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 \(\displaystyle d = |(23, 29) - (3, 4)| = \sqrt{(23 - 3)^2 + (29 - 4)^2} \approx 32.0\). In a similar fashion we can get the magnitude of the velocity (another vector!) to find \(\displaystyle v = \sqrt{400^2 + 500^2} \approx 640.3 \text{ km/hr}\).

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

-Dan