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a-1-2x-2-with-x-1-and-v-0-at-t-0-find-time-that-particle-takes-to-reach-x-0-25m-




Question Number 37579 by ajfour last updated on 15/Jun/18
a=((−1)/(2x^2 ))  with x=1 and v=0 at t=0  find time that particle takes to  reach x=0.25m .
$${a}=\frac{−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:\:{with}\:{x}=\mathrm{1}\:{and}\:{v}=\mathrm{0}\:{at}\:{t}=\mathrm{0} \\ $$$${find}\:{time}\:{that}\:{particle}\:{takes}\:{to} \\ $$$${reach}\:{x}=\mathrm{0}.\mathrm{25}{m}\:. \\ $$
Answered by ajfour last updated on 15/Jun/18
since a < 0 and v_0 =0  this means negative velocity  starts developing as t increases  from zero.     a=((vdv)/dx) = −(1/(2x^2 ))  ⇒   (v^2 /2) = (1/2)((1/x)−1)  ⇒    v=−(√((1/x)−1))     ∫_1 ^(  x) (dx/( (√((1/x)−1)))) = −t  let     x=cos^2 θ            dx = −sin 2θdθ  −t = −∫_0 ^(  cos^(−1) (√x))   ((2sin θcos θdθ)/(tan θ))  t=∫_0 ^(  cos^(−1) (√x))   (1+cos 2θ)dθ    =(θ+((sin 2θ)/2)+c)∣_0 ^(cos^(−1) (√x))     t =cos^(−1) (√x)+((√x))((√(1−x)))  And to reach  x=0.25       t = (𝛑/3)+((√3)/4) .
$${since}\:{a}\:<\:\mathrm{0}\:{and}\:{v}_{\mathrm{0}} =\mathrm{0} \\ $$$${this}\:{means}\:{negative}\:{velocity} \\ $$$${starts}\:{developing}\:{as}\:{t}\:{increases} \\ $$$${from}\:{zero}. \\ $$$$\:\:\:\boldsymbol{{a}}=\frac{\boldsymbol{{vdv}}}{\boldsymbol{{dx}}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\:\:{v}=−\sqrt{\frac{\mathrm{1}}{{x}}−\mathrm{1}} \\ $$$$\:\:\:\int_{\mathrm{1}} ^{\:\:{x}} \frac{{dx}}{\:\sqrt{\frac{\mathrm{1}}{{x}}−\mathrm{1}}}\:=\:−{t} \\ $$$${let}\:\:\:\:\:{x}=\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:{dx}\:=\:−\mathrm{sin}\:\mathrm{2}\theta{d}\theta \\ $$$$−{t}\:=\:−\int_{\mathrm{0}} ^{\:\:\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}} \:\:\frac{\mathrm{2sin}\:\theta\mathrm{cos}\:\theta{d}\theta}{\mathrm{tan}\:\theta} \\ $$$${t}=\int_{\mathrm{0}} ^{\:\:\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}} \:\:\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta \\ $$$$\:\:=\left(\theta+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}+{c}\right)\mid_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}} \\ $$$$\:\:{t}\:=\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}+\left(\sqrt{{x}}\right)\left(\sqrt{\mathrm{1}−{x}}\right) \\ $$$${And}\:{to}\:{reach}\:\:{x}=\mathrm{0}.\mathrm{25} \\ $$$$\:\:\:\:\:\boldsymbol{{t}}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:. \\ $$

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