Chapter 9.6, Problem 11OE

To determine

To find:The representation of the rate of an expression and the rate is in km per hour.

Answer to Problem 11OE

The equation is stating that the rate of the swimmer with current is $\text{3}\text{km}/\text{h}$ .

Explanation of Solution

Given:

  $\text{s+c=3}$ .

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:

If $\text{s}$ is the rate of the swimmer in still water and $\text{c}$ is the rate of the current, then the rate of swimmer with the current will be:

  $\text{s+c=3}$ .

Since the direction of both current and the swimmer is in the same direction.

There resultant will be sum up.

Formula for the resultant will be:

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

Since the direction of flow is same the angle between them will be ${\text{0}}^0$ .

So, $\cos 0^0=1$ and the resultant will be:

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

  $\Rightarrow {\text{R}}^{\text{2}}\text{=}{\left(\text{A+B}\right)}^{\text{2}}$ .

  $\text{R=}\left(\text{A+B}\right)$ .

Hence, the equation is stating that the rate of the swimmer with current is $\text{3}\text{km}/\text{h}$ .