Particle moves along a straight line with the acceleration of a = 2 / sqrt (v) where v = dx/dt is the velocity. When t=2s then x = 64/3m with speed v = 16m/s. Determine the location and acceleration when t= 3s

Question

irishboy123 · Answer

break it down

you have 

$a = \dfrac{2}{\sqrt{v}}$

which means that 

$\dfrac{dv}{dt} = \dfrac{2}{\sqrt{v}}$

So

$\int \sqrt{v} \, dv  =\int  2  dt $

irishboy123 · Answer

$v(2) = 16$

beggi9 · Answer

@Irishboy123 ah okay I see, thank you so much!