An airplane flying into a headwind travels 1160 miles in 4 hours and 50 minutes. On the return flight, the distance is traveled in 4 hours. Find the airspeed of the plane and the speed of the wind, assuming that both remain constant.

Question

anonymous · Answer

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nurali · Answer

Let S = the airplane's speed

Let W = the wind speed

4 hours and 50 minutes = 4.8 hours

Into the wind,

(S)(4.8 hours) - (W)(4.8 hours) = 1160 miles

With the wind,

(S)(4.0) + (W)(4.0) = 1160 miles Using this equation first, divide both sides by 4.0, to yield:

S + W = 1160/4.0 = 290

Therefore, S = 290 - W

Substituting this value for S into the first equation yields:

(290 - W)(4.8) - (W)(4.8) = 1160

Again, divide both sides by 4.8 to yield:

290 - W - W = 242

Rearranging gives us:

290 - 242 = 2W

Therefore; W = 24

Because S = 290 - W, S = 290 - 24 = 266

whpalmer4 · Answer

$$d = rt$$where $d$ is distance, $r$ is rate or speed, and $t$ is time.

Let the airplane's speed be $v_a$ and the wind $v_w$.  When the airplane flies with the wind (tailwind) the speed over the ground will be $v_a+v_w$.  When the airplane flies against the wind (headwind) the speed over the ground will be $v_a-v_w$.

We know that the distance in both cases is the same, 1160 miles.  

Flying into the headwind:
1160 miles = 4 hours 50 minutes * $(v_a-v_w)$

Flying with the wind:
1160 miles = 4 hours * $(v_a+v_w)$

Convert the times to minutes (or decimal hours), expand the two equations, and solve.