average plane speed

Question

A plane travels 800 miles from Dallas, Texas, to Atlanta, Georgia, with a prevailing wind of 49 mph. The return trip against the wind takes 1/2 hour longer. Find the average speed of the plane in still air.

Answer 1

R is the speed of the plane.

T is the time in hours

 

800 = (R+49)*t

 

800 = (R-49)*(T+1/2)

 

(R+49)t = (R-49)(t + 1/2) <--- they both equal 800 miles, so they are equal by transitive property

 

rt + 49t = rt - 49t + R/2 - 49/2 <-- distributive left side; FOIL on right side

 

49t = -49t + R/2 - 49/2 <--  subtracts rt from both sides

 

98t = r/2 - 49/2

 

t = r/196 - 49/196  = (r-49)/196

 

Substituting into the first equation:

800 = (r+49)*(r-49)/196

 

800 * 196 = r^2 - 2401

 

156800 = r^2 - 2401

 

0 = r^2 - 159201

 

0 = ( r    +  399   )( r    -     399  )

 

R + 399 = 0 results in negative speeds

 

r - 399 = 0 ---> r = 399

 

The speed of the plane is 399.

 

With the wind it travels 399 + 49 = 448 mph

800/448= 1 and 352/448 = 1 and 11/14  (reduces by gcf of 32 found using Euclid's method)

   or just over 1 hour and 47 minutes and 8.5 seconds

 

 

Against the wind, the plane travels 399 - 49 = 350 mph.

800/350 = 2 hours 17 minutes and 8.5 seconds

 

This is indeed 30 minutes or 1/2 longer, which shows that the

speed of the plane is 399 mph