Answer
The maximum occurs at $t=0$ and the highest speed is $1$.
Work Step by Step
$v(t)=\dfrac{dr}{dt}=\dfrac{-t \ i +j}{(t^2+1)^{3/2}}$
$\implies |v(t)|=|\dfrac{-it+j}{(t^2+1)^{3/2}}|=\dfrac{1}{(t^2+1)}$
$\implies \dfrac{d|v(t)|}{dt}=\dfrac{-2t}{(t^2+1)^2}$
$\implies \dfrac{d|v(t)|}{dt}=\dfrac{-2t}{(t^2+1)^2}=0$
For $t \gt 0$, we notice that $ \dfrac{-2t}{(t^2+1)^2} \lt 0$.
This means that the maximum occurs at $t=0$ and the highest speed is $1$.
