how fast was each plane flying?

Question

two planes left the airport at noon. one flew east and the other flew west at twice the speed. after 3 hours the planes were 2700 mi apart. how fast was each plane flying?

Answer 1

R₁ = speed of first plane;   2R₁ = speed of 2nd plane

 

both planes flew for 3 hours

 

R times time = distance

 

R₁(3) = distance flown by first plane

 

2R₁(3) = distance flown by 2nd plane

 

Add the two distances:

 

              3R₁ + 3(2R₁) = 2700 miles

 

                            9R₁ = 2700

 

                              R₁ = 300 

 

one plane flew at 300 m.p.h. and the other flew at 600 m.p.h.

Answer 2

Taylor,

 

We have two planes.  Let's say the distance traveled by the 1st plane is d₁ and the distance traveled by the 2nd plane is d₂.

 

Then:

 

(1) d₁ + d₂ = 2700miles

 

Now, the general form for the distance equation is:

 

d = st (where s is the airplane's speed in miles/hr and t is the number of hours flown)

 

For plane 1 then:

 

(2) d₁ = 3s₁ (since we know they each flew 3 hours)

 

For plane 2:

 

(3) d₂ = 3s₂

 

Right now, we have 3 equations and 4 variables; we need one more equation.

 

We also know that the speed of the 2nd plane is twice that of the 1st.  So:

 

(4) s₂ = 2s₁

 

Now, given (1), (2), and (3), we can write;

 

(5) 3s₁ + 3s₂ = 2700miles 

 

To find s₁, substitute the value of s₂ in (4) for s₂ in (5):

 

3s₁ + 3(2s₁) = 2700

 

3s₁ + 6s₁ = 2700

 

9s₁ = 2700

 

s₁ = 2700/9 = 300miles/hour for plane 1.

 

To find the speed of plane 2, substitute 300 for s₁ in (4):

 

s₂ = 2(300) = 600miles/hour for plane 2.

 

Interesting problem!