The woman will arrive in New York City at 11 AM.
Plane takeoff angle.
Angel
During take off, a plane leaves the ground and travels in a straight line until it reaches a height of 10 km. The distance the plane flies during take off should be in the range 57 km to 62 km. What is the smallest possible angle that the path of the plane could make with the ground? Give your answer in degrees to 1 d. p.
Let's assume that the plane travels a distance of x km during take off and reaches a height of 10 km. Then, using trigonometry, we can find the angle θ between the ground and the path of the plane:
tan(θ) = 10/x
We want to find the smallest possible angle θ, which means we need to maximize x. From the given information, we know that x must be in the range 57 km to 62 km. Therefore, to maximize x, we choose x = 62 km.
Plugging this into the equation above, we get:
tan(θ) = 10/62
θ = arctan(10/62) ≈ 8.8°
Therefore, the smallest possible angle that the path of the plane could make with the ground is approximately 8.8 degrees.
Angel
Xochitl spots an airplane on radar that is currently approaching in a straight line, and that will fly directly overhead. The plane maintains a constant altitude of 7425 feet. Xochitl initially measures an angle of elevation of 19° to the plane at point A. At some later time, she measures an angle of elevation of 37° to the plane at point B.
Find the distance the plane traveled from point A to point B. Round your answer to the nearest foot if necessary
Let's assume that the distance from Xochitl to point A is d1 and the distance from Xochitl to point B is d2. We want to find the distance the plane traveled from point A to point B, which is the difference between d1 and d2.
From the given information, we can set up the following two equations:
tan(19°) = 7425/d1
tan(37°) = 7425/d2
Solving for d1 and d2, we get:
d1 = 7425/tan(19°) ≈ 22977.6 feet
d2 = 7425/tan(37°) ≈ 13060.2 feet
Therefore, the distance the plane traveled from point A to point B is:
d1 - d2 ≈ 9917.4 feet ≈ 3021 meters
Rounding to the nearest foot, we get that the distance the plane traveled from point A to point B is approximately 9917 feet.
Angel
Plane B is flying 75mph 75 mph faster than Plane A. Find the time it takes for Plane A to travel 2300 2300 miles if it takes Plane B the same amount of time to travel 2600 2600 miles.
Let's start by using the formula for distance, rate, and time:
distance = rate x time
Let's assume that Plane A's speed is r mph. Then we know that Plane B's speed is r + 75 mph.
We also know that Plane A travels 2300 miles and Plane B travels 2600 miles. Since they take the same amount of time to travel their respective distances, we can set up the following equation:
2300/r = 2600/(r + 75)
To solve for r, we can cross-multiply and simplify:
2300(r + 75) = 2600r
2300r + 172500 = 2600r
300r = 172500
r = 575 mph
Now that we know Plane A's speed, we can use the formula for distance, rate, and time to find the time it takes for Plane A to travel 2300 miles:
distance = rate x time
2300 = 575 x time
time = 4 hours
Therefore, it takes Plane A 4 hours to travel 2300 miles.
Angel
A woman leaves at 7 AM and drive to New York City 200 miles away. If she averages 50 mph what time will she arrive at the city?
To determine the arrival time of the woman in New York City, we need to use the formula for time:
time = distance / rate
In this case, the distance is 200 miles and the rate (or speed) is 50 mph. Plugging these values into the formula, we get:
time = 200 miles / 50 mph = 4 hours
Since the woman left at 7 AM, we can simply add the travel time of 4 hours to the departure time to get the arrival time:
7 AM + 4 hours = 11 AM
Therefore, the woman will arrive in New York City at 11 AM.