Quadratic Equation word problem

Question

It took a small plane 2 hr longer to fly 375 mi against the wind than it took the plane to fly the same distance with the wind. The rate of the wind was 25 mph. Find the rate of the plane in calm air.

Answer 1

Let V = Velocity of plane in still air.

Let v = Velocity of the air = 25 mph

The time of flight can be found from Rate * Time = Distance formula:

Time = Distance / Rate.

Let D = the common distance traveled = 375 mi.

Going against the wind, the net speed of the plane is V - v. The time required to travel a distance D is

D / (V - v)

Going with the wind, the net speed of the plane is V + v. The time required to travel a distance D is

D / (V + v)

The first time is 2 hours longer than the first:

2 = D / (V-v) - D / (V+v)

We seek V.

Factor out the common factor D:

2 = D [ 1 / (V-v) - 1 / (V+v) ]

Multiply both sides by (V-v)(V+v):

2 (V-v) (V+v) = D [ (V+v) - (V-v) ] = 2 v D

Divide both sides by 2 and note that (V-v) (V+v) = V² - v²

V² - v² = Dv

Solving for V:

V = sqrt( v² + Dv )

We were given that D = 375 mph and v = 25 mph:

V = sqrt( 25² + 375*25)

= sqrt ( 25² + 375*25 )

= sqrt ( 25² + 15*25*25 )

= sqtr[ 25²( 1 + 15) ]

= sqrt [ 25²(16) ]

= 25 * 4 Note: Negative are not allowed because V > 0.

= 100 mph

Check:

2 ?=? 375 / (100 - 25) - 375 / (100 + 25)

= 375 / 75 - 375 / 125

= 15 / 3 - 15 / 5

= 5 - 3

= 2