Chapter 9.6, Problem 5P

To determine

Tofind:the rate of current and the equation that flows.

Answer to Problem 5P

The rate of current is $\text{4}\text{km}/\text{h}$ .

Explanation of Solution

Given:

  $\text{r}$ is rate in $\text{km}/\text{h}$

  $\text{c}$ is rate in $\text{km}/\text{h}$

Travelling downs stream rate:

  $\text{15}\text{km}/\text{h}$ .

Traveling upstream rate:

  $7\text{km}/\text{h}$ .

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 downstream at a rate of $\text{15}\text{km}/\text{h}$ .

Travelling downstream is given by $\text{r+c}$ .

  $\text{r+c=15}$ ………… $\left(\text{i}\right)$ .

Traveling upstream at a rate of $7\text{km}/\text{h}$

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

  $\text{r-c=7}$ …………. $\left(\text{ii}\right)$

Adding the equation $\left(\text{i}\right)$ and $\left(\text{ii}\right)$ :

  $\text{2r=22}$

Diving both side by $2$ :

  $\text{r=11}$ .

Substituting the value $\text{r=11}$ in equation $\left(\text{ii}\right)$ :

  $\text{r-c=7}$ .

  $\text{11-c=7}$ .

  $\text{c=4}$

Hence, the rate of current is $\text{4}\text{km}/\text{h}$ .