Mathematics

Mathematics

Problem

For a particle moving in one dimension, the acceleration is related to velocity by dvdt=−kv2\displaystyle \frac{dv}{dt}=-k v^{2}, where k>0k>0 is a constant and v(0)=v0>0v(0)=v_{0}>0. Which of the following gives v(t)v(t) for t≥0t\ge 0? a) v(t)=v0e−ktv(t)=v_{0}e^{-k t} b) v(t)=v01−kv0t\displaystyle v(t)=\frac{v_{0}}{1-k v_{0} t} c) v(t)=v01+kv0t\displaystyle v(t)=\frac{v_{0}}{1+k v_{0} t} d) v(t)=v01+kt\displaystyle v(t)=\frac{v_{0}}{1+k t}

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