Question

The velocity of a particle is given by v = V0 sin er where v0 is constant and w = 2π/T find the average velocity in time interval t = 0 to t = T/2​

Answer 1

Answer :

If the velocity is variable and depends on time time t, to find the average value of velocity for time interval t = t₁ to t = t₂

${ \leadsto{ \sf{ \bar{v} = \dfrac{ \int \limits_{t_2}^{t_1} vdt }{ \int\limits_{t_2}^{t_1}dt} }}}$

v = v$\sf{_0}$sinωt

t = t₁ = T/2

t = t₂ = 0

${ \leadsto{ \sf{ \bar{v} = \dfrac{ \int \limits_{ 0 }^{T/2} v _{0} \sin( \omega t) dt }{ \int\limits_{0}^{ T/2 }dt} }}}$

${ \leadsto{ \sf{ \bar{v} = \dfrac{ v _{0} \dfrac{ | - \cos ( \omega t)|_{ \tiny{ 0 }}^{ \tiny{ T/2 }} }{ \omega} }{ |t|_{ \tiny{ 0 }}^{ \tiny{ T/2 }} = \bigg( \dfrac{T}{2} - 0 \bigg)} }}}$

${ \leadsto{ \sf{ \bar{v} = \dfrac{2v _{0} }{ \omega tT} \bigg[ \bigg \{ - \cos \bigg( \dfrac{ \omega T }{2} \bigg) \bigg \} - \big \{ - \cos(0) \big \} \bigg]}}}$

${ \leadsto{ \sf{ \bar{v} = \dfrac{2 v_{o}}{2\pi} \big( - \cos\pi + 1 \big) }}}$

${ \leadsto{ \sf{ \bar{v} = \dfrac{ v_{o}}{\pi} \big( - ( - 1) + 1 \big) }}} \: \: \: \: ( \because \cos\pi = \cos180 = - 1 )$

${ \leadsto{ \sf{ \bar{v} = \dfrac{2 v_{o}}{\pi} }}}$