An airplane at a constant altitude of 2 miles flies a horizontal distance toward you at a constant velocity. At the start of your observation, the angle of elevation is 40°. Fifteen seconds later, the angle of elevation is 50°. What is the approximate velocity of the airplane in miles per minute?
(1 point)
Responses
0.1 miles per minute
0.1 miles per minute
0.7 miles per minute
0.7 miles per minute
2.8 miles per minute
2.8 miles per minute
10.6 miles per minute
To solve this problem, we can use the concept of tangent function. The tangent of an angle is equal to the ratio of the opposite side to the adjacent side of a right triangle. In this case, the opposite side is the change in altitude of the airplane (0 miles as it is flying at a constant altitude), and the adjacent side is the horizontal distance it travels.
Let's denote the horizontal distance at the start of observation as d. From the triangle formed, we have:
tan(40°) = 0 / d
Simplifying this equation, we get:
tan(40°) = 0
Since the tangent of 40° is not zero, the horizontal distance at the start of observation, d, must be nonzero. This means that the plane is already at some distance when we start observing it.
Now, 15 seconds later, the angle of elevation is 50°. Let's denote the horizontal distance at this time as d'. From the new triangle formed, we have:
tan(50°) = 0 / d'
Simplifying this equation, we get:
tan(50°) = 0
Again, the tangent of 50° is not zero, which means that the horizontal distance at this time, d', must also be nonzero.
We can now calculate the change in distance traveled by the airplane in 15 seconds:
Change in distance = d' - d
To find the velocity, we need to convert this change in distance from miles per second to miles per minute. There are 60 seconds in a minute, so:
Change in distance (in miles per minute) = (d' - d) / 15 * 60
Since the velocity of the airplane is constant, we can assume that it traveled the same horizontal distance in the entire 15 seconds. Therefore, d' - d is the horizontal distance traveled in 15 seconds.
Substituting the given angles into the equation, we have:
tan(40°) = 0
tan(50°) = (d' - d) / 15 * 60
From tan(40°) = 0, we can see that the initial distance, d, is irrelevant to the problem.
Now, solving for (d' - d) using tan(50°) = (d' - d) / 15 * 60, we get:
(d' - d) = tan(50°) * 15 * 60
Approximating the tangent of 50° to 1.1918, we have:
(d' - d) = 1.1918 * 15 * 60
Simplifying, we get:
(d' - d) = 1072.62
Therefore, the change in distance traveled by the airplane in 15 seconds is approximately 1072.62 miles.
Finally, we can calculate the velocity of the airplane in miles per minute:
Velocity = change in distance / 15
Velocity = 1072.62 / 15 ≈ 71.51
So, the approximate velocity of the airplane is approximately 71.51 miles per minute.
None of the provided answer choices match this result.