Math Application Air Search Problem

Question

A search plane takes off from an airport at 6:00 A.M. and travels due north at 200 miles per hour. A second plane takes off at 6:30 A.M. and travels due east at 170 miles per hour. The planes carry radios with a maximum range of 500 miles. When (to the nearest minute) will these planes no longer be able to communicate with each other?

Answer 1

Hi Ana.

 

The planes in your problem are flying at right angles to each other, and you want to know when the hypotenuse of this triangle is 500 miles long.

 

You know a few things:

 

1. The first plane travels at 200 miles/ hr and leaves at 6:00 (t = 0)

2. The second plane travels at 170 miles/ hr and leaves at 6:30 (t = 1/2)

3. Distance = Rate * Time

 

Using the Pythagorean Theorem:

 

(Distance of Plane 1)^2 + (Distance of Plane 2)^2 = 500^2

 

Distance of Plane 1 = 200*t

Distance of Plane 2 = 170*(t - 1/2)

 

The (t - 1/2) term comes from the fact that Plane 2 leaves 30 minutes after Plane 1. Since our unit of time in mph is hours, we need to convert 30 minutes to hours.

 

Here's where it gets ugly.

 

Substituting:

 

(200t)^2 + (170*(t-1/2))^2 = 500^2

 

40000*t^2 + 28900*(t^2 - t + 1/4) = 250000

 

40000*t^2 + 28900*t^2 - 28900*t + 7225 = 250000

 

68900*t^2 - 28900*t - 242775  = 0

 

You can now use the quadratic formula to solve for t.

 

t1 = 2.10

t2 = -1.68

 

Since you can't have negative time, we'll toss out t2.

 

The planes will no longer be in contact 2.10 hours after the first plane takes off.

 

Always good to verify your results:

 

200*2.10 = 420

170*(2.10 - 1/2) = 272

 

420^2 + 272^2 = 250384

 

sqrt(250384) = 500.4

 

Close enough with the rounding errors.

 

0.10 hours = 6 minutes

 

2 hours and 6 minutes after Plane 1 takes off is 8:06am.

 

Hope this helps.