A steamboat travels between two cities on the same river.…

A steamboat travels between two cities on the same river. The trip upstream takes $t_1 = 9\ \mathrm{h}$, and the trip downstream takes $t_2 = 4\ \mathrm{h}$. The distance between the cities is $d = 72\ \mathrm{km}$. Find the average speed $u$ of the river current relative to the shore and the speed $v$ of the steamboat relative to the water.

Explanation of Method

A steamboat going upstream and downstream.

Step 1: Speeds relative to the shore

Downstream the current helps the boat; upstream it works against it: $v_{\text{down}} = v + u, \qquad v_{\text{up}} = v - u$.

$\text{Downstream: } t_2 = 4\ \mathrm{h}$ $v = 13$ $u = 5$ $v + u = \tfrac{72}{4} = 18\ \mathrm{km/h}$ $\text{Upstream: } t_1 = 9\ \mathrm{h}$ $v = 13$ $u = 5$ $v - u = \tfrac{72}{9} = 8\ \mathrm{km/h}$ $\text{river flow}$

boat relative to water
current
boat relative to shore

The current always flows the same way. Going downstream it adds to the boat's own speed; going upstream it subtracts. (Arrows are drawn to scale, speeds in km/h.)

Step 2: Use the distance and the times

From speed = distance / time for each trip: \[v + u = \frac{d}{t_2} = \frac{72\ \mathrm{km}}{4\ \mathrm{h}} = 18\ \mathrm{km/h}, \qquad v - u = \frac{d}{t_1} = \frac{72\ \mathrm{km}}{9\ \mathrm{h}} = 8\ \mathrm{km/h}.\]

Step 3: Solve the system

Add the two equations to eliminate $u$:

\[ (v + u) + (v - u) = 18 + 8 \;\Longrightarrow\; 2v = 26\ \mathrm{km/h} \;\Longrightarrow\; v = 13\ \mathrm{km/h}. \]

Subtract them to eliminate $v$:

\[ (v + u) - (v - u) = 18 - 8 \;\Longrightarrow\; 2u = 10\ \mathrm{km/h} \;\Longrightarrow\; u = 5\ \mathrm{km/h}. \]

\[v = 13\ \mathrm{km/h}, \qquad u = 5\ \mathrm{km/h}\]

Check

$ \text{upstream: } \dfrac{72\ \mathrm{km}}{(13 - 5)\ \mathrm{km/h}} = \dfrac{72}{8}\ \mathrm{h} = 9\ \mathrm{h}\ \checkmark $

$ \text{downstream: } \dfrac{72\ \mathrm{km}}{(13 + 5)\ \mathrm{km/h}} = \dfrac{72}{18}\ \mathrm{h} = 4\ \mathrm{h}\ \checkmark $

Result

\[\boxed{\,v = 13\ \mathrm{km/h}, \qquad u = 5\ \mathrm{km/h}\,}\]

Remark

Handy shortcut. The boat's own speed is the average of the two shore speeds, $v = \tfrac{18 + 8}{2}$, and the current is half their difference, $u = \tfrac{18 - 8}{2}$. Compare with Translational Kinematics 009, where only the downstream trip was known and the boat's own speed was given.