Question

1. helicopter J starts at a elevation of 1400 feet and descends at 8 feet per second at the same time a second helicopter K flying at 1500 ft starts to descend at 10 ft per a second a.) write an equation to model the elevation of each helicopter b.) what are the helicopters elevations after 20 seconds?after 60 seconds c. when does one helicopter pass the other what is the elevation when that happens

Answer 1

Step-by-step explanation:

To model the elevation of each of the helicopter, the equation will be as follows respectively.

Elevation of J = 1400 - 8t

Elevation of K = 1500 - 10t

B) The helicopter elevation after 20s and 60s be realised by substituting each of the value into the equations above.

For Elevation of J = 1400 - 8t

J = 1400 - 8(20)

J = 1400-160

J = 1240ft

Also at 60s

J= 1400 - 8(60)

J = 1400-480

J = 920ft

For Elevation of K = 1500 - 10t

K = 1500- 10(20)

K = 1500-200

K= 1300ft

At 60s

K = 1500- 10(60)

K = 1500-600

K= 900ft

C) Elevation of J = Elevation of K

1400 - 8t = 1500 - 10t

Collecting like terms

1400-1500 = -10t+8t

-100 = -2t

t = -100/-2

t = 50 seconds

Elevation = 1400 - 400 = 1000 ft