#### Answer

$2 \sqrt 5 i+\sqrt 5 j$

#### Work Step by Step

We are given that the velocity vector is tangent to the graph of $y^2=2x$ at the point$(2,2)$ with length $5$.
We need to find the direction of the velocity vector.
$y^2=2x \implies 2y\dfrac{dy}{dx}=2$
Thus, the slope at the point$(2,2)$ is $\dfrac{dy}{dx}=\dfrac{1}{2}$
So, the tangent vector lies in the direction of the vector $i+\dfrac{1}{2} y$
Hence, the velocity $=5(\dfrac{i+\dfrac{1}{2} y}{\sqrt{1+1/4}})=2 \sqrt 5 i+\sqrt 5 j$