(a) Walking 73.2 m at 1.22 m/s would take
[tex]\dfrac{73.2\,\rm m}{1.22 \frac{\rm m}{\rm s}} = 60 \,\rm s[/tex]
and running 13.2 m at 3.02 m/s would take
[tex]\dfrac{13.2\,\rm m}{3.02\frac{\rm m}{\rm s}} \approx 4.37\,\rm s[/tex]
You've undergone a total displacement of 73.2 + 13.2 = 86.4 m in a matter of approximatly 64.37 s, so your average velocity is
[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{86.4\,\mathrm m}{64.37\,\rm s} \approx \boxed{1.34\dfrac{\rm m}{\rm s}}[/tex]
(b) In the first 1.00 min = 60 s, you undergo a displacement of
[tex](60\,\mathrm s) \left(1.22 \dfrac{\rm m}{\rm s}\right) = 73.2 \,\rm m[/tex]
and in the second minute, you undergo a displacement of
[tex](60\,\mathrm s) \left(3.05\dfrac{\rm m}{\rm s}\right) = 183 \,\rm m[/tex]
Your total displacement is then 73.2 + 183 = 256.2 m in a matter of 2.00 min = 120 s, so your average velocity is
[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{256.2\,\mathrm m}{120\,\rm s} \approx \boxed{2.14\dfrac{\rm m}{\rm s}}[/tex]
(c) For part (a), your displacement [tex]x(t)[/tex] (in m) at time [tex]t[/tex] (in s) is given by
[tex]x(t) = \begin{cases}1.22t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.02 (t-60) & \text{for } t > 60\end{cases}[/tex]
and for part (b), your displacement is given by the very similar
[tex]x(t) = \begin{cases}1.22 t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.05(t-60) & \text{for } t > 60 \end{cases}[/tex]
See the attached plots. The average velocity for the given situation is the slope of the dotted line.
