Consider plane A with respect to B,
initial relative Altitude [tex] h_{a, b}=h_a-h_b=-972 [/tex]
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:
Step-by-step 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.
