A plane flies 720 mi against a steady 30 mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air?
To find the plane's speed in still air, we can use the concept of relative velocity.
Let's assume the plane's speed in still air is 'x' mph. Since there is a headwind on the outbound journey, the effective speed of the plane (relative to the ground) will be (x - 30) mph. Similarly, on the return journey (with the wind), the effective speed of the plane (relative to the ground) will be (x + 30) mph.
Now, we know that the plane covers a distance of 720 miles on the outbound journey, so the time taken for the outbound leg of the trip can be calculated as:
Time = Distance / Speed = 720 / (x - 30)
On the return journey, the plane covers the same distance and takes the same time. Therefore, the time taken for the return leg of the trip can be calculated as:
Time = Distance / Speed = 720 / (x + 30)
We are also given that the total trip time is 10 hours. Therefore, we can write the equation:
720 / (x - 30) + 720 / (x + 30) = 10
To solve this equation and find the value of 'x', we can multiply through by (x - 30)(x + 30) to eliminate the denominators:
720(x + 30) + 720(x - 30) = 10(x - 30)(x + 30)
Expanding and rearranging the equation, we get:
720x + 21600 + 720x - 21600 = 10(x^2 - 900)
Simplifying further:
1440x = 10x^2 - 9000
Rearranging the terms to form a quadratic equation:
10x^2 - 1440x - 9000 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Here, a = 10, b = -1440, and c = -9000. Plugging in the values, we get:
x = (1440 ± √((-1440)^2 - 4 * 10 * -9000)) / (2 * 10)
Simplifying further, we find two possible solutions for 'x':
x1 ≈ 144
x2 ≈ -63
Since speed can only be positive, the plane's speed in still air is approximately 144 mph.