MHB Calculating the Time an Airplane Will Reach an Airport

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To determine when the airplane will reach the airport, the initial position is given as (3,4,5) and the airport's position is (23,29,0). The airplane travels at a velocity of 400i + 500j - k km/h. The displacement in the horizontal plane is calculated as the distance between the two points, which is approximately 32 km. The velocity's magnitude is about 640.3 km/h, allowing for the calculation of time to reach the airport, while the vertical component is ignored since the plane is descending to ground level. The discussion emphasizes understanding displacement and the use of vector calculations in this context.
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Hello! (Wave)

An airplane is at the position $(3,4,5)$ at noon and travels with velocity $400 i+500 j-k$ kilometers per hour.
The pilot detects an airport at the position $(23, 29,0)$.
How can we find the time at which the airplane will pass exactly over the airport?
We suppose that the Earth is flat and the vector $k$ shows upwards.
 
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evinda said:
Hello! (Wave)

An airplane is at the position $(3,4,5)$ at noon and travels with velocity $400 i+500 j-k$ kilometers per hour.
The pilot detects an airport at the position $(23, 29,0)$.
How can we find the time at which the airplane will pass exactly over the airport?
We suppose that the Earth is flat and the vector $k$ shows upwards.
x = v t. where x is the displacement, v is the velocity, and t is the time it takes to travel x.

Since the plane is flying with a constant velocity and is always flying in the same direction we can take the magnitude of the above equation to get x = vt, where x = |(23, 29) - (3, 4)| and v = |(400, 500)|.

A question for you: Why can we ignore the k coordinate? And check to make sure the plane doesn't plow itself into the ground before it gets to the airport!

-Dan
 
topsquark said:
x = v t. where x is the displacement, v is the velocity, and t is the time it takes to travel x.

Since the plane is flying with a constant velocity and is always flying in the same direction we can take the magnitude of the above equation to get x = vt, where x = |(23, 29) - (3, 4)| and v = |(400, 500)|.

A question for you: Why can we ignore the k coordinate? And check to make sure the plane doesn't plow itself into the ground before it gets to the airport!

-Dan

Could you explain to me what you mean by displacement?
Why do we pick $ x = |(23, 29) - (3, 4)|$ and don't use the formula $x=vt$ twice , once for $x=(3,4,5)$ and once for $x=(23,29,0)$ ?
 
evinda said:
Could you explain to me what you mean by displacement?
Why do we pick $ x = |(23, 29) - (3, 4)|$ and don't use the formula $x=vt$ twice , once for $x=(3,4,5)$ and once for $x=(23,29,0)$ ?
The displacement is a vector in the direction from initial point (typically a point fixed in space) to a final point. Loosely speaking it is a "directed distance." A common displacement is the displacement between the position of an object measured from a fixed origin...This is called the "position vector" and is used quite often.

Sorry, I'm using a vector format to write the equation. For example, if we want to find the distance from the point (3, 4) to (23, 29) then we can use the distance formula [math]d = |(23, 29) - (3, 4)| = \sqrt{(23 - 3)^2 + (29 - 4)^2} \approx 32.0[/math]. In a similar fashion we can get the magnitude of the velocity (another vector!) to find [math]v = \sqrt{400^2 + 500^2} \approx 640.3 \text{ km/hr}[/math].

If that's not clear just let me know and I'll give you more info.

-Dan
 
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