The particle's velocity in the [tex]y[/tex] direction at time [tex]t[/tex] is
[tex]v_y(t)=\left(7.1\,\dfrac{\mathrm m}{\mathrm s^2}\right)t[/tex]
so that after 0.7 s, its velocity in the [tex]y[/tex] direction is
[tex]v_y=4.97\,\dfrac{\mathrm m}{\mathrm s}[/tex]
At this point, the particle has a velocity vector [tex]v=\langle v_x,v_y\rangle=\langle4.5,4.97\rangle\,\dfrac{\mathrm m}{\mathrm s}[/tex]. Its speed will be the magnitude of this vector, which is given by
[tex]\|v\|=\sqrt{\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2+\left(4.97\,\dfrac{\mathrm m}{\mathrm s}\right)^2}\approx6.7\,\dfrac{\mathrm m}{\mathrm s}[/tex]
