an airplane flying due east at 400 km/h pasess over an airport 12 minutes before a second plane flying south 30 degrees west at 500 km/h.if the airplanes are at the same altitude how fast will they by seprating when the seccong plane is over the airport.?When will they be the nearest.
i tried the question but because i just started calculus its a hard questions necause it talk about degree left and right so can somebody please help me ?The answer is suppose to be 650 km/h,5.11 min before second plane reaches the airport.Thank you
To solve this problem, we will break it down into different components and use trigonometry. Let's start with calculating the distance between the two planes when the second plane is over the airport.
First, let's convert the time into hours. Since there are 60 minutes in an hour, 12 minutes is equal to 12/60 = 1/5 hours.
Now, we can find the distance traveled by the first plane in that time. The distance traveled by an object can be calculated using the formula: distance = speed × time.
The first plane is flying due east at a speed of 400 km/h. Therefore, the distance traveled by the first plane in 1/5 hours is 400 × 1/5 = 80 km.
Next, let's find the distance of the second plane from the airport at that time. We will use trigonometry to calculate this distance.
The second plane is flying south 30 degrees west. We know that the angle between the second plane's direction and due south is 30 degrees.
Using trigonometry, we can find the distance traveled by the second plane (opposite side) based on the given angle and speed.
sin(30) = opposite / hypotenuse
sin(30) = opposite / 500
opposite = 500 × sin(30) = 250 km
Now, we need to find the distance between the first and second planes when the second plane is over the airport. We can use the Pythagorean theorem to calculate this distance.
Distance^2 = (80 km)^2 + (250 km)^2
Distance^2 = 6400 km^2 + 62500 km^2
Distance^2 = 68900 km^2
Distance ≈ 262.45 km
At this point, we have found the distance between the two planes when the second plane is over the airport.
Next, let's find the rate at which the distance is changing. This can be calculated by taking the derivative of the distance equation with respect to time.
Differentiating the equation Distance^2 = (80 km)^2 + (250 km)^2 and using the chain rule, we get:
2 * Distance * (dDistance / dt) = 0 + 0
2 * Distance * (dDistance / dt) = 0
dDistance/dt = 0 / (2 * Distance)
dDistance/dt = 0
The rate at which the distance is changing is 0 km/h. This means that the planes are not moving closer or farther apart at this moment.
Finally, let's find the time when the two planes are nearest to each other. To find this, we need to find when the rate at which the distance is changing (dDistance/dt) becomes zero.
dDistance/dt = 0
Using the previous equation, we know that dDistance/dt = 0. The distance between the planes is constant, and there is no change in their separation. Therefore, the planes are always the nearest to each other.
In conclusion, the rate at which the planes are separating when the second plane is over the airport is 0 km/h. The planes are never nearest to each other because they maintain a constant distance of approximately 262.45 km throughout their flight.
To solve this problem, we can use vector addition and trigonometry. Let's break it down step by step:
1. First, we need to determine the velocities of each plane in terms of their horizontal (east) and vertical (south) components.
- The first plane is flying due east, so its velocity can be written as: v1 = (400 km/h, 0 km/h). This means its velocity in the east direction is 400 km/h and its velocity in the south direction is 0 km/h.
- The second plane is flying south 30 degrees west. We can split its velocity into horizontal and vertical components using trigonometry. Since the angle is given with respect to the west direction, we need to subtract the angle from 180 degrees to get the angle with respect to the east direction.
Let's call the angle with respect to the east direction θ. θ = 180 degrees - 30 degrees = 150 degrees.
Now, we can find the horizontal and vertical components of the velocity using trigonometric functions:
Horizontal component: v2_horizontal = 500 km/h * cos(θ)
Vertical component: v2_vertical = 500 km/h * sin(θ)
So, v2 = (v2_horizontal , v2_vertical) = (500 km/h * cos(150 degrees), 500 km/h * sin(150 degrees))
2. Next, we need to determine the relative velocity of the second plane with respect to the first plane when the second plane is over the airport.
To find the relative velocity, we simply subtract the velocity of the first plane from the velocity of the second plane:
Relative velocity = v2 - v1
3. Now that we have the relative velocity, we can find the speed at which the planes are separating.
The speed of separation (or relative speed) is the magnitude of the relative velocity. We can calculate it using the formula:
Speed of separation = sqrt((v2_horizontal - v1)^2 + (v2_vertical - 0)^2)
4. To find the time when the planes are nearest, we can use distance-speed-time relationship.
Since we know the speed of separation, we need to determine the time it takes for the second plane to reach the airport after the first plane passed overhead.
We can calculate this time by dividing the horizontal distance between the planes (traveled by the second plane) by the horizontal component of the relative velocity.
Assuming the horizontal distance between the planes is constant during this time, we can find the time using the formula:
Time = distance / v2_horizontal
5. Plug the values into the equations and calculate:
Let's calculate the values step by step:
v2_horizontal = 500 km/h * cos(150 degrees) = -433.01 km/h (negative because it's westward)
v2_vertical = 500 km/h * sin(150 degrees) = -250 km/h (negative because it's going south)
Relative velocity = v2 - v1 = (-433.01 km/h - 400 km/h , -250 km/h - 0 km/h) = (-833.01 km/h, -250 km/h)
Speed of separation = sqrt((-833.01 km/h)^2 + (-250 km/h)^2) ≈ 869.39 km/h
Now, let's calculate the time:
Time = distance / v2_horizontal = 0 km / (-433.01 km/h) ≈ 0 hours (since distance is 0 at the airport)
Lastly, let's convert the time to minutes:
0 hours * 60 = 0 minutes
So, the planes are separating at a speed of approximately 869.39 km/h, and they will be nearest to each other 0 minutes before the second plane reaches the airport (since they passed over the airport 12 minutes apart). This means they are closest to each other when the second plane is directly above the airport.