# Velocity Vector Problem

• Jan 19th 2009, 10:39 AM
Frostking
Velocity Vector Problem
There is a plane that just took off and is climbing northwest through still air at an airspeed of 200 km/hr, and rising at a ate of 300 m/min. Resolve its velocity vector into components.

First, I converted 200 m/min to 18 km/hr so all values have the same units. Then I said the velocity of the plane would =

(-) 200 cos135degrees i + 200 sin135degrees j + 18 k

= - 141.42 i + 141.42 j + 18 k

However, the correct answer is - 140.8 i + 140.8 j + 18 k

Can someone please explain where I made my mistake? Thanks very much!
• Jan 19th 2009, 10:56 AM
skeeter
Quote:

There is a plane that just took off and is climbing northwest through still air at an airspeed of 200 km/hr, and rising at a rate of 300 m/min. Resolve its velocity vector into components.
300 m/min = 18 km/hr

angle of climb relative to the x-y plane ... $\theta = \arcsin\left(\frac{18}{200}\right) \approx 5.16^{\circ}$

speed along the x-y plane is $200\cos{\theta}$

x component = $-(200\cos{\theta})\cos(45^{\circ}) = -140.8 \, km/hr$

y component = $(200\cos{\theta})\sin(45^{\circ}) = 140.8 \, km/hr$
• Jan 19th 2009, 12:06 PM
Soroban
Hello, Frostking!

Quote:

There is a plane that just took off and is climbing northwest through still air
at an airspeed of 200 km/hr, and rising at a rate of 300 m/min.
Resolve its velocity vector into components.

The answer is: . $- 140.8 i + 140.8 j + 18 k$

Look at the side view of the plane climbing.
Code:

B *
|  *
|      *  200
18 |          *
|              *
|                  *
C * - - - - - - - - - - - * A
x

The hypotenuse $AB = 200,\;BC = 18.$

Hence: . $x\:=\:\sqrt{200^2-18^2} \:=\:\sqrt{39,\!676} \quad\Rightarrow\quad x \:\approx\: 199.19$

Now look down at the ground.
Code:

|
C * - - - - + E
: *      |
:  * x  |
:    *45°|
:  45° * |
- - + - - - - * - - -
D        |A
|

We see that: . $AD \:=\:AE \:=\:\frac{x}{\sqrt{2}} \:\approx\:140.8$