Question

An air traffic controller is tracking two planes. To start, Plane A is at an altitude of 3028 feet and Plane B is at an altitude of 4000 feet. Plane A is gaining altitude at 55.75 feet per second and Plane B is gaining altitude at 35.5 feet per second. How many seconds will pass before the planes are at the same altitude? seconds What will their altitude be when they're at the same altitude? feet

Answer 1

Consider plane A with respect to B,

initial relative Altitude
`h_(a, b)=h_a-h_b=-972`

relative rate of altitude(speed) $r_{a,b}= 55.75-35.5=20.25$

To get at same altitude, they'll take same time, and their relative height will be $0$ So,

$H_{a,b}=h_{a,b}+r_{a,b}t$

$0=-972+20.25t$

$t=48$

and altitude will be, $3028+55.75\cdot 48=5704$

Answer 2

• 48 seconds
• 5704 feet

Explanation:

We can write equations for the altitude of each plane:

A(t) = 3028 +55.75t . . . . . initially at 3028 ft; gaining at 55.75 ft/s

B(t) = 4000 +35.5t . . . . . initially at 4000 ft; gaining at 35.5 ft/s

The two altitudes will be equal when ...

A(t) = B(t)

3028 +55.75t = 4000 +35.5t . . . . substitute the expressions for A and B

20.25t = 972 . . . . . . subtract 3028+35.5t

t = 48 . . . . . . . . . . . . divide by 20.25

The common altitude will be ...

B(48) = 4000 +35.5(48) = 5704 . . . . feet

The planes will both be at an altitude of 5704 feet after 48 seconds.