Vector V1 is 9.00 units long and points along the negative x axis. Vector V2 is 4.72 units long and points at +50.0° to the +x axis.
(a) What are the x and y components of each vector?
V1x =
V1y =
V2x =
V2y =
(b) Determine the sum V1 + V2. Magnitude
AND
Direction (counterclockwise from the +x axis is positive)
Go back and answer my reply to the previous question.
(a) What are the x and y components of each vector?
V1x = -9.00
V1y = 0
V2x = 4.72 cos 50
V2y = 4.72 sin 50 (if there is no z component)
(b) Add the components
To find the x and y components of each vector, we can use trigonometry.
For vector V1, since it points along the negative x-axis, its x-component is negative and its y-component is zero. Therefore:
V1x = -9.00
V1y = 0
For vector V2, we need to find the x and y components based on its length and direction.
Given that the vector makes an angle of 50.0° with the positive x-axis, we can use sine and cosine functions to find the x and y components.
V2x = 4.72 * cos(50.0°)
V2y = 4.72 * sin(50.0°)
Now, we can calculate these values:
V2x = 4.72 * cos(50.0°) ≈ 3.59
V2y = 4.72 * sin(50.0°) ≈ 3.63
Moving on to part (b), to determine the sum of V1 and V2, we need to add their respective x and y components.
Sum of x components:
V1x + V2x = -9.00 + 3.59 = -5.41
Sum of y components:
V1y + V2y = 0 + 3.63 = 3.63
To find the magnitude of the sum, we can use the Pythagorean theorem:
Magnitude = √((Sum of x components)^2 + (Sum of y components)^2)
Magnitude = √((-5.41)^2 + (3.63)^2) ≈ 6.42
Lastly, to find the direction of the sum, we can use inverse tangent (arctan) to find the angle relative to the positive x-axis.
Direction = arctan((Sum of y components)/(Sum of x components))
Direction = arctan(3.63/-5.41) ≈ -33.4°
Therefore, the sum of V1 and V2 has a magnitude of approximately 6.42 units and points at an angle of -33.4° counterclockwise from the positive x-axis.