#### Answer

highest speed $=1$

#### Work Step by Step

$v(t)=\dfrac{dr}{dt}=\dfrac{-it+j}{(t^2+1)^{3/2}}$
and $|v(t)|=|\dfrac{-it+j}{(t^2+1)^{3/2}}|=\dfrac{1}{t^2+1}$
and $\dfrac{d|v(t)|}{dt}=\dfrac{-2t}{(t^2+1)^2}$
Set $\dfrac{d|v(t)|}{dt}=\dfrac{-2t}{(t^2+1)^2}=0$
For $t \gt 0$, we can see that $ \dfrac{-2t}{(t^2+1)^2} \lt 0$.
This implies that the maximum occurs at $t=0$ with highest speed $=1$.