Calculating the Time an Airplane Will Reach an Airport

[evinda]

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.

 

[topsquark]

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

 

[evinda]

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 $x=(23,29,0)$ ?

 

[topsquark]

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