Let \( t_1 \) be the time (in hours) flown at 150 mph, and \( t_2 \) be the time (in hours) flown at 200 mph. According to the problem, we have the following two equations:
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The total flight time is 4 hours: \[ t_1 + t_2 = 4 \]
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The total distance flown is 650 miles. The distance for each part of the trip can be calculated as speed multiplied by time: \[ 150 t_1 + 200 t_2 = 650 \]
Now we can substitute \( t_2 \) from the first equation into the second equation. From \( t_1 + t_2 = 4 \), we can express \( t_2 \) as: \[ t_2 = 4 - t_1 \]
Substituting \( t_2 \) into the distance equation gives: \[ 150 t_1 + 200(4 - t_1) = 650 \]
Expanding this: \[ 150 t_1 + 800 - 200 t_1 = 650 \]
Combining like terms: \[ -50 t_1 + 800 = 650 \]
Subtracting 800 from both sides: \[ -50 t_1 = -150 \]
Dividing by -50: \[ t_1 = 3 \]
Now, substituting \( t_1 \) back into the equation for \( t_2 \): \[ t_2 = 4 - t_1 = 4 - 3 = 1 \]
Thus, the times flown at each speed are: \[ \text{Time at 150 mph: } t_1 = 3 \text{ hours} \] \[ \text{Time at 200 mph: } t_2 = 1 \text{ hour} \]
To check, we calculate the distances:
- Distance at 150 mph: \[ 150 \times 3 = 450 \text{ miles} \]
- Distance at 200 mph: \[ 200 \times 1 = 200 \text{ miles} \]
Total distance: \[ 450 + 200 = 650 \text{ miles} \]
This confirms the calculations are correct. Therefore, the plane flew for:
- 3 hours at 150 mph
- 1 hour at 200 mph.
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