# The current of a river is 2 miles per hour. A boat travels to a point 8 miles upstream and back again in 3 hours. What is the speed of the boat in still water?

Nov 8, 2015

$3 , 737$ miles/hour.

#### Explanation:

Let the speed of the boat in still water be $v$.

Therefore total trip is the sum of the upstream part and the downstream part.

Total distance covered is hence ${x}_{t} = 4 m + 4 m = 8 m$

But since speed = distance/time, $x = v t$, so we may conclude that
${v}_{T} = {x}_{T} / {t}_{T} = \frac{8}{3}$mi/hr
and hence write :

${x}_{T} = {x}_{1} + {x}_{2}$

$\therefore {v}_{T} {t}_{T} = {v}_{1} {t}_{1} + {v}_{2} {t}_{2}$

$\therefore \frac{8}{3} \cdot 3 = \left(v - 2\right) {t}_{1} + \left(v + 2\right) {t}_{2}$

Also, ${t}_{1} + {t}_{2} = 3$.

Furthermore, ${t}_{1} = \frac{4}{v - 2} \mathmr{and} {t}_{2} = \frac{4}{v + 2}$

$\therefore \frac{4}{v - 2} + \frac{4}{v + 2} = 3$

$\therefore \frac{4 \left(v + 2\right) + 4 \left(v - 2\right)}{\left(v + 2\right) \left(v - 2\right)} = 3$

This leads to the quadratic equation in v, $3 {v}^{2} - 8 v - 12 = 0$, which upon solving yields $v = 3 , 737 \mathmr{and} v = - 1 , 07$.
Clearly the latter is impossible and so hence $v = 3 , 737$ is the only feasible solution.