Chapter 9.6, Problem 5OE

To determine

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

Answer to Problem 5OE

The resultant of the vector is sum up if the direction of the boat and current are same.

Explanation of Solution

Given:

  $\text{r+c}$ .

Where

  $\text{r=}$ rate of rowboat instill water.

  $\text{s=}$ rate of a swimmer in still water.

  $\text{c}$ =rate of current of silver river.

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 $\text{r}$ is the rate of the boat in still water and $\text{c}$ is the rate of the current, then the rate of boat with the current will be $\text{r+c}$ .

Since the direction of both current and the boat is 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 resultant of the vector is sum up if the direction of the boat and current are same.