Since acceleration is constant, the average and instantaneous accelerations are the same, so that
[tex]a = a_{\rm ave} = \dfrac{\Delta v}{\Delta t} = -\dfrac{v_i}{10.0\,\rm s}[/tex]
By the same token, we have the kinematic relation
[tex]v^2 - {v_i}^2 = 2a\Delta x[/tex]
where [tex]v[/tex] is final speed, [tex]v_i[/tex] is initial speed, [tex]a[/tex] is acceleration, and [tex]\Delta x[/tex] is displacement.
Substitute everything you know and solve for [tex]v_i[/tex] :
[tex]0^2 - {v_i}^2 = 2\left(-\dfrac{v_i}{10.0\,\rm s}\right)(75.0\,\mathrm m)[/tex]
[tex]\implies {v_i}^2 - \left(15.0\dfrac{\rm m}{\rm s}\right) v_i = 0[/tex]
[tex]\implies v_i \left(v_i - 15.0\dfrac{\rm m}{\rm s}\right) = 0[/tex]
[tex]\implies v_i = \boxed{15.0\dfrac{\rm m}{\rm s}}[/tex]
