Chapter 9.6, Problem 6OE

To determine

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

Answer to Problem 6OE

The resultant of the vector is difference of them if the direction of the boat and current are opposite to each other.

Explanation of Solution

Given:

  $\text{r-c}$ .

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 against the current will be $\text{r-c}$ .

Since the direction of both current and the boat is opposite in direction.

There resultant will be difference of them.

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 ${\text{180}}^0$ .

So, $\cos {180}^0=-1$ 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 difference of them if the direction of the boat and current are opposite to each other.