# A paddle boat can move at a speed of 2 ​km/h in still water. The boat is paddled 6 km downstream in a river in the same time it takes to go 3 km upstream. What is the speed of the​ river?

Nov 21, 2017

${v}_{r} = {v}_{b} / 3 = \frac{2 k m . h {r}^{- 1}}{3} = \frac{2}{3}$ $k m . h {r}^{- 1}$

#### Explanation:

${v}_{b}$ : Speed of the boat in still water,
${v}_{r}$ : Speed of the river current.

${v}_{\setminus \uparrow}$ : Speed of the boat upstream,
${v}_{\setminus \downarrow}$ : Speed of the boat downstream.

v_{\uarr} = v_b - v_r; \qquad v_{\darr} = v_b + v_r;

${S}_{\setminus \uparrow}$ : Distance travelled upstream in time $\setminus \Delta t$
${S}_{\setminus \downarrow}$ : Distance travelled downstream in time $\setminus \Delta t$

Given: $\setminus q \quad {v}_{b} = 2$ km.hr^{-1}; \qquad S_{\uarr} = 3 km; \qquad S_{\darr} = 6 $k m$

For the same time interval, calculate the distance travelled upstream and downstream,

Upstream: \qquad S_{\uarr} = v_{uarr}.\Delta t;

$\setminus \Delta t = \setminus \frac{{S}_{\setminus \uparrow}}{{v}_{\setminus \uparrow}} = {S}_{\setminus \uparrow} / \left({v}_{b} - {v}_{r}\right)$ ...... (1)

Downstream:\qquad S_{\darr} = v_{darr}.\Delta t;

$\setminus \Delta t = \setminus \frac{{S}_{\setminus \downarrow}}{{v}_{\setminus \downarrow}} = {S}_{\setminus \downarrow} / \left({v}_{b} + {v}_{r}\right)$ ...... (2)

Comparing (1) and (2) we get,
\frac{S_{\uarr}}{v_b - v_r} = \frac{S_{\darr}}{v_b + v_r}; \qquad (v_b + v_r) = (\frac{S_{\darr}}{S_{\uarr}}) (v_b - v_r)

${S}_{\setminus \uparrow} = 3$ km; \qquad S_{\darr} = 6 km; \qquad \frac{S_{\darr}}{S_{\uarr}} = (6 km)/(3 km) = 2

(v_b + v_r) = 2 (v_b - v_r);
${v}_{r} = {v}_{b} / 3 = \frac{2 k m . h {r}^{- 1}}{3} = \frac{2}{3}$ $k m . h {r}^{- 1}$