Chapter 9.6, Problem 12P

To determine

Tofind:the team’s rate in still water and the rate of current.

Answer to Problem 12P

The rate in still water $\text{7}$

  $\text{m}/\text{sec}$ and the rate of the current is $\text{1}\text{m}/\text{sec}$ .

Explanation of Solution

Given:

A team paddle downstream $\text{480m}$ in $60\sec$ .

The same team make the upstream in $\text{80sec}$ .

Concept used:

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

In the elimination method:

Either add or subtract equation to get an equation in one variable.

When the coefficients in one variable are opposites then add the equations to eliminate the variable and when the coefficients of one variable are equal and subtract the equation to eliminate a variable.

Calculation:

Let $\text{r}$ be the rate ins till water and $\text{c}$ be the rate of current. Then the canoe’s rate going downstream is $\text{r+c}$ and upstream is $\text{r-c}$ .

Using formula:

  $\text{distance=rate×time}$ .

  $\text{480=}\left(\text{r+c}\right)\left(\text{60}\right)$ .

Dividing both side by $\text{60}$ :

  $\text{r+c=8}$ …………… $\left(\text{i}\right)$

Similarly,

  $\text{480=}\left(\text{r-c}\right)\left(\text{80}\right)$ .

Dividing both side by $\text{80}$ :

  $\text{r-c=6}$ ……………….. $\left(\text{ii}\right)$

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

  $\text{r=7}$ and $\text{c=1}$

Hence, the rate in still water $\text{7}$

  $\text{m}/\text{sec}$ and the rate of the current is $\text{1}\text{m}/\text{sec}$ .