Question

An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. One airplane is 150 miles from the point and has a speed of 300 miles per hour. The other is 200 miles from the point and has a speed of 400 miles per hour.(a) At what rate is the distances between the planes decreasing?(b) How much time does the air traffic controller have to get one of the planes on a different flight path?

Answer 1

Answer:

The answer to this question can be defined as follows:

In option A, the answer is "- 357.14 miles per hour".  

In option B, the answer is "-0.98".

Step-by-step explanation:

Given:

$\frac{dx}{dt} =- 300 \text{ miles per hour}$

$\frac{dy}{dt} =- 400 \text{ miles per hour}$

find:

$\frac{ds}{dt} =?$ when

$x= 150 \\y= 200\\s=x+y$

  $= 150+200 \\\\=350$

$\to s^2=x^2+y^2$

differentiate the above value:

$\to 2s\frac{ds}{dt}= 2x \frac{dx}{dt}+2y \frac{dy}{dt}$

$\to 2s\frac{ds}{dt}= 2(x \frac{dx}{dt}+y \frac{dy}{dt})\\\\\to \frac{ds}{dt}= \frac{(x \frac{dx}{dt}+y \frac{dy}{dt})}{s}$

        $= \frac{(150 \times -300 +200 \times -400 )}{350}\\\\= \frac{-45000+ (-80000) }{350}\\\\= \frac{- 125000 }{350}\\\\= - 357.14 \ \text{miles per hour}$

In option B:

$\to d=rt\\\\ \to t= \frac{d}{r}$

$\to \ \ d= 350 \ \ \ \ \ \ r= -357.14$

$\to t= - \frac{350}{357.14}\\\\\to t= - 0.98$