Chapter 9.6, Problem 14OE

To determine

To find: the corresponding statement of the equation in option and the rate is in km per hour.

Answer to Problem 14OE

The option $\text{b}$ is correct the rowboat is travelled $\text{30}$ km upstream in $\text{2}$ hours.

Explanation of Solution

Given:

  $\frac{\text{30}}{\text{r-c}}\text{=2}$ .

Concept used:

If the thing is going against the gravity is taken as $\text{-ive}$ .

Vector formula:

  ${\text{R}}^2{\text{=A}}^{\text{2}}{\text{+B}}^{\text{2}}\text{+2ABcosθ}$ .

If the direction of any against the gravity means it make ${\text{180}}^0$ with the gravity.

  ${\text{R}}^2{\text{=A}}^{\text{2}}{\text{+B}}^{\text{2}}\text{-2AB}$ .

  $\text{R=A-B}$ .

  $\text{distance=rate}\times \text{time}$ .

Calculation:

According to the given equation:

  $\frac{\text{30}}{\text{r-c}}\text{=2}$ .

Here $\text{r-c}$ is the rate of the boat against the current so the boat is travelling upstream for $\text{2 hours}$ which matches with option $\text{b}$ .

Hence the option $\text{b}$ is correct the rowboat is travelled $\text{30}$ km upstream in $\text{2}$ hours.