Question

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

Answer 1

break it down

you have

\(a = \dfrac{2}{\sqrt{v}}\)

which means that

\(\dfrac{dv}{dt} = \dfrac{2}{\sqrt{v}}\)

So

\(\int \sqrt{v} \, dv =\int 2 dt \)

Answer 2

\(v(2) = 16\)

Answer 3

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