Problem 9
Easy Translational Kinematics

A boat travels downstream a distance s = 24 km with a…

A boat travels downstream a distance $s = 24\ \mathrm{km}$ with a constant speed $v_B = 8\ \mathrm{km/h}$ relative to the water. The trip takes $t = 2.5\ \mathrm{h}$. Find the speed of the river current $v_R$.

Explanation of Method

A boat moving downstream.

Step 1: Speed relative to the shore

Going downstream, the current pushes the boat forward, so the boat's speed relative to the shore is the sum: $v_{\text{down}} = v_B + v_R$.

Step 2: Use the measured distance and time

The 24 km and 2.5 h are measured from the shore, so they give the shore speed directly: \[v_{\text{down}} = \frac{s}{t} = \frac{24\ \mathrm{km}}{2.5\ \mathrm{h}} = 9.6\ \mathrm{km/h}.\]

$v_B = 8\ \mathrm{km/h}\ \ (\text{boat relative to water})$ $v_R = \,?$ $\text{river flow}$ $v_B + v_R = \tfrac{24\ \mathrm{km}}{2.5\ \mathrm{h}} = 9.6\ \mathrm{km/h}\ \ (\text{relative to shore})$
Downstream, the current carries the boat forward, so the two velocities add. The speed relative to the shore comes from the measured distance and time.

Step 3: Speed of the current

\[ v_R = v_{\text{down}} - v_B = 9.6\ \mathrm{km/h} - 8\ \mathrm{km/h}. \]

\[v_R = 1.6\ \mathrm{km/h}\]

Check

$ (8 + 1.6)\ \mathrm{km/h} \cdot 2.5\ \mathrm{h} = 9.6 \cdot 2.5\ \mathrm{km} = 24\ \mathrm{km}\ \checkmark $


Result

\[\boxed{\,v_R = 1.6\ \mathrm{km/h}\,}\]

Remark

Note. Without the current, the boat would need $24/8 = 3\ \mathrm{h}$. The current saves half an hour on the way down, but on the way back upstream it would cost time: the speed would be only $8 - 1.6 = 6.4\ \mathrm{km/h}$.