Chapter 9.6, Problem 22P

To determine

To find: the speed of the plane in still air and the speed of the wind.

Answer to Problem 22P

The rate at still air is $\text{770}\text{km}/\text{h}$ and the rate of the wind is $\text{70}\text{km}/\text{h}$ .

Explanation of Solution

Given:

The speed of an airplane in still air is $\text{760}\text{km}/\text{h}$ .

It travels $\text{2000}$ km with the wind $\text{1800}$ km against the wind in the same amount of time.

Concept used:

Speed and the rate have same meaning here.

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

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 $\text{w}$ be the speed of the wind and $\text{t}$ be the time takes for both directions.

If the speed in still air is $760\text{km}/\text{h}$ ,then the speed with the wind is $\text{760+w}$ and against the wind is $\text{760-w}$ .

Using formula:

  $\text{distance=rate×time}$ :

  $\text{2000=}\left(\text{760+w}\right)\text{t}$ .

Dividing both side by $\text{760+w}$ :

  $\text{t=}\frac{\text{2000}}{\text{760+w}}$ ………………. $\left(\text{i}\right)$

Similarly,

  $\text{1800=}\left(\text{760-w}\right)\text{t}$ .

Dividing both side by $\text{760-w}$ :

  $\text{t=}\frac{\text{180}}{\text{760-w}}$ ……………….. $\left(\text{ii}\right)$

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

  $\frac{\text{2000}}{\text{760+w}}\text{=}\frac{\text{180}}{\text{760-w}}$ .

  $\text{2000}\left(\text{760-w}\right)\text{=1800}\left(\text{760+w}\right)$ .

  $\text{20}\left(\text{760-w}\right)\text{=18}\left(\text{760+w}\right)$ .

  $\text{15200-20w=13680+18w}$ .

  $\text{w=40}$ .

  $\text{24r+72=36r-108}$ .

  $\text{r=15}$ .

Hence, the rate or speed of the wind is $15$ .