Question

The relation between time and position of a particle is given by t = 3x² + ax where a and B are constant, then the acceleration of the particle is
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Answer 1

Explanation:

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Answer 2

Given that , The relation between time and position of a particle is given by ❝ t = βx² + αx  ❞  where α & β are constant .

Exigency To Find : Acceleration of the Particle ?

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding ❝  Velocity , v  ❞  of Particle :

$\qquad \dashrightarrow \sf \:t \:=\:\beta \:x^2\:+\:\alpha x \:$

$\qquad \bigstar \:\underline {\purple {\pmb{\sf \:By \:\:Differentiating \:\:the \:Given \:Equation \:w.r.t\:\:x \:\::\:}}}$

$\qquad \dashrightarrow \sf \dfrac{dt}{dx} \\\\\\ \qquad \dashrightarrow \sf \dfrac{dt}{dx}\: =\: 2\beta x\:+\:\alpha \:\qquad \pmb{\sf \bigg\lgroup Eq^n\:1\:\bigg\rgroup} \:$

$\dag\:\underline {\frak{ As \:We \:Know \:that \::}}$

$\qquad \dashrightarrow \pmb{\pink{\sf Velocity\;(\:or \:v\:) \:=\:\dfrac{dx}{dt} }}$

$\qquad \qquad \underline {\dag\:\frak{ \:By \:Using \:the \:Eq^n\:1\:\::\:}}$

$\qquad \dashrightarrow \sf \: Velocity \:=\:\dfrac{dx}{dt} \\\\\\\qquad \dashrightarrow \sf Velocity \:=\:\dfrac{dx}{dt}\:=\:\dfrac{1}{2\beta x\:+\:\alpha}\:\\\\\\ \qquad \dashrightarrow \underline {\boxed {\pmb{\purple { \frak{ \:Velocity \:,\:v \:= \:\big\{ 2\beta x\:+\:\alpha\:\big\}^{-1}\:\:}}}}}\:\:\bigstar \:$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding ❝  Acceleration , a  ❞  of Particle :

$\dag\:\underline {\frak{ As \:We \:Know \:that \::}}$

$\qquad \dashrightarrow \pmb{\pink{\sf Acceleration \;\: (\:or\:a\:) \:=\:\dfrac{dv}{dt} }}$

$\qquad \bigstar \:\underline {\purple {\pmb{\sf \:By \:\:Differentiating \:\:the \:Velocity \:,v\:\::\:}}}$

$\qquad \dashrightarrow \sf Acceleration \;,\: \:a\: \:=\:\dfrac{dv}{dt} \\\\\\\qquad \dashrightarrow \sf Acceleration \;,\: \:a\: \:=\:\dfrac{d}{dt}\big\{ 2\beta x\:+\:\alpha\big\}^{-1} \\\\\\ \qquad \dashrightarrow \sf Acceleration \;,\: \:a\: \:=\:\dfrac{d}{dt}\big\{ 2\beta x\:+\:\alpha\big\}^{-1} \\\\\\ \qquad \dashrightarrow\sf Acceleration \;,\: \:a\: \:=\:-\:\Bigg[ \dfrac{1}{\big\{2\beta x + \alpha \big\} ^2 }\Bigg] \: 2\beta \: \times \:\dfrac{dx}{dt}\: \\\\\\ \qquad \dashrightarrow \sf Acceleration \;,\: \:a\: \:=\:\:\Bigg[ \dfrac{-2\beta }{\big\{2\beta x + \alpha \big\} ^2 }\Bigg] \: \: \times \:\dfrac{dx}{dt}\:$

$\qquad \underline {\dag \:\frak{ Putting \:Value \:of \:\: \pmb{\frak{ dx/dt}}\:\::}}$

$\qquad \dashrightarrow \sf Acceleration \;,\: \:a\: \:=\:\:\Bigg[ \dfrac{-2\beta }{\big\{2\beta x + \alpha \big\} ^2 }\Bigg] \: \: \times \:\dfrac{dx}{dt}\: \\\\\\ \qquad \dashrightarrow\sf Acceleration \;,\: \:a\: \:=\:\:\Bigg[ \dfrac{-2\beta }{\big\{2\beta x + \alpha \big\} ^2 } \Bigg] \: \: \times \Bigg[ \:\dfrac{1}{2\beta x\:+\:\alpha}\:\Bigg] \\\\\\ \qquad \dashrightarrow \sf Acceleration \;,\: \:a\: \:=\:\:\Bigg[ \dfrac{-2\beta }{\big\{2\beta x + \alpha \big\} ^3} \Bigg] \: \:\\\\\\ \qquad \dashrightarrow \underline {\boxed {{\purple { \frak{ \:Acceleration \;,\: \:a\: \:=\:\:\Bigg[ \dfrac{-2\beta }{\big\{2\beta x + \alpha \big\} ^3} \Bigg] \:\:\:}}}}}\:\:\bigstar \:$

$\therefore \sf Hence ,\:The \:Acceleration \:of \:Particle \:is \:\purple {\pmb{\sf \:Option \:A\:)}}\:\red{\sf \:\Bigg[ \dfrac{-2\beta }{\big\{2\beta x + \alpha \big\} ^3} \Bigg] \:}\:.\:$