A boat moves with a constant speed relative to the water…
A boat moves with a constant speed relative to the water between two cities on a river. The trip upstream takes $t_1 = 6\ \mathrm{h}$ and the trip downstream takes $t_2 = 3\ \mathrm{h}$. How long would the boat need to travel downstream with the engine turned off?
Explanation of Method
A boat drifting downstream with the engine off.
Step 1: Write the two trips
Let $v_1$ be the boat's speed relative to the water, $v_2$ the speed of the current, and $d$ the distance between the cities (not given, and not needed): $v_1 - v_2 = \frac{d}{t_1}$.
\[ v_1 + v_2 = \frac{d}{t_2}. \]
With the engine off the boat just drifts with the current, so its speed is $v_2$ and the time is $t_3 = \dfrac{d}{v_2}$.
Step 2: Find the current
Subtract the first equation from the second to eliminate $v_1$: \[2v_2 = \frac{d}{t_2} - \frac{d}{t_1} = d\,\frac{t_1 - t_2}{t_1 t_2} \;\Longrightarrow\; v_2 = \frac{d\,(t_1 - t_2)}{2\,t_1 t_2}.\]
Step 3: Time with the engine off
\[ t_3 = \frac{d}{v_2} = \frac{2\,t_1 t_2}{t_1 - t_2} = \frac{2 \cdot 6\ \mathrm{h} \cdot 3\ \mathrm{h}}{6\ \mathrm{h} - 3\ \mathrm{h}}. \]
\[t_3 = 12\ \mathrm{h}\]
Check
From $6(v_1 - v_2) = 3(v_1 + v_2)$ we get $v_1 = 3v_2$, so $d = 6 \cdot 2v_2 = 12\,v_2$: upstream: $ \dfrac{12 v_2}{3v_2 - v_2} = 6\ \mathrm{h}\ \checkmark $ downstream: $ \dfrac{12 v_2}{3v_2 + v_2} = 3\ \mathrm{h}\ \checkmark $ drifting: $ \dfrac{12 v_2}{v_2} = 12\ \mathrm{h} $
Result
\[\boxed{\,t_3 = 12\ \mathrm{h}\,}\]
Remark
Note. Compare with Translational Kinematics 010, where the distance was known and we found the speeds themselves. Here only times are given, so we can find the answer as a time, but not the speeds.