Question

a motor boat covers the distance between two spots on the river in t1=8 hr and t2=12 hr downstream and upstream respectively.the time required for the boat to cover this distance in still water will be

Answer 1

1. A motor boat covers the distance between two spots on the river in $t_1 = 8$ hr and $t_2 = 12$ hr downstream and upstream respectively. The time required for the boat to cover this distance in still water will be...

$$\begin{array}{l} t_1 = 8\, \text{hr} \\ t_2 = 12\, \text{hr} \\ \hline t_0 - ? \end{array}$$

Solution.

Let denote the speed of the boat in still water by $v$ and the speed of the water by $v_0$. The speed of boat downstream and upstream is $v + v_0$, $v - v_0$, respectively.

Let the distance between the spots be equal to $l$.

The time, which is spent for covering this distance downstream and upstream respectively:

$$t_1 = \frac{l}{v + v_0}, \quad t_2 = \frac{l}{v - v_0}.$$

The time required for the boat to cover the distance in still water is

$$t_0 = \frac{l}{v},$$

but we do not know both quantities $l$ and $v$. So, we have to express the ratio $\frac{l}{v}$ from the system of the equations (1). Let do some transformations with these equations.

$$\begin{array}{l} t_1 = \frac{1}{\frac{v}{l} + \frac{v_0}{l}}, \quad t_2 = \frac{1}{\frac{v}{l} - \frac{v_0}{l}}, \\ \frac{v}{l} + \frac{v_0}{l} = \frac{1}{t_1}, \quad \frac{v}{l} - \frac{v_0}{l} = \frac{1}{t_2}. \end{array}$$

Let write the sum of the last two expressions:

$$\frac{2v}{l} = \frac{1}{t_1} + \frac{1}{t_2}.$$

Dividing by 2, we obtain the required ratio:

$$\frac{v}{l} = \frac{1}{2} \left( \frac{1}{t_1} + \frac{1}{t_2} \right).$$

So, the time required for the boat to cover this distance in still water is

$$t_0 = \frac{l}{v} = \frac{1}{\frac{v}{l}} = \frac{1}{\frac{1}{2} \left( \frac{1}{t_1} + \frac{1}{t_2} \right)}, \quad \boxed{t_0 = \frac{2}{\frac{1}{t_1} + \frac{1}{t_2}}}.$$

Let check the dimension.

$$\left[ t_0 \right] = \frac{1}{\frac{1}{hr}} = hr.$$

Let evaluate the quantity.

$$t_0 = \frac{2}{\frac{1}{8} + \frac{1}{12}} = 9.6\, \text{(hr)}, \quad \text{so} \quad t_0 = 9.6\, \text{hr} \cdot 36\, \text{min}.$$

Answer: $9.6\, \text{hr} \cdot 36\, \text{min}$.