An airplane flies 500 mi/h in still air. Flying with the jet stream, the plane travels 1200 mi from City A to City B. The plane then returns to City A, flying against the jet stream. The round-trip flight time is 5 h. What is the speed of the jet stream in miles per hour?
A: 25 mi/h
B: 100 mi/h
C: 150 mi/h
D: 223 mi/h
THIS IS A STUPID QUESTION FOR ALGEBTA
To find the speed of the jet stream, we can set up an equation using the information given. Let's call the speed of the jet stream "S", and let's call the time it takes the plane to fly from City A to City B "T".
When flying with the jet stream from City A to City B, the plane's effective speed will be the sum of the airplane's speed in still air (500 mi/h) and the speed of the jet stream (S). So the equation for the distance traveled from City A to City B is:
(500 + S) * T = 1200
Similarly, when flying against the jet stream from City B to City A, the plane's effective speed will be the difference between the airplane's speed in still air (500 mi/h) and the speed of the jet stream (S). So the equation for the distance traveled from City B to City A is:
(500 - S) * (5 - T) = 1200
Notice that in the second equation, we use (5 - T) instead of T because the round-trip flight time is 5 hours.
Now we have a system of two equations:
1) (500 + S) * T = 1200
2) (500 - S) * (5 - T) = 1200
We can solve this system of equations to find the value of S, the speed of the jet stream.
Solving equation 2) for S:
500 - S = (1200)/(5 - T)
S = 500 - (1200)/(5 - T)
Substituting this value of S into equation 1):
(500 + 500 - (1200)/(5 - T)) * T = 1200
(1000 - (1200)/(5 - T)) * T = 1200
1000T - (1200T)/(5 - T) = 1200
(1000T(5 - T) - 1200T)/(5 - T) = 1200
(5000T - 1000T^2 - 1200T)/(5 - T) = 1200
2000T + 1000T^2 = 1200(5 - T)
Expanding and rearranging:
1000T^2 + 2000T + 1200T - 6000 = 1200 - 1200T
1000T^2 + 3200T - 6000 = 0
Now we have a quadratic equation. We can solve it by factoring or by using the quadratic formula. Factoring does not give us integer values, so we'll use the quadratic formula:
T = (-b ± sqrt(b^2 - 4ac))/(2a)
For our equation:
a = 1000, b = 3200, c = -6000
Plugging these values into the quadratic formula:
T = (-3200 ± sqrt(3200^2 - 4*1000*(-6000)))/(2*1000)
Simplifying:
T = (-3200 ± sqrt(10240000))/(2000)
T = (-3200 ± 3200)/(2000)
Now we have two possible values for T: T = 0 or T = 3.2
Since T cannot be 0 (this would mean the plane did not fly at all), we discard that solution.
Therefore, T = 3.2 (hours).
Now we can substitute this value of T back into our equation for S:
S = 500 - (1200)/(5 - 3.2)
S = 500 - 2400/(1.8)
S = 500 - 1333.33
S = 166.67
So, the speed of the jet stream is approximately 166.67 mi/h.
The closest option listed is D: 223 mi/h, but the calculated value is not an exact match.