Question

An airliner maintaining a constant elevation of 2 miles passes over an airport at noon traveling 500 mi/hr due west. At 1:00 PM, another airliner passes over the same airport at the same elevation traveling due north at 550 mi/h. Assuming both airliners maintain the (equal) elevations, how fast is the distance between them changing at 2:30 PM

Answer 1

Answer:

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

Step-by-step explanation:

Let suppose that airliners travel at constant speed. The equations for travelled distance of each airplane with respect to origin are respectively:

First airplane

$r_{A} = 500\,\frac{mi}{h}\cdot t\\r_{B} = 550\,\frac{mi}{h}\cdot t$

Where t is the time measured in hours.

Since north and west are perpendicular to each other, the staight distance between airliners can modelled by means of the Pythagorean Theorem:

$s=\sqrt{r_{A}^{2}+r_{B}^{2}}$

Rate of change of such distance can be found by the deriving the expression in terms of time:

$\frac{ds}{dt}=\frac{r_{A}\cdot \frac{dr_{A}}{dt}+r_{B}\cdot \frac{dr_{B}}{dt}}{\sqrt{r_{A}^{2}+r_{B}^{2}} }$

Where $\frac{dr_{A}}{dt} = 500\,\frac{mi}{h}$ and $\frac{dr_{B}}{dt} = 550\,\frac{mi}{h}$, respectively. Distances of each airliner at 2:30 PM are:

$r_{A}= (500\,\frac{mi}{h})\cdot (1.5\,h)\\r_{A} = 750\,mi$

$r_{B}=(550\,\frac{mi}{h} )\cdot (1.5\,h)\\r_{B} = 825\,mi$

The rate of change is:

$\frac{ds}{dt}=\frac{(750\,mi)\cdot (500\,\frac{mi}{h} )+(825\,mi)\cdot(550\,\frac{mi}{h})}{\sqrt{(750\,mi)^{2}+(825\,mi)^{2}} }$

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$