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Question Number 209404 by Tawa11 last updated on 09/Jul/24
Answered by mr W last updated on 09/Jul/24
s=4t+at^2 +bt^3 v=(ds/dt)=4+2at+3bt^2 a=(dv/dt)=2a+6bt at v_(max) : a=0=2a+6bt ⇒t=−(a/(3b))=6 ⇒a=−18b s(6)=4×6+a×6^2 +b×6^3 =48 ⇒3a+18b=2 ⇒3a−a=2 ⇒a=1 ⇒b=−(1/(18)) v_(max) =4+2×6−(6^2 /6)=10 yd/s s=4t+t^2 −(t^3 /(18)) (ds/dt)=4+2t−(t^2 /6)=0 ⇒t=6(1+(√(5/3))) s_(max) =4t+t^2 −(t/3)(2t+4)=(t^2 /3)+((8t)/3) =2(2t+4)+((8t)/3)=((20t)/3)+8 =((20)/3)×6(1+(√(5/3)))+8 =48+40(√(5/3))≈99.6 yd<100 yd that means the boy can never reach 100 yards. ⇒question is wrong or question is not clear enough to be understood.
$${s}=\mathrm{4}{t}+{at}^{\mathrm{2}} +{bt}^{\mathrm{3}} \\ $$$${v}=\frac{{ds}}{{dt}}=\mathrm{4}+\mathrm{2}{at}+\mathrm{3}{bt}^{\mathrm{2}} \\ $$$${a}=\frac{{dv}}{{dt}}=\mathrm{2}{a}+\mathrm{6}{bt} \\ $$$${at}\:{v}_{{max}} :\:{a}=\mathrm{0}=\mathrm{2}{a}+\mathrm{6}{bt}\:\Rightarrow{t}=−\frac{{a}}{\mathrm{3}{b}}=\mathrm{6}\:\Rightarrow{a}=−\mathrm{18}{b} \\ $$$${s}\left(\mathrm{6}\right)=\mathrm{4}×\mathrm{6}+{a}×\mathrm{6}^{\mathrm{2}} +{b}×\mathrm{6}^{\mathrm{3}} =\mathrm{48} \\ $$$$\Rightarrow\mathrm{3}{a}+\mathrm{18}{b}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}{a}−{a}=\mathrm{2}\:\Rightarrow{a}=\mathrm{1}\:\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{18}} \\ $$$${v}_{{max}} =\mathrm{4}+\mathrm{2}×\mathrm{6}−\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{6}}=\mathrm{10}\:{yd}/{s} \\ $$$${s}=\mathrm{4}{t}+{t}^{\mathrm{2}} −\frac{{t}^{\mathrm{3}} }{\mathrm{18}} \\ $$$$\frac{{ds}}{{dt}}=\mathrm{4}+\mathrm{2}{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{6}}=\mathrm{0}\:\Rightarrow{t}=\mathrm{6}\left(\mathrm{1}+\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right) \\ $$$${s}_{{max}} =\mathrm{4}{t}+{t}^{\mathrm{2}} −\frac{{t}}{\mathrm{3}}\left(\mathrm{2}{t}+\mathrm{4}\right)=\frac{{t}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{8}{t}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{2}{t}+\mathrm{4}\right)+\frac{\mathrm{8}{t}}{\mathrm{3}}=\frac{\mathrm{20}{t}}{\mathrm{3}}+\mathrm{8} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{20}}{\mathrm{3}}×\mathrm{6}\left(\mathrm{1}+\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)+\mathrm{8} \\ $$$$\:\:\:\:\:=\mathrm{48}+\mathrm{40}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\approx\mathrm{99}.\mathrm{6}\:{yd}<\mathrm{100}\:{yd} \\ $$$${that}\:{means}\:{the}\:{boy}\:{can}\:{never}\:{reach} \\ $$$$\mathrm{100}\:{yards}. \\ $$$$\Rightarrow{question}\:{is}\:{wrong}\:{or}\:{question}\:{is} \\ $$$${not}\:{clear}\:{enough}\:{to}\:{be}\:{understood}. \\ $$
Commented by Tawa11 last updated on 09/Jul/24
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by mr W last updated on 09/Jul/24
a strange boy! he runs forwards at first. when he has run 99.6 yd, he begins to run backwards.
$${a}\:{strange}\:{boy}!\:{he}\:{runs}\:{forwards}\:{at} \\ $$$${first}.\:{when}\:{he}\:{has}\:{run}\:\mathrm{99}.\mathrm{6}\:{yd},\:{he} \\ $$$${begins}\:{to}\:{run}\:{backwards}. \\ $$
Commented by Tawa11 last updated on 09/Jul/24
Hahahahahaha
$$\mathrm{Hahahahahaha} \\ $$
Commented by A5T last updated on 09/Jul/24
s=10t after 6s; s=4t+t^2 −(t^3 /(18)) was upto 6s and not after.
$${s}=\mathrm{10}{t}\:{after}\:\mathrm{6}{s};\: \\ $$$${s}=\mathrm{4}{t}+{t}^{\mathrm{2}} −\frac{{t}^{\mathrm{3}} }{\mathrm{18}}\:{was}\:{upto}\:\mathrm{6}{s}\:{and}\:{not}\:{after}. \\ $$
Commented by mr W last updated on 09/Jul/24
thanks. you are right. it makes sinse too. but i still think the question is not clear enough.
$${thanks}.\:{you}\:{are}\:{right}.\:{it}\:{makes}\:{sinse}\:{too}. \\ $$$${but}\:{i}\:{still}\:{think}\:{the}\:{question}\:{is}\:{not} \\ $$$${clear}\:{enough}. \\ $$
Answered by A5T last updated on 09/Jul/24
s=f(t)=4t+at^2 +bt^3 f(6)=48⇒3a+18b=2...(i) v=(ds/dt)=4+2at+3bt^2 maximum when t=6 v=4+12a+108b at maximum point...(ii) (dv/dt)=^? 0⇒2a+6bt=0⇒2a+36b=0⇒a=−18b (i)⇒−2×18b=2⇒b=((−1)/(18))⇒a=1 (ii)⇒maximum speed=16−6=10yd/s After 6s; s=10×t⇒t=(s/(10))=((100−48)/(10))=5.2s The time taken to complete the rest is 5.2sec ⇒Total time taken=6s+5.2s=11.2seconds
$${s}={f}\left({t}\right)=\mathrm{4}{t}+{at}^{\mathrm{2}} +{bt}^{\mathrm{3}} \\ $$$${f}\left(\mathrm{6}\right)=\mathrm{48}\Rightarrow\mathrm{3}{a}+\mathrm{18}{b}=\mathrm{2}…\left({i}\right) \\ $$$${v}=\frac{{ds}}{{dt}}=\mathrm{4}+\mathrm{2}{at}+\mathrm{3}{bt}^{\mathrm{2}} \:{maximum}\:{when}\:{t}=\mathrm{6} \\ $$$${v}=\mathrm{4}+\mathrm{12}{a}+\mathrm{108}{b}\:{at}\:{maximum}\:{point}…\left({ii}\right) \\ $$$$\frac{{dv}}{{dt}}\overset{?} {=}\mathrm{0}\Rightarrow\mathrm{2}{a}+\mathrm{6}{bt}=\mathrm{0}\Rightarrow\mathrm{2}{a}+\mathrm{36}{b}=\mathrm{0}\Rightarrow{a}=−\mathrm{18}{b} \\ $$$$\left({i}\right)\Rightarrow−\mathrm{2}×\mathrm{18}{b}=\mathrm{2}\Rightarrow{b}=\frac{−\mathrm{1}}{\mathrm{18}}\Rightarrow{a}=\mathrm{1} \\ $$$$\left({ii}\right)\Rightarrow{maximum}\:{speed}=\mathrm{16}−\mathrm{6}=\mathrm{10}{yd}/{s} \\ $$$${After}\:\mathrm{6}{s};\:{s}=\mathrm{10}×{t}\Rightarrow{t}=\frac{{s}}{\mathrm{10}}=\frac{\mathrm{100}−\mathrm{48}}{\mathrm{10}}=\mathrm{5}.\mathrm{2}{s} \\ $$$${The}\:{time}\:{taken}\:{to}\:{complete}\:{the}\:{rest}\:{is}\:\mathrm{5}.\mathrm{2}{sec} \\ $$$$\Rightarrow{Total}\:{time}\:{taken}=\mathrm{6}{s}+\mathrm{5}.\mathrm{2}{s}=\mathrm{11}.\mathrm{2}{seconds} \\ $$
Commented by Tawa11 last updated on 09/Jul/24
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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