# 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.