An electric current is $\vec{J}=y^2 \hat{j} + z \hat{k}$. At time = 0, a charged particle is at (0, 0.25, 1). Where will it be at $t=11$ and $t=26$?
The divergence of the current is $\nabla\cdot \vec{J} =2y+1$. I don't understand how this relates to time.
This problem has nothing to do with divergence. Just solve it as a bunch of equations:
$$\frac{dx}{dt}=0~~~~\rightarrow ~~~~x=0$$ $$\frac{dy}{dt}=y^2~~~~\rightarrow ~~~~\int\frac{1}{y^2}dy=\int dt ~~~~\rightarrow~~~~ y=\frac{-1}{t+c_1}$$ $$\frac{dz}{dt}=z~~~~\rightarrow ~~~~\int\frac{1}{z}dz=\int dt ~~~~\rightarrow ~~~~z=e^{(t+c_2)}$$
Then find the constants $c_1, c_2$ with initial conditions:
$$y(t=0)=0.25=\frac{-1}{c_1} ~~~~\rightarrow ~~~~c_1=-4 ~~~~\rightarrow ~~~~y(t)=\frac{1}{4-t}$$ $$z(t=0)=e^{c_2}=1 ~~~~\rightarrow ~~~~c_2=0~~~~\rightarrow ~~~~z(t)=e^t$$
That gets you a position equations:
$$r(t)=\frac{1}{4-t}\hat{y} + e^t \hat{z}$$
Then you can plug your times into this equations.