The motion of a body falling from rest in a resistive medium is described by the equation (dv)/(dt) = a - bv, where a and b are constants. The velocity at any time t is

Correct answer: B. (a)/(b) ( 1 - e^ - bt ) (dv)/(dt) = a - bv or ∫ _ 0 ^ v (dv)/(a - bv) = ∫ _ 0 ^ t dt or - (1)/(b) [ (a - bv)] _ 0 ^ v = t or ((a - bv)/(a)) = e^ - bt or a - (b)/(a)v = e^ - bt v = (a)/(b) ( 1 - e^ - bt )

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

The motion of a body falling from rest in a resistive medium is...

The motion of a body falling from rest in a resistive medium is described by the equation

\dfrac{dv}{dt} = a - bv,

where

a

and

b

are constants. The velocity at any time

t

Options

A $a ( 1 - b^{2t} )$
B $\dfrac{a}{b} ( 1 - e^{- bt} )$
C $abe^{- t}$
D $ab^{2} ( 1 - t )$

Correct answer

B. $\dfrac{a}{b} ( 1 - e^{- bt} )$

Step-by-step solution

\dfrac{dv}{dt} = a - bv

<br/><br/>or

\mathop{∫}_{0}^{v}\dfrac{dv}{a - bv} = \mathop{∫}_{0}^{t}dt

<br/><br/>or

- \dfrac{1}{b}{\left[\log \left(a - bv\right)\right]}_{0}^{v} = t

<br/><br/>or \left(\dfrac{a - bv}{a}\right) = e^{- bt}

a - \dfrac{b}{a}v = e^{- bt}

v = \dfrac{a}{b} ( 1 - e^{- bt} )

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