t₀=0, x₀=0, v(x)=√(p+qx). x(t)=?

t₀=0
x₀=0
v(x)=√(p+qx)
x(t)=?

Calculus-based solution:

dx/dt=q⁰·⁵(p/q+x)⁰·⁵

(p/q+x)⁻⁰·⁵dx = q⁰·⁵dt

(p/q+x)⁻⁰·⁵d(p/q+x) = q⁰·⁵dt

2(p/q+x)⁰·⁵ = q⁰·⁵t + c

2(p/q+x)⁰·⁵ -2(p/q+x₀)⁰·⁵= q⁰·⁵(t-t₀)

2(p/q+x)⁰·⁵ -2(p/q)⁰·⁵= q⁰·⁵t

(p/q+x)⁰·⁵ -(p/q)⁰·⁵= ½q⁰·⁵t

(p/q+x)⁰·⁵ = ½q⁰·⁵t+(p/q)⁰·⁵

p/q+x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵ + p/q

x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵

x = p⁰·⁵t + ¼qt²

x = v₀ t + ½at²
v₀=p⁰·⁵
a=½q

Physics-based and algebra-based solution
If a=const then v²=v₀²+2a·Δx, so
v=√(v₀²+2a·Δx)

As it is given:
v=√(p+qx)
so p=v₀², v₀=√p=p⁰·⁵
as x₀=0 then 2a·Δx=2ax
2ax=qx, 2a=q, a=½q

as x = v₀ t + ½at²
then x = √p t + ¼qt²

⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ⁱⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ·ΔΘΣΦΧΨΩαβγδεζηθικλμνξπρςστφχψωϑϓϕ√∛∜∞∠∣∥∫∬∭∮∯≤≥⊥℃ℇ℉ℏ℔№⅀⅏⎛⎞⎜⍊⎝½