word problem algebra 2?
Matt and Ann Killian are frequent fliers on Fast-n-go airlines. They often fly from cities that are 800 miles apart. On one particular trip, they flew into the wind and the trip took 2.5 hours. The return trip with the wind behind them, only took 2 hours. Find the speed of the wind and the speed of the plane in still air.
help me please?
Comments
An unspoken assumption in this problem, is that the wind has the same magnitude and direction at both flight times.
Velocity equals distance over time
V = d/t
The first trip is actually clocking the time it took when the plane was flying against the wind: the calculated velocity will be the plane's velocity if it was flying in still wind, minus the wind's velocity
Vp - Vw = d/t
Vp - Vw = 800mi/2.5h
Vp - Vw = 320mi/h
The reutrn trip is clocking the time it took when the plane is flying with the wind: the calculated velocity will be the plane`s velocity if it was flying in still wind, plus the wind`s velocity
Vp + Vw = d/t
Vp + Vw = 800mi/2h
Vp + Vw = 400mi/h
We now have two similar equations. Look what happens if we subtract the second one from the first one:
Vp - Vw - (Vp + Vw) = 320mi/h - 400mi/h
This simplifies to
-2Vw = -80mi/h
Divide each side by -2 to solve for the wind velocity:
Vw = 40mi/h
Now that we know the wind`s velocity, we can use either of the two equations to find the speed of the plane in still air:
Vp + Vw = 400mi/h
Vp + (40mi/h) = 400mi/h
Vp = 360mi/h
x - y = 800/2.5 = 320
x + y = 800/2 = 400
2x = 720
x = 360
The plane travels at 360mph
360 - y = 320
y = 40
The wind is 40mph.
2.5(s - w) = 800
2(s + w) = 800 or s = 400 - w
substitute second equation into first equation:
2.5(400 - w) - 2.5w = 800
1000 - 2.5w - 2.5w = 800
-5w = -200
w = -200/-5 = 40mi/hr
using second equation:
s = 400 - 40 = 360mi/hr
check using first equation:
2.5(360 - 40) = 800
800 = 800
wind speed = 40mi/hr
plane speed in still air = 360mi/hr
- .--
a - w = 320
a + w = 400
2a = 720
a = 360 mph
w = 40 mph