The runner has initial velocity vector
[tex]\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath[/tex]
and acceleration vector
[tex]\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)[/tex]
so that her velocity at time [tex]t[/tex] is
[tex]\vec v=\vec v_0+\vec at[/tex]
She runs directly east when the vertical component of [tex]\vec v[/tex] is 0:
[tex]2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s[/tex]
It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time [tex]t[/tex] would be
[tex]\vec x=\vec v_0t+\dfrac12\vec at^2[/tex]
so that after 10.4 s, her position would be
[tex]\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath[/tex]
which is 19.9 m away from her starting position.
