Northern Component of velocity

[anglum]

Components of velocity

Homework Statement

An airplane travels at 146 km=h toward the
northeast.
What is the northern component of its ve-
locity? Answer in units of km=h.

Homework Equations

asquared + bsquared = csquared

The Attempt at a Solution

however i am not sure if u use the pythagorean formula to solve this?

 

[drpizza]

Yes, you're on the right track. (Assuming by "North East" they mean that 45 degree angle between north and east. That means NorthEast would be the hypotenuse/resultant of a two vectors: 1 north and 1 east of equal magnitudes.)

 

[berkeman]

First, the plane speed is 146 km/h (not km=h).

Next, you find the components in the north and east directions by multiplying the hypoteneuse by the sine or cosine of the appropriate angles...

 

[anglum]

ooo i have to do sin/cosine of the angle

 

[anglum]

so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h

 

[berkeman]

so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h

Yes. That's an unusual way to write it, however. More like this (I'll use latex):

$$v_y = 146 km/h * sin(45)$$

 

[berkeman]

so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h

Well, except for the math you did. The sin and cos of 45 degrees should be the same...

 

[anglum]

ok thanks guys i got it i was doing some bad math... thank you