Question

two planes start at 11 am from two airports 1800 miles apart, and fly toward each other at average rates of 200 and 250 miles per hour. the time at which they pass each other will be...

Answer 1

Let the airports be airport A and airport B.

Let the plane with average rate 200mi/h leave airport A, at the same time the plane with average rate 250 mi/h leaves airport B.

Let |AC|=x, then |CB|=1800-x

Let the 2 planes pass each other at point C, after t hours.

Recall the formula: Distance= Rate*Time

so Time=Distance /Rate

(i) t=$\frac{x}{200}$ hours

(ii) t=$\frac{1800-x}{250}$ hours

equalizing:

$\frac{x}{200}=\frac{1800-x}{250}$

250x=200(1800-x)

2.5x=2(1800-x)

2.5x=3600-2x

2.5x+2x=3600

4.5x=3600

x=3600/4.5=800 (mi)

Having found x, now we can apply the formula distance=rate*time again to find the time t:

200*t=800

t=800/200=4 (hours)


we add 4 hours to 11 am, so the time will be 3pm




Answer: 3pm