Question

please answer this question !

Proper explanation is very much required !! ​

Answer 1

$\frak{Given}\:\begin{cases} \:\quad \sf Force \:acting \:on \:particle \:is \:\pmb{\frak{ Be^{-Ct} }}\:\\ \:\quad \sf Mass \:is\:\pmb{\frak{ m }}\:.\\ \:\quad \sf Velocity \:is \:\pmb{\frak{0\:}}\:at\:t\:=\:\pmb{\frak{0\:s}}\:. \:\end{cases}\:$

Need To Find : It's Terminal Velocity ( Velocity after a long times ) ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad\twoheadrightarrow \sf \: F \:=\: Be^{-Ct}\:$

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

$\qquad \dashrightarrow \:\pmb{\pink{\sf \: F\:=\: m a\:}}$

Where ,

• F = Force acting on Particle ,
• m = Mass of Particle &
• a = Acceleration of Particle.

$\qquad \therefore\sf \: ma \: =\: Be^{-Ct}\:\:$

$\qquad \twoheadrightarrow \:\sf ma \: =\: Be^{-Ct}\:\:$

$\qquad \twoheadrightarrow \: \sf a \: =\:\dfrac{ Be^{-Ct}}{m}\:\:$

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

$\qquad \dashrightarrow \:\pmb{\pink{\sf \: Acceleration \:=\: \dfrac{dv}{dt}\:}}$

$\qquad \therefore \: \sf \dfrac{dv}{dt} \: =\: \dfrac{1}{m} \:Be^{-Ct}\:\:$

$\qquad \twoheadrightarrow \sf \: \dfrac{dv}{dt} \: =\: \dfrac{1}{m} \:Be^{-Ct}\:\:$

$\qquad \twoheadrightarrow \sf \: dv \: =\: \dfrac{1}{m} \:Be^{-Ct}\:dt\:$

$\qquad \bigstar \:\underline {\pmb{\purple { \sf By \:integrating \:both \:sides\:,\:we\:get\:,}}}$

$\qquad \twoheadrightarrow \sf \: dv \: =\: \dfrac{1}{m} \:Be^{-Ct}\:dt\:$

$\qquad \twoheadrightarrow \sf \:\int dv \: =\:\int \dfrac{1}{m} \:Be^{-Ct}\:dt\:$

$\qquad \twoheadrightarrow \sf \:\int dv \: =\:\int \dfrac{1}{m} \:Be^{-Ct}\:dt\:$

$\qquad \twoheadrightarrow \sf \:v(t) \: =\:- \dfrac{B}{mC} \:e^{-Ct}\:+\:K\:$

$\frak{At}\:\begin{cases} \:\quad \sf t\:(\:or \:Time \:) \:\pmb{\frak{ 0\:s }}\:\&\:\\ \:\quad \sf Velocity \:is \:\pmb{\frak{0\:}}\: \:\end{cases}\:$

$\qquad \twoheadrightarrow \sf K \:=\:\dfrac{B}{mC}$

$\qquad \twoheadrightarrow \sf\pmb{\pink{ K \:=\:\dfrac{B}{mC}}}$

✇ Terminal Velocity [ Velocity after long time ] :

$\qquad \twoheadrightarrow \sf \:v(t) \: =\:- \dfrac{B}{mC} \:e^{-Ct}\:+\:K\:$

$\qquad \twoheadrightarrow \sf \:v(t) \: =\:- \dfrac{B}{mC} \:e^{-Ct}\:+\:\dfrac{B}{mC}\:$

$\qquad \twoheadrightarrow \sf \:v(t) \: =\:- \dfrac{B}{mC} \:\big\{ 1 - \:e^{-Ct} \:\big\}\:\:$

$\qquad \twoheadrightarrow \sf \:v(t) \: =\: \dfrac{B}{mC}\:$

$\qquad \twoheadrightarrow \sf\underline{\pmb{\boxed{\pink{\frak{ Terminal \:Velocity \:=\:v_t\:\:=\:\dfrac{B}{mC}}}}}}\:\:\bigstar$

$\qquad \therefore \:\sf Hence, \:Terminal \:Velocity \:is \:\: \pmb{\sf Option \:B \:)\:\dfrac{B}{mC}\:}.$

Answer 2

Answer:

Given⎩⎨⎧ForceactingonparticleisBe−CtBe−CtMassismm.Velocityis00att=0s0s.

Need To Find : It's Terminal Velocity ( Velocity after a long times ) ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

↠�=��−��↠F=Be−Ct

†���������ℎ��:‾†AsWeknowthat:

⇢�=��⇢F=maF=ma

Where ,

F = Force acting on Particle ,m = Mass of Particle &a = Acceleration of Particle.

∴��=��−��∴ma=Be−Ct

↠��=��−��↠ma=Be−Ct

↠�=��−���↠a=mBe−Ct

†���������ℎ��:‾†AsWeknowthat:

⇢������������=����⇢Acceleration=dtdvAcceleration=dtdv

∴����=1���−��∴dtdv=m1Be−Ct

↠����=1���−��↠dtdv=m1Be−Ct

↠��=1���−����↠dv=m1Be−Ctdt

★����������������ℎ�����,�����,‾★Byintegratingbothsides,weget,Byintegratingbothsides,weget,

↠��=1���−����↠dv=m1Be−Ct

Explanation:

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