airplane rate of change problem

Question

Flying against the wind, an airplane travels 4200 km in 7 hours. Flying with the wind, the same plane travels 4000 km in 4 hours. What is the rate of the plane in still air and what is the rate of the wind?

Answer 1

Let V_a = the velocity of the airplane

Let V_w = the velocity of the wind

 

When flying with the wind:

 

(V_a+V_w)*(4 hours) = 4000

4V_a + 4V_w = 4000

4V_w = 4000 - 4V_a

V_w = 1000 - V_a

 

When flying against the wind:

(V_a-V_w)*(7 hours) = 4200 km
7V_a - 7V_w = 4200

 

Substitute 1000-V_a for V_w and solve for V_a:

 

7V_a - 7(1000-V_a) = 4200          

7V_a -7000 + 7V_a = 4200

14V_a = 11200

 

V_a = 800 km/hr

 

Rate of wind:

 

Vw = 1000 - Va = 1000 - 800 = 200 km/hour

Answer 2

distance=rate x time

d=rt

4200=r*7

r=600 kph

4000=r*4

r=1000 kph

r=rate of plane in still air

w=rate of wind

600=r-w

1000=r+w

add the two equations

1600=2r

r=800 kph (rate of plane)

1000=800+w

w=200 kph (rate of wind)