The velocity of the particle [tex]\mathbf v(\xi)[/tex] is given by the derivative of the position vector [tex]\mathbf r(\xi)=x(\xi)\,\mathbf i+y(\xi)\,\mathbf j[/tex]. We have
[tex]\mathbf r(\xi)=k\xi\,\mathbf i+2k(1-e^\xi)\,\mathbf j[/tex]
with derivative
[tex]\dfrac{\mathrm d\mathbf r}{\mathrm d\xi}=\mathbf v(\xi)=k\,\mathbf i-2ke^\xi\,\mathbf j[/tex]
At [tex]\xi=0.5[/tex], we get a velocity vector of
[tex]\mathbf v(0.5)=k\,\mathbf i-2k\sqrt e\,\mathbf j[/tex]
Not important as far as I can tell, but the particle's speed at [tex]\xi[/tex] is [tex]v(\xi)=\beta\xi[/tex], which satisfies
[tex]\|\mathbf v(0.5)\|=v(\xi)\implies\sqrt{k^2+4k^2e}=\dfrac\beta2[/tex]
(Perhaps you may be required to solve for [tex]k[/tex] in terms of [tex]\beta[/tex], then report the velocity/acceleration vectors with respect to [tex]\beta[/tex], but I'll not do so because the original question makes no mention of needing to do that.)
Acceleration is given by the derivative of the velocity vector:
[tex]\dfrac{\mathrm d\mathbf v}{\mathrm d\xi}=\mathbf a(\xi)=-2ke^\xi\,\mathbf j[/tex]
and at [tex]\xi=0.5[/tex], the acceleration vector is
[tex]\mathbf a(0.5)=-2k\sqrt e\,\mathbf j[/tex]
