Assuming the deer's acceleration was constant, we have
[tex]\Delta x=\dfrac{v_f+v_0}2t[/tex]
where [tex]\Delta x[/tex] is the deer's total displacement, [tex]v_f,v_0[/tex] are the deer's final and initial velocities, and [tex]t[/tex] is time. Then
[tex]63.5\,\mathrm m=\dfrac{13.38\,\frac{\mathrm m}{\mathrm s}+3.29\,\frac{\mathrm m}{\mathrm s}}2t[/tex]
[tex]\implies t=7.62\,\mathrm s[/tex]
