chahiye $t$, aur $a$ na diya gaya hai na chahiye. Missing letter: $a$ → use $s=\frac{1}{2}(u+v)t$. Kyun? Isme $a$ nahi aata, toh hume kabhi uski zaroorat nahi padti. $60=\frac{1}{2}(4+16)t=10t\Rightarrow t=6\text{s}$

List: $u=30$, $v=0$, $s=90$, chahiye $a$, time $t$ nahi diya. Missing letter: $t$ → use $v^2=u^2+2as$. $0=30^2+2a(90)\Rightarrow 0=900+180a\Rightarrow a=-5{\text{m/s}}^2$ Minus ka matlab hai acceleration motion ke opposite direction mein point kar raha hai — braking, bilkul waisi hi jaisi expect thi.

List: $u=20$, $a=-9.8$, top pe $v=0$, chahiye $s$, time $t$ nahi diya. Missing letter: $t$ → use $v^2=u^2+2as$. $0=20^2+2(-9.8)s\Rightarrow 0=400-19.6s\Rightarrow s=\frac{400}{19.6}=20.41\text{m}$ $a$ negative kyun? Upar hamari positive direction hai; gravity neeche kheenchti hai, toh woh velocity se ghataati hai. Agar $a=+9.8$ likhte toh negative height milti — ek bakwaas sign jo galti batata.

List: $u=20$, $a=-9.8$, $s=0$ (same height pe wapas), chahiye $t$, final $v$ nahi chahiye. Missing letter: $v$ → use $s=ut+\frac{1}{2}a{t}^2$. $0=20t+\frac{1}{2}(-9.8)t^2=20t-4.9t^2=t(20-4.9t)$ Do solutions: $t=0$ (phenkne ka waqt) aur $20-4.9t=0\Rightarrow t=\frac{20}{4.9}=4.08\text{s}$. Do answers kyun? Equation poori height vs time parabola describe karti hai; woh $s=0$ ko do baar cross karti hai — jaate waqt aur aate waqt. Hume doosra crossing chahiye.

List: $u=25$, $v=0$, $t=20$, chahiye $s$, $a$ nahi chahiye. Missing letter: $a$ → use $s=\frac{1}{2}(u+v)t$. $s=\frac{1}{2}(25+0)(20)=\frac{1}{2}(25)(20)=250\text{m}$ Yeh bas $v$–$t$ graph ke neeche triangle ka area hai — dekho Velocity-time graphs.

List: $u=5$, $a=9.8$, $s=40$, chahiye $v$, time $t$ nahi diya. Missing letter: $t$ → use $v^2=u^2+2as$. $v^2=5^2+2(9.8)(40)=25+784=809$ $v=\sqrt{809}=28.44\text{m/s}$ Hum positive root lete hain kyunki pathar poori journey mein positive (neeche wali) direction mein move karta hai — koi turning point nahi aata. (Negative root ek aise paththar ko describe karta jo upar se launch hua aur us speed se aaya — yeh woh scenario nahi hai.)

List: $u=0$ (rest se), $v=24$, $t=12$, chahiye $a$ phir $s$. (a) Rest se straight line: slope $a=\frac{v-u}{t}=\frac{24-0}{12}=2{\text{m/s}}^2$. (b) Pehle 6 s mein: $u=0$, $a=2$, $t=6$, chahiye $s$ → missing $v$ → $s=ut+\frac{1}{2}a{t}^2$. $s=0+\frac{1}{2}(2)(6^2)=36\text{m}$ Graph par yeh $t=6$ tak line ke neeche shaded triangle hai: base $6$, height $v(6)=2\times 6=12$, area $\frac{1}{2}(6)(12)=36\text{m}$. ✓ Dono routes agree karte hain.

List: $u=-8$, $a=2$; part (a) chahta hai time jab $v=0$ ho; part (b) chahta hai $s$ at $t=8$. (a) Rest matlab $v=0$. Use $v=u+at$: $0=-8+2t\Rightarrow t=4\text{s}$. Object pehle slow hota hai, $t=4$ par rukta hai, phir reverse karta hai. (b) $s=ut+\frac{1}{2}a{t}^2=(-8)(8)+\frac{1}{2}(2)(8^2)=-64+64=0\text{m}$. $t=8$ par displacement zero hai: woh sabse door ke point tak gaya (at $t=4$, $s=-8(4)+\frac{1}{2}(2)(16)=-32+16=-16\text{m}$) phir bilkul start par wapas aa gaya. Distance travelled $32\text{m}$ hai lekin displacement $0$ hai — parent note se distance/displacement distinction.

List: $u=10$; 5th second mein doori $22$ hai; chahiye $a$. $n$-th second mein doori = ($n$ tak doori) minus ($n-1$ tak doori). $s=ut+\frac{1}{2}a{t}^2$ use karo $t=5$ aur $t=4$ par: $s_5=u(5)+\frac{1}{2}a(5{)}^2=5u+\frac{25}{2}a$ $s_4=u(4)+\frac{1}{2}a(4{)}^2=4u+\frac{16}{2}a$ Ab term by term subtract karo — $u$-terms dete hain $5u-4u=u$, aur $a$-terms dete hain $\frac{1}{2}a(25-16)=\frac{9}{2}a$: $s_5-s_4=(5u-4u)+\frac{1}{2}a(25-16)=u+\frac{9}{2}a$ Isko given $22$ m ke barabar set karo aur solve karo: $22=10+4.5a\Rightarrow 4.5a=12\Rightarrow a=\frac{12}{4.5}=2.667{\text{m/s}}^2$

List (phase 1): $u=0$, $a_1=3$, $t_1=4$, chahiye top speed $v_1$ aur distance $s_1$. Top speed $v_1=u+a_1{t}_1=0+3(4)=12\text{m/s}$. Distance: $s_1=\frac{1}{2}(u+v_1)t_1=\frac{1}{2}(0+12)(4)=24\text{m}$. List (phase 2): speed constant $v_1=12$ par, toh $a=0$, $t_2=6$, chahiye $s_2$. $s_2=v_1{t}_2=12\times 6=72\text{m}$. Total: $s=24+72=96\text{m}$. Split kyun kiya? SUVAT ko $a$ chahiye jo ek phase ke andar constant ho. Acceleration $t=4$ s par change hota hai, toh hum har stretch alag alag treat karte hain aur phase 1 ki top speed phase 2 ki (constant) speed ban jaati hai.

List (window ke across): $s=1.5$, $t=0.2$, $a=9.8$, unknown entry speed $u=v_{\text{top}}$. $s=ut+\frac{1}{2}a{t}^2$ use karo $u=v_{\text{top}}$ ke saath: $1.5=v_{\text{top}}(0.2)+\frac{1}{2}(9.8)(0.2^2)=0.2v_{\text{top}}+0.196$ $0.2v_{\text{top}}=1.304\Rightarrow v_{\text{top}}=6.52\text{m/s}$ List (release se window top tak): $u=0$ (rest se), $v=6.52$, $a=9.8$, chahiye $h$, time unknown. Missing letter: $t$ → use $v^2=u^2+2as$: $6.52^2=0^2+2(9.8)h\Rightarrow h=\frac{42.5104}{19.6}=2.169\text{m}$ Toh yeh window ke top se lagbhag $2.17\text{m}$ upar release hua tha.

List: height $y$ ground se measure karo, upar positive. Dono balls share karte hain $a=-9.8$.

• Ball A: $y_A=25t-4.9t^2$ ($0$ se shuru, upar jaati hai).
• Ball B: $y_B=40+0\cdot t-4.9t^2=40-4.9t^2$ ($40$ se shuru, drop ki gayi). Woh milte hain jab $y_A=y_B$: $25t-4.9t^2=40-4.9t^2$ $-4.9t^2$ terms cancel ho jaate hain — dono ek jaisi gravity feel karte hain, toh unka gap constant approach rate $25\text{m/s}$ se close hota hai: $25t=40\Rightarrow t=1.6\text{s}$ Height: $y_A=25(1.6)-4.9(1.6^2)=40-12.544=27.456\text{m}$. Insight (Figure 2 dekho): kyunki gravity dono ko equally affect karti hai, relative motion bas $40\text{m}$ ka gap $25\text{m/s}$ se close karna hai — ek khoobsurat simple hidden structure. (Dekho kaise yeh thinking Projectile motion mein kaam aati hai.)

List (reaction phase): $u=20$, $a=0$, $t_r=0.8$, chahiye $s_r$. $s_r=u{t}_r=20(0.8)=16\text{m}$. List (braking phase): $u=20$, $v=0$ (rukti hai), $a=-6$, chahiye $s_b$, time nahi chahiye. Missing letter: $t$ → use $v^2=u^2+2as$: $0=20^2+2(-6)s_b\Rightarrow s_b=\frac{400}{12}=33.33\text{m}$ Total: $s=16+33.33=49.33\text{m}$. Reaction par split kyun kiya? Reaction time ke dauran acceleration $0$ hai (constant speed); braking mein alag constant $a$ hai. Do phases, har ek ka apna constant acceleration — SUVAT do baar apply kiya.

List: release par packet balloon ki velocity share karta hai, toh $u=+6\text{m/s}$ (upar), zero nahi. Ground $y=0$ hai, toh release se displacement $s=-50\text{m}$ hai; $a=-9.8$; chahiye $t$ phir $v$. Pehle time, $s=ut+\frac{1}{2}a{t}^2$ use karke: $-50=6t-4.9t^2\Rightarrow 4.9t^2-6t-50=0$ $t=\frac{6\pm \sqrt{36+4(4.9)(50)}}{2(4.9)}=\frac{6\pm \sqrt{1016}}{9.8}=\frac{6\pm 31.875}{9.8}$ Positive time lo: $t=\frac{37.875}{9.8}=3.865\text{s}$ (negative root ek kalpanic "release se pehle" time hai). Impact speed via $v=u+at=6+(-9.8)(3.865)=6-37.88=-31.88\text{m/s}$. Minus dikhata hai ki woh neeche ja raha hai; impact par speed $31.88\text{m/s}$ hai.

Paanch letters likho, jo pata hai bharo, jo letter ki parwah nahi usse cross out karo, aur woh equation use karo jisme woh nahi hai. Agar motion beech mein apna rule change kare — reaction phir braking, upar phir neeche, ek phase phir doosra — usse pieces mein kaato jahan $a$ constant rahe, har piece ko apni List se solve karo, aur ek piece ki end velocity ko agle ki start pe de do. Woh ek adat — List, cross out, pick, aur har rule change par split — is page ke har problem mein kaam aati hai.