Chapter 9.6, Problem 17P

To determine

To find: the time taken and distance toward the downstream.

Answer to Problem 17P

Len can spend $\frac{1}{2}$ hours going downstream at a rate of $\text{5}\text{mi}/\text{h}$ ,his distance downstream would is $\text{5}\text{.}\frac{\text{1}}{\text{2}}\text{=2}\text{.5miles}$ .

Explanation of Solution

Given:

Len can paddle in still water at $\text{3}\text{mi}/\text{h}$ and the rate of current is $2\text{mi}/\text{h}$ .

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:

Let $\text{t}$ be the time taken to travel downstream. Since the trip is $3$ hours long, the time to travel upstream is $\text{3-t}$ .

His rate in still water is $\text{3+2=5}$

  $\text{mi}/\text{h}$ .letting $\text{d}$ be the distance each way and the formula:

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

  $\text{d=t}\left(\text{5}\right)$

  $\text{d=5t}$ ........ $\left(\text{i}\right)$

  $\text{d=}\left(\text{3-t}\right)\left(\text{1}\right)$ .

  $\text{d=3-t}$ ........ $\left(\text{ii}\right)$

Substitute $\text{d}$ by the equation $\left(\text{i}\right)$ and $\left(\text{ii}\right)$ get:

  $\text{5t=3-t}$ .

  $\text{6t=3}$ .

Dividing both side by $\text{3}$ :

  $\text{t=}\frac{1}{2}$ .

Hence, Len can spend $\frac{1}{2}$ hours going downstream at a rate of $\text{5}\text{mi}/\text{h}$ ,his distance downstream would is $\text{5}\text{.}\frac{\text{1}}{\text{2}}\text{=2}\text{.5miles}$ .