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The-distance-x-m-have-used-by-a-particle-in-time-t-sec-is-described-by-the-equation-x-10-12t-2-Find-the-average-speed-of-the-particle-between-the-interval-t-2-sec-and-t-5-sec-

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Question Number 9636 by tawakalitu last updated on 22/Dec/16
The distance x m have used by a particle in time t sec is described by the equation x = 10 + 12t^2 Find the average speed of the particle between the interval t = 2 sec and t = 5 sec
$$\mathrm{The}\:\mathrm{distance}\:\mathrm{x}\:\mathrm{m}\:\mathrm{have}\:\mathrm{used}\:\mathrm{by}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{in} \\ $$$$\mathrm{time}\:\mathrm{t}\:\mathrm{sec}\:\mathrm{is}\:\mathrm{described}\:\mathrm{by}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{x}\:=\:\mathrm{10}\:+\:\mathrm{12t}^{\mathrm{2}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{average}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{interval}\:\mathrm{t}\:=\:\mathrm{2}\:\mathrm{sec}\:\mathrm{and}\:\mathrm{t}\:=\:\mathrm{5}\:\mathrm{sec} \\ $$
Commented by ridwan balatif last updated on 22/Dec/16
x=10+12t^2 v=(dx/dt)=24t v_1 (t=2s)=48m/s v_2 (t=5s)=120m/s v^− =((120−48)/(5−2))=((72)/3)=24m/s
$$\mathrm{x}=\mathrm{10}+\mathrm{12t}^{\mathrm{2}} \\ $$$$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{24t} \\ $$$$\mathrm{v}_{\mathrm{1}} \left(\mathrm{t}=\mathrm{2s}\right)=\mathrm{48m}/\mathrm{s} \\ $$$$\mathrm{v}_{\mathrm{2}} \left(\mathrm{t}=\mathrm{5s}\right)=\mathrm{120m}/\mathrm{s} \\ $$$$\overset{−} {\mathrm{v}}=\frac{\mathrm{120}−\mathrm{48}}{\mathrm{5}−\mathrm{2}}=\frac{\mathrm{72}}{\mathrm{3}}=\mathrm{24m}/\mathrm{s} \\ $$
Commented by tawakalitu last updated on 22/Dec/16
thanks sir. but the answer options here are. a) 62m/s (b) 72 m/s (c) 84 m/s
$$\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{options}\:\mathrm{here}\:\mathrm{are}. \\ $$$$\left.\mathrm{a}\right)\:\mathrm{62m}/\mathrm{s}\:\left(\mathrm{b}\right)\:\mathrm{72}\:\mathrm{m}/\mathrm{s}\:\left(\mathrm{c}\right)\:\mathrm{84}\:\mathrm{m}/\mathrm{s} \\ $$
Commented by ridwan balatif last updated on 22/Dec/16
ohh i′m so sorry, i wrong use formula v^− =((x_2 −x_1 )/(t_2 −t_1 )) x_1 =10+12(2)^2 =10+48=58m x_2 =10+12(5)^2 = 310m v^− =((310−58)/(5−2)) =((252)/3) =84m/s if we use this formula ((v_2 −v_1 )/(t_2 −t_1 )),its means we find average accelaration
$$\mathrm{ohh}\:\mathrm{i}'\mathrm{m}\:\mathrm{so}\:\mathrm{sorry},\:\mathrm{i}\:\mathrm{wrong}\:\mathrm{use}\:\mathrm{formula} \\ $$$$\overset{−} {\mathrm{v}}=\frac{\mathrm{x}_{\mathrm{2}} −\mathrm{x}_{\mathrm{1}} }{\mathrm{t}_{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} } \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{10}+\mathrm{12}\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{10}+\mathrm{48}=\mathrm{58m} \\ $$$$\mathrm{x}_{\mathrm{2}} =\mathrm{10}+\mathrm{12}\left(\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\:\mathrm{310m} \\ $$$$\overset{−} {\mathrm{v}}=\frac{\mathrm{310}−\mathrm{58}}{\mathrm{5}−\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{252}}{\mathrm{3}} \\ $$$$\:\:\:=\mathrm{84m}/\mathrm{s} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula}\:\frac{\mathrm{v}_{\mathrm{2}} −\mathrm{v}_{\mathrm{1}} }{\mathrm{t}_{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} },\mathrm{its}\:\mathrm{means}\:\mathrm{we}\:\mathrm{find} \\ $$$$\mathrm{average}\:\mathrm{accelaration} \\ $$
Commented by tawakalitu last updated on 22/Dec/16
I really appreciate your effort sir. God bless you.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Answered by mrW last updated on 22/Dec/16
distance in the interval t=2 sec to t=5 sec: Δs=[10+12×5^2 ]−[10+12×2^2 ]=12×(5^2 −2^2 )=252 m average speed v^� =((Δs)/(Δt))=((252)/3)=84 m/s
$$\mathrm{distance}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\mathrm{t}=\mathrm{2}\:\mathrm{sec}\:\mathrm{to}\:\mathrm{t}=\mathrm{5}\:\mathrm{sec}: \\ $$$$\Delta\mathrm{s}=\left[\mathrm{10}+\mathrm{12}×\mathrm{5}^{\mathrm{2}} \right]−\left[\mathrm{10}+\mathrm{12}×\mathrm{2}^{\mathrm{2}} \right]=\mathrm{12}×\left(\mathrm{5}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \right)=\mathrm{252}\:\mathrm{m} \\ $$$$\mathrm{average}\:\mathrm{speed}\:\bar {\mathrm{v}}=\frac{\Delta\mathrm{s}}{\Delta\mathrm{t}}=\frac{\mathrm{252}}{\mathrm{3}}=\mathrm{84}\:\mathrm{m}/\mathrm{s} \\ $$

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