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A-particle-moves-in-a-linear-scare-such-that-acceleration-after-t-seconds-is-a-ms-2-where-a-2t-2-t-If-its-initial-velocity-was-3ms-1-find-an-expression-for-S-the-distance-in-meters-travele

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Question Number 47967 by Rio Michael last updated on 17/Nov/18
A particle moves in a linear scare such that acceleration after t seconds is a ms^(−2) where a= 2t^2 + t.If its initial velocity was 3ms^(−1) find an expression for S,the distance in meters traveled from start t seconds.
$${A}\:{particle}\:{moves}\:{in}\:{a}\:{linear}\:{scare}\:{such}\:{that}\:{acceleration} \\ $$$${after}\:{t}\:{seconds}\:{is}\:{a}\:{ms}^{−\mathrm{2}} \:{where}\:{a}=\:\mathrm{2}{t}^{\mathrm{2}} +\:{t}.{If}\:{its}\:{initial}\: \\ $$$${velocity}\:{was}\:\mathrm{3}{ms}^{−\mathrm{1}} \:{find}\:{an}\:{expression}\:{for}\:{S},{the}\:{distance}\:{in}\:{meters} \\ $$$${traveled}\:{from}\:{start}\:{t}\:{seconds}. \\ $$
Answered by ajfour last updated on 17/Nov/18
a= (dv/dt) = 2t^2 +t ⇒ ∫_3 ^( v) dv = ∫_0 ^( t) (2t^2 +t)dt ⇒ v =(dx/dt) = (2/3)t^3 +(t^2 /2)+3 ⇒ ∫_x_9 ^( x) dx = ∫_0 ^( t) ((2/3)t^3 +(t^2 /2)+3)dt s = x−x_0 = t((t^3 /6)+(t^2 /6)+3) ⇒ s = (t/6)(t^3 +t^2 +18) .
$$\:\:{a}=\:\frac{{dv}}{{dt}}\:=\:\mathrm{2}{t}^{\mathrm{2}} +{t} \\ $$$$\Rightarrow\:\:\:\int_{\mathrm{3}} ^{\:\:{v}} {dv}\:=\:\int_{\mathrm{0}} ^{\:{t}} \left(\mathrm{2}{t}^{\mathrm{2}} +{t}\right){dt} \\ $$$$\Rightarrow\:\:{v}\:=\frac{{dx}}{{dt}}\:=\:\:\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3} \\ $$$$\Rightarrow\:\int_{{x}_{\mathrm{9}} } ^{\:\:{x}} {dx}\:=\:\int_{\mathrm{0}} ^{\:\:{t}} \left(\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}\right){dt} \\ $$$$\:{s}\:=\:{x}−{x}_{\mathrm{0}} \:=\:{t}\left(\frac{{t}^{\mathrm{3}} }{\mathrm{6}}+\frac{{t}^{\mathrm{2}} }{\mathrm{6}}+\mathrm{3}\right) \\ $$$$\Rightarrow\:\:\:\:\:\boldsymbol{{s}}\:=\:\frac{\boldsymbol{{t}}}{\mathrm{6}}\left(\boldsymbol{{t}}^{\mathrm{3}} +\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{18}\right)\:. \\ $$
Commented by Rio Michael last updated on 17/Nov/18
please sir i don′t understand the point where ∫_x_9 ^x dx=∫_0 ^t ((2/3)t^3 +(t^2 /2)+3)dt
$${please}\:{sir}\:{i}\:{don}'{t}\:{understand}\:{the}\:{point}\:{where}\: \\ $$$$\int_{{x}_{\mathrm{9}} } ^{{x}} {dx}=\int_{\mathrm{0}} ^{{t}} \left(\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}\right){dt} \\ $$
Commented by ajfour last updated on 17/Nov/18
x is the position coordinate at time t, x_0 the initial position coordinate; s the displacement is always = △x = x−x_0 .
$$\:\:{x}\:{is}\:{the}\:{position}\:{coordinate} \\ $$$${at}\:{time}\:{t},\:{x}_{\mathrm{0}} \:{the}\:{initial}\:{position} \\ $$$${coordinate};\:\:{s}\:{the}\:{displacement} \\ $$$${is}\:{always}\:=\:\bigtriangleup{x}\:=\:{x}−{x}_{\mathrm{0}} \:. \\ $$
Commented by Rio Michael last updated on 18/Nov/18
thanks sir
$${thanks}\:{sir} \\ $$

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