For a particle moving in one dimension

MATHEMATICS · Single correct · CBSE · Mathematics · Differential Equations · Differential Equations

Problem

For a particle moving in one dimension, the acceleration is related to velocity by

\displaystyle \frac{dv}{dt}=-k v^{2}

, where

k>0

is a constant and

v(0)=v_{0}>0

. Which of the following gives

v(t)

for

t\ge 0

? (a)

v(t)=v_{0}e^{-k t}

(b)

\displaystyle v(t)=\frac{v_{0}}{1-k v_{0} t}

(c)

\displaystyle v(t)=\frac{v_{0}}{1+k v_{0} t}

(d)

\displaystyle v(t)=\frac{v_{0}}{1+k t}