Determine the x and y components of the following three vectors in the xy plane.
1) A 19-m/s velocity vector that makes an angle of 40° counterclockwise from the -x direction.
x:
2) y:
Explain please!
well, the x component is 19costheta, and the y component is 19sintheta.
draw the axis, measure the angle counterclockwise from x.
This is true for any angle, say 104. If the angle had been 104 deg
x=magnitude*cos104
y=magnitude*sin104
Caution: On geographical coordinates, a common convention is to measure angles from N, clockwise, even if the angle is given as xxx deg W of S. Convert that to degrees clockwise from N
N=magnitude*cosTheta
E=magnitude*sinTheta
and it will work regardless of the angle.
1. V = 19m/s[40o]S. of W. = 19[220o]CCW from +x-axis.
X = 19*Cos220 =
Y = 18*sin220 =
Correcti0n: Y = 19*sin220 =
To determine the x and y components of a vector, we can use trigonometry.
For the given vector with magnitude 19 m/s and an angle of 40° counterclockwise from the -x direction, we can break it down into its x and y components.
The x-component, denoted as x, is calculated using the cosine function. The formula is:
x = magnitude * cos(angle)
In this case,
x = 19 m/s * cos(40°)
x ≈ 14.51 m/s
Therefore, the x-component of the vector is approximately 14.51 m/s.
The y-component, denoted as y, is calculated using the sine function. The formula is:
y = magnitude * sin(angle)
In this case,
y = 19 m/s * sin(40°)
y ≈ 12.17 m/s
Therefore, the y-component of the vector is approximately 12.17 m/s.
So, the x and y components of the given vector are approximately:
x: 14.51 m/s
y: 12.17 m/s