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Question Number 57735 by Tawa1 last updated on 10/Apr/19

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Apr/19

s=2t^3 −13t^2 +20t s=t(2t^2 −13t+20) s=t(2t^2 −8t−5t+20) =t{2t(t−4)−5(t−4)} =t(t−4)(2t−5) s=0 when t=0,t=4 and t=2.5 when t=4 s=0 that means in 4 seconds it cover same distance twice but in opposite direction .hence displace ment is zero. S_(t=2) =2(2−4)(2×2−5)=2×−2×−1=4 so total distance covered in 4 seconds=2×4=8 meter S=2t^3 −13t^2 +20t (dS/dt)=6t^2 −26t+20 (d^2 s/dt^2 )=12t−26 given accelaration −ve (d^2 s/dt^2 )<0 12t−26<0 →t<((26)/(12)) so in time duration ((13)/6)> t>0 acc is negetive pls check is it correect answer.. then maximum speed to[be calculated (ds/dt)=6t^2 −26t+20 when t=0 (ds/dt)=20 that means initial velocity=20. now (d^2 s/dt^2 )=12t−26 now in time interval ((26)/(12))>t>0 accelaration is −ve so in time interval [0,((26)/(12))] vel_(max) =speed_(max) =20 in time interval [((26)/(12)),4] acc is +ve (ds/dt)=6t^2 −26t+20 6t^2 −26t+20=0 3t^2 −13t+10=0 3t^2 −3t−10t+10=0 3t(t−1)−10(t−1)=0 (t−1)((3t−10)=0 when t=1 velocity=0 when t=((10)/3) velocity=0 so in time interval 4>t>0 max speed is=20 pls check

$$ \\ $$$$ \\ $$$${s}=\mathrm{2}{t}^{\mathrm{3}} −\mathrm{13}{t}^{\mathrm{2}} +\mathrm{20}{t} \\ $$$${s}={t}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{13}{t}+\mathrm{20}\right) \\ $$$${s}={t}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{8}{t}−\mathrm{5}{t}+\mathrm{20}\right) \\ $$$$\:\:\:={t}\left\{\mathrm{2}{t}\left({t}−\mathrm{4}\right)−\mathrm{5}\left({t}−\mathrm{4}\right)\right\} \\ $$$$\:\:\:\:={t}\left({t}−\mathrm{4}\right)\left(\mathrm{2}{t}−\mathrm{5}\right) \\ $$$${s}=\mathrm{0}\:\:{when}\:{t}=\mathrm{0},{t}=\mathrm{4}\:\:{and}\:{t}=\mathrm{2}.\mathrm{5} \\ $$$$\:{when}\:{t}=\mathrm{4}\:\:{s}=\mathrm{0} \\ $$$${that}\:{means}\:{in}\:\mathrm{4}\:{seconds}\:{it}\:{cover}\:{same}\:{distance} \\ $$$${twice}\:{but}\:{in}\:{opposite}\:{direction}\:.{hence}\:{displace} \\ $$$${ment}\:{is}\:{zero}. \\ $$$${S}_{{t}=\mathrm{2}} =\mathrm{2}\left(\mathrm{2}−\mathrm{4}\right)\left(\mathrm{2}×\mathrm{2}−\mathrm{5}\right)=\mathrm{2}×−\mathrm{2}×−\mathrm{1}=\mathrm{4} \\ $$$${so}\:{total}\:{distance}\:{covered}\:{in}\:\mathrm{4}\:{seconds}=\mathrm{2}×\mathrm{4}=\mathrm{8}\:{meter} \\ $$$$ \\ $$$${S}=\mathrm{2}{t}^{\mathrm{3}} −\mathrm{13}{t}^{\mathrm{2}} +\mathrm{20}{t} \\ $$$$\frac{{dS}}{{dt}}=\mathrm{6}{t}^{\mathrm{2}} −\mathrm{26}{t}+\mathrm{20} \\ $$$$\:\:\:\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} }=\mathrm{12}{t}−\mathrm{26} \\ $$$${given}\:{accelaration}\:−{ve} \\ $$$$\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} }<\mathrm{0} \\ $$$$\mathrm{12}{t}−\mathrm{26}<\mathrm{0}\:\:\:\rightarrow{t}<\frac{\mathrm{26}}{\mathrm{12}} \\ $$$${so}\:{in}\:\:{time}\:{duration}\:\:\:\frac{\mathrm{13}}{\mathrm{6}}>\:{t}>\mathrm{0}\:{acc}\:{is}\:{negetive} \\ $$$${pls}\:{check}\:{is}\:{it}\:{correect}\:{answer}.. \\ $$$${then}\:{maximum}\:{speed}\:{to}\left[{be}\:{calculated}\right. \\ $$$$\frac{{ds}}{{dt}}=\mathrm{6}{t}^{\mathrm{2}} −\mathrm{26}{t}+\mathrm{20}\:\:\:{when}\:{t}=\mathrm{0}\:\:\:\:\frac{{ds}}{{dt}}=\mathrm{20}\:\:{that}\:{means}\:{initial} \\ $$$${velocity}=\mathrm{20}.\:\: \\ $$$${now}\:\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} }=\mathrm{12}{t}−\mathrm{26}\:\:{now}\:{in}\:{time}\:{interval}\: \\ $$$$\frac{\mathrm{26}}{\mathrm{12}}>{t}>\mathrm{0}\:\:{accelaration}\:{is}\:−{ve}\: \\ $$$${so}\:{in}\:{time}\:{interval}\:\left[\mathrm{0},\frac{\mathrm{26}}{\mathrm{12}}\right]\:{vel}_{{max}} ={speed}_{{max}} =\mathrm{20} \\ $$$${in}\:{time}\:{interval}\:\:\left[\frac{\mathrm{26}}{\mathrm{12}},\mathrm{4}\right]\:\:{acc}\:{is}\:+{ve} \\ $$$$\frac{{ds}}{{dt}}=\mathrm{6}{t}^{\mathrm{2}} −\mathrm{26}{t}+\mathrm{20} \\ $$$$\mathrm{6}{t}^{\mathrm{2}} −\mathrm{26}{t}+\mathrm{20}=\mathrm{0} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{13}{t}+\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{10}{t}+\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{3}{t}\left({t}−\mathrm{1}\right)−\mathrm{10}\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left(\left(\mathrm{3}{t}−\mathrm{10}\right)=\mathrm{0}\right. \\ $$$${when}\:{t}=\mathrm{1}\:\:\:{velocity}=\mathrm{0} \\ $$$${when}\:{t}=\frac{\mathrm{10}}{\mathrm{3}}\:\:\:{velocity}=\mathrm{0} \\ $$$${so}\:{in}\:{time}\:{interval}\:\:\mathrm{4}>{t}>\mathrm{0} \\ $$$${max}\:{speed}\:{is}=\mathrm{20} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa1 last updated on 11/Apr/19

God bless you sir. It should be right sir. I will study your solution.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{It}\:\mathrm{should}\:\mathrm{be}\:\mathrm{right}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{will}\:\mathrm{study}\:\mathrm{your}\:\mathrm{solution}. \\ $$

Commented by Tawa1 last updated on 11/Apr/19

God bless you sir. I appreciate your time.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Apr/19

3)a) when they meet x_p =x_Q 5t^2 (t+7)=t^3 (t+3) t^2 (5t+35)−t^3 (t+3)=0 t^2 (5t+35−t^2 −3t)=0 when t=0 they meet at gas station −t^2 +2t+35=0 t^2 −2t−35=0 t^2 −7t+5t−35=0 t(t−7)+5(t−7)=0 (t−7)(t+5)=0 t≠−5 so after t=7 they meet. b)v_p =(dx_p /dt)=((d(5t^3 +35t^2 ))/dt)=15t^2 +70t v_Q =((d(t^4 +3t^3 ))/dt) =4t^3 +9t^2 so velocity of p w.r.t Q=v_p −v_Q =(15t^2 +70t)−(4t^3 +9t^2 ) =6t^2 +70t−4t^3 c)v_p >v_Q v_p −v_Q >0 (15t^2 +70t)−(4t^3 +9t^2 )>0 6t^2 +70t−4t^3 >0 t(4t^2 −6t−70)<0 2t(2t^2 −3t−35)<0 2t(2t^2 −10t+7t−35)<0 2t{2t(t−5)+7(t−5)}<0 2t(t−5)(2t+7)<0 t≠negetive so[critical value of t =0,5 f(t)=2t(t−5)(2t+7)<0 when t>5 f(t)>0 so when t>0 but t<5 then f(t)<0 so in time interval 5>t>0 v_p >v_Q

$$\left.\mathrm{3}\left.\right){a}\right)\:{when}\:{they}\:{meet}\:{x}_{{p}} ={x}_{{Q}} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} \left({t}+\mathrm{7}\right)={t}^{\mathrm{3}} \left({t}+\mathrm{3}\right) \\ $$$${t}^{\mathrm{2}} \left(\mathrm{5}{t}+\mathrm{35}\right)−{t}^{\mathrm{3}} \left({t}+\mathrm{3}\right)=\mathrm{0} \\ $$$${t}^{\mathrm{2}} \left(\mathrm{5}{t}+\mathrm{35}−{t}^{\mathrm{2}} −\mathrm{3}{t}\right)=\mathrm{0} \\ $$$${when}\:{t}=\mathrm{0}\:{they}\:{meet}\:{at}\:{gas}\:{station} \\ $$$$−{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{35}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{35}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{7}{t}+\mathrm{5}{t}−\mathrm{35}=\mathrm{0} \\ $$$${t}\left({t}−\mathrm{7}\right)+\mathrm{5}\left({t}−\mathrm{7}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{7}\right)\left({t}+\mathrm{5}\right)=\mathrm{0} \\ $$$${t}\neq−\mathrm{5}\:\:{so}\:{after}\:{t}=\mathrm{7}\:{they}\:{meet}. \\ $$$$\left.{b}\right){v}_{{p}} =\frac{{dx}_{{p}} }{{dt}}=\frac{{d}\left(\mathrm{5}{t}^{\mathrm{3}} +\mathrm{35}{t}^{\mathrm{2}} \right)}{{dt}}=\mathrm{15}{t}^{\mathrm{2}} +\mathrm{70}{t} \\ $$$${v}_{{Q}} =\frac{{d}\left({t}^{\mathrm{4}} +\mathrm{3}{t}^{\mathrm{3}} \right)}{{dt}}\:=\mathrm{4}{t}^{\mathrm{3}} +\mathrm{9}{t}^{\mathrm{2}} \\ $$$${so}\:{velocity}\:{of}\:{p}\:\:{w}.{r}.{t}\:{Q}={v}_{{p}} −{v}_{{Q}} \\ $$$$\:\:\:=\left(\mathrm{15}{t}^{\mathrm{2}} +\mathrm{70}{t}\right)−\left(\mathrm{4}{t}^{\mathrm{3}} +\mathrm{9}{t}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\mathrm{6}{t}^{\mathrm{2}} +\mathrm{70}{t}−\mathrm{4}{t}^{\mathrm{3}} \\ $$$$\left.{c}\right){v}_{{p}} >{v}_{{Q}} \:\:\:{v}_{{p}} −{v}_{{Q}} >\mathrm{0} \\ $$$$\left(\mathrm{15}{t}^{\mathrm{2}} +\mathrm{70}{t}\right)−\left(\mathrm{4}{t}^{\mathrm{3}} +\mathrm{9}{t}^{\mathrm{2}} \right)>\mathrm{0} \\ $$$$\mathrm{6}{t}^{\mathrm{2}} +\mathrm{70}{t}−\mathrm{4}{t}^{\mathrm{3}} >\mathrm{0} \\ $$$${t}\left(\mathrm{4}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{70}\right)<\mathrm{0} \\ $$$$\mathrm{2}{t}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{35}\right)<\mathrm{0} \\ $$$$\mathrm{2}{t}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{10}{t}+\mathrm{7}{t}−\mathrm{35}\right)<\mathrm{0} \\ $$$$\mathrm{2}{t}\left\{\mathrm{2}{t}\left({t}−\mathrm{5}\right)+\mathrm{7}\left({t}−\mathrm{5}\right)\right\}<\mathrm{0} \\ $$$$\mathrm{2}{t}\left({t}−\mathrm{5}\right)\left(\mathrm{2}{t}+\mathrm{7}\right)<\mathrm{0} \\ $$$${t}\neq{negetive}\:\:{so}\left[{critical}\:{value}\:{of}\:{t}\:=\mathrm{0},\mathrm{5}\right. \\ $$$${f}\left({t}\right)=\mathrm{2}{t}\left({t}−\mathrm{5}\right)\left(\mathrm{2}{t}+\mathrm{7}\right)<\mathrm{0} \\ $$$${when}\:{t}>\mathrm{5}\:\:{f}\left({t}\right)>\mathrm{0} \\ $$$${so}\:{when}\:{t}>\mathrm{0}\:\:{but}\:{t}<\mathrm{5}\:\:{then}\:{f}\left({t}\right)<\mathrm{0} \\ $$$${so}\:{in}\:{time}\:{interval}\:\:\:\mathrm{5}>{t}>\mathrm{0} \\ $$$${v}_{{p}} >{v}_{{Q}} \\ $$

Commented by Tawa1 last updated on 11/Apr/19

Wow, God bless you sir. Thanks for your time sir.

$$\mathrm{Wow},\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$

Commented by peter frank last updated on 13/Apr/19

thank you

$${thank}\:{you} \\ $$

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