A 650 mile, 4 hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 150 mph. The remainder of the trip was flown at an average speed of 200 mph. For how long did the plane fly at each speed?

1 answer

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:

  1. The total flight time is 4 hours: \[ t_1 + t_2 = 4 \]

  2. 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.