Let's denote the speed of the eastbound plane as \( x \) mph. Therefore, the speed of the westbound plane, which is flying 140 mph faster, will be \( x + 140 \) mph.
Both planes are flying for 2 hours, so we can set up the following equations based on the distance traveled:
The distance traveled by the eastbound plane in 2 hours is: \[ \text{Distance} = \text{Speed} \times \text{Time} = x \times 2 = 2x \text{ miles} \]
The distance traveled by the westbound plane in 2 hours is: \[ \text{Distance} = \text{Speed} \times \text{Time} = (x + 140) \times 2 = 2(x + 140) = 2x + 280 \text{ miles} \]
Since the total distance between the two planes after 2 hours is 1520 miles, we can set up the equation: \[ 2x + (2x + 280) = 1520 \]
Now, simplify the equation: \[ 2x + 2x + 280 = 1520 \] \[ 4x + 280 = 1520 \]
Next, solve for \( x \): \[ 4x = 1520 - 280 \] \[ 4x = 1240 \] \[ x = \frac{1240}{4} = 310 \]
So the speed of the eastbound plane is \( 310 \) mph.
Now, to find the speed of the westbound plane: \[ \text{Speed of westbound plane} = x + 140 = 310 + 140 = 450 \text{ mph} \]
In summary:
- Speed of the eastbound plane: 310 mph
- Speed of the westbound plane: 450 mph
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