Question

A particle is moving in a straight line such that it's velocity varies as v= vo e^-lambda t ,where lambda and t are constant . Find the average velocity during the time interval in which the velocity decreases from vo to vo/2

Answer 1

Given that,

Velocity $v=v_{0}e^{-\lambda t}$

Where, $\lambda$ and t are constant

v = v₀ at t = 0

$v=\dfrac{V_{0}}{2}$ at t = t₀

We need to calculate the value of  t₀

Using formula of velocity

$v=v_{0}e^{-\lambda t}$

v = v₀ at t = 0

$v=\dfrac{V_{0}}{2}$ at t = t₀

Put the value of v in given equation

$\dfrac{v_{0}}{2}=v_{0}e^{-\lambda t_{0}}$

$\dfrac{1}{2}=e^{-\lambda t_{0}}$

$e^{\lambda t_{0}}=2$

Taking log on base e both side

$\lambda t_{0}=\log_{e}2$

$t_{0}=\dfrac{\log_{e}2}{\lambda}$

We need to calculate the average velocity for time duration t = 0 to t = t₀

Using formula of average velocity

$\bar{v}=\dfrac{\int_{0}^{t_{0}}{v dt}}{\int_{0}^{t_{0}}{dt}}$

$\bar{v}=\dfrac{\int_{0}^{t_{0}}{v_{0}e^{-\lambda t}}}{\int_{0}^{t_{0}}{dt}}$

$\bar{v}=\dfrac{v_{0}\int_{0}^{t_{0}}{e^{-\lambda t}}}{(t)_{0}^{t_{0}}}$

$\bar{v}=\dfrac{v_{0}}{t_{0}}(\dfrac{e^{-\lambda t}}{-\lambda})_{0}^{t_{0}}$

$\bar{v}=\dfrac{v_{0}}{-\lambda t_{0}}(e^{-\lambda t_{0}}-e^{0})$

$\bar{v}=\dfrac{v_{0}}{\lambda t_{0}}(1-e^{-\lambda t_{0}})$

$\bar{v}=\dfrac{v_{0}}{\lambda t_{0}}(1-\dfrac{1}{2})$

$\bar{v}=\dfrac{v_{0}}{2\lambda t_{0}}$

Put the value of t₀

$\bar{v}=\dfrac{v_{0}}{2\lambda\times\dfrac{\log_{e}2}{\lambda}}$

$\bar{v}=\dfrac{v_{0}}{2\log_{e}2}$

Hence, The average velocity is $\dfrac{v_{0}}{2\log_{e}2}$