airplanes travels 500miles against wind in 4 hrs. makes return trip with same wind in 1 hrs. find the rate of the wind

Question

need help again please 

Answer 1

Michelle,

 

Let the speed of the plane = V_p and let the speed of the wind = V_w.  Distance = velocity x time.

 

For the trip against the wind 500 mi = 4 hrs·V mi/hr where V = V_p - V_w, so,

 

Outbound trip:  500 = 4 (V_p - V_w) ⇒ 4·Vp = 500 + 4·V_w

 

For the return trip V = V_p + V_w so, 500 mi = 1 hr·V gives:

 

Return trip:  500 = 1·(V_p + V_w) ⇒ V_p = 500 - V_w

 

Replacing this expression for the planes velocity in the equation for the outbound trip gives:

 

4·(500 - V_w) = 500 + 4·V_w

 

Expanding the left side:  2000 - 4·V_w = 500 + 4·V_w

 

Collecting terms:  8·V_w = 2000 - 500 = 1500

 

Dividing by 8 gives:  V_w = 1500/8 = 187.5 mi/hr