Recall that
[tex]{v_f}^2={v_i}^2+2a\Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are the skateboarder's velocity, respectively; [tex]a[/tex] is her acceleration; and [tex]\Delta x[/tex] is the change in her position.
Substitute everything you know and solve for [tex]v_f[/tex]:
[tex]{v_f}^2=\left(2.5\dfrac{\rm m}{\rm s}\right)^2+2\left(1.7\dfrac{\rm m}{\mathrm s^2}\right)(130\,\mathrm m)\implies v_f=448.25\dfrac{\mathrm m^2}{\mathrm s^2}\implies\boxed{v_f\approx21\dfrac{\rm m}{\rm s}}[/tex]
