Answer
a) $t=\frac{18}{5u}$
b) $t=\frac{9}{2d}$
c) Yes
Work Step by Step
The distance ($d$), rate ($r$), and time ($t$) of uniform motion problems is given by
$$
d=rt
.$$
Since the rowing team rows upstream for $2$ miles with a rate of $u$, then
$$\begin{aligned}
2&=ut_1
\\
t_1&=\frac{2}{u}
,\end{aligned}$$ where $t_1$ is the time to row upstream.
Since the rowing team rows downstream for $2$ miles with a rate of $25\%$ faster than the rate upstream, then
$$\begin{aligned}
2&=(u+0.25u)t_2
\\
2&=1.25ut_2
\\
t_2&=\frac{2}{1.25u}
\\&=
\frac{200}{125u}
\\&=
\frac{8}{5u}
,\end{aligned}$$ where $t_2$ is the time to row downstream.
a) Adding the two time expressions above, then the time, $t$, that the team spent rowing is
$$\begin{aligned}
t&=t_1+t_2
\\&=
\frac{2}{u}+\frac{8}{5u}
\\&=
\frac{10}{5u}+\frac{8}{5u}
\\&=
\frac{18}{5u}
.\end{aligned}$$Hence, the total time spent rowing is $t=\frac{18}{5u}$.
b) If $d$ represents the team's rate rowing downstream (which is $25\%$ faster than the rate upstream), then
$$\begin{aligned}
d&=u+0.25u
\\
d&=1.25u
\\
\frac{d}{1.25}&=u
\\
u&=0.8d
.\end{aligned}$$
Since $t=\frac{18}{5u}$ and $u=0.8d$, then
$$
t=\frac{18}{5(0.8d)}=\frac{18}{4d}=\frac{9}{2d}
.$$ Hence, the total time spent rowing is $t=\frac{9}{2d}$.
c) Yes, the expressions in (a) and (b) represent the same time as long as the relationship between $u$ and $d$ (i.e. $u=0.8d$) is assumed.
