Chapter 9.6, Problem 4P

To determine

Tofind:the equation of the row boat if its travelling upstream.

Answer to Problem 4P

The equation can be written as $\text{r=10+c}$ .

Explanation of Solution

Given:

  $\text{r}$ is rate in $\text{km}/\text{h}$ of the row boat in still water.

  $\text{c}$ is rate in $\text{km}/\text{h}$ of the current.

The row boat take $\text{2hours}$ to go $\text{20}$ km upstream.

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:

Travelling upstream is $\text{r-c}$ .

  $\text{distance=rate}\times \text{time}$ ……. $\left(\text{i}\right)$

Since distance is given as $\text{20km}$ .

Since the rowboat is travelling upstream it’s rate will be $\text{r-c}$ .

Time here is also given as $2$ hours.

By putting those given in equation ……. $\left(\text{i}\right)$

  $\text{20=2}\left(\text{r-c}\right)$

Which can be also simplify as:

  $\text{10=r-c}$ or $\text{r=10+c}$ .

Hence the equation can be written as $\text{r=10+c}$ .