# Thread: Expressing velocity as a vector.

1. ## Expressing velocity as a vector.

I am confused at how to express the velocity as a vector.

An airplane is flying on a bearing of 23 degrees east of north at 650 mph. Express the velocity of the airplane as a vector.

If its flying 23 degrees east of north, does that mean its flying 650 mph at 67 degrees?

2. Originally Posted by Såxon
I am confused at how to express the velocity as a vector.

An airplane is flying on a bearing of 23 degrees east of north at 650 mph. Express the velocity of the airplane as a vector.

If its flying 23 degrees east of north, does that mean its flying 650 mph at 67 degrees?
The vector would be expressed as
$\displaystyle <|v|\cos \theta, |v|\sin \theta>$

The first component would represent the speed in the easterly direction, and the second component would represent the speed in the northerly direction.

You're right that 23°E of N corresponds to 67° in the Cartesian plane. So $\displaystyle \theta = 67^{\circ}$. |v| = 650 is the magnitude of this vector. Therefore your vector would be
$\displaystyle <650\cos 67^{\circ}, 650\sin 67^{\circ}>$
or
$\displaystyle <253.98, 598.32>$.

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3. Originally Posted by Såxon
I am confused at how to express the velocity as a vector.

An airplane is flying on a bearing of 23 degrees east of north at 650 mph. Express the velocity of the airplane as a vector.

If its flying 23 degrees east of north, does that mean its flying 650 mph at 67 degrees?
Ok, I'll try to explain this without drawing a pic as I'm far too lazy to draw one. Draw a line to represent north from some origin, then from the same origin draw a line at 23 degrees from this line (on the right hand side of the north line). Now join the top of the two lines with a horizontal line. What you should now have is a right angle triangle for which you should be able to use trig to work out what the lengths of the north and east facing lines should be given that the hyp is 650. Once you have worked out these other lengths they form your vector with the horizontal velocity first then the vertical velocity. Hope this helps. Try drawing the pic and post it on the forum if you want someone to check it.

4. Originally Posted by Såxon
I am confused at how to express the velocity as a vector.

An airplane is flying on a bearing of 23 degrees east of north at 650 mph. Express the velocity of the airplane as a vector.

If its flying 23 degrees east of north, does that mean its flying 650 mph at 67 degrees?
The "standard" way of expressing direction mathematically (as opposed to on a map) is to measure the angle counter-clockwise, from the positive x-axis (which, by convention points 'east' on a map). So, yes, "23 degrees east of north" is 90- 23= 67 degrees "north of east" which is counter-clockwise from the positive x-axis.

A little bit more complicated: 55 degrees "West of South" would be 270- 55= 215 degrees because "south" corresponds to 270 degrees (counting by right angles counter-clockwise from east, north is 90 degrees, west is 180 degrees, and south is 270 degrees).

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