Vector V1 is 8.58 units long and points along the -x axis. Vector V2 is 4.45 units long and points at +40.0° 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
Direction
° (counterclockwise from the +x axis is positive)
(a) To find the x and y components of each vector, we can use trigonometry.
For Vector V1:
Since it points along the -x axis, its x component will be negative. The magnitude of V1 is 8.58 units, and since it points along the -x axis, its y component will be 0.
Therefore, the components of V1 are:
V1x = -8.58
V1y = 0
For Vector V2:
The magnitude of V2 is 4.45 units, and it points at an angle of +40.0° to the +x axis. To find its x and y components, we can use the trigonometric functions cosine and sine.
V2x = magnitude of V2 * cos(angle)
V2y = magnitude of V2 * sin(angle)
V2x = 4.45 * cos(40.0°)
V2y = 4.45 * sin(40.0°)
Calculating these values, we get:
V2x ≈ 3.396
V2y ≈ 2.836
Therefore, the components of V2 are:
V2x ≈ 3.396
V2y ≈ 2.836
(b) To determine the sum V1 + V2, we can add their respective components.
Sum of x components: V1x + V2x = -8.58 + 3.396
Sum of y components: V1y + V2y = 0 + 2.836
Calculating these values, we get:
Sum of x components ≈ -5.184
Sum of y components ≈ 2.836
To find the magnitude of the sum, we can use the Pythagorean theorem:
Magnitude of the sum = √(sum of x components^2 + sum of y components^2)
Magnitude of the sum ≈ √((-5.184)^2 + (2.836)^2)
Calculating this value, we get:
Magnitude of the sum ≈ 5.81 units
To find the direction of the sum, we can use the inverse tangent function:
Direction = arctan(y component sum / x component sum)
Direction ≈ arctan(2.836 / -5.184)
Calculating this value, we get:
Direction ≈ -28.0° (counterclockwise from the +x axis is positive)
Therefore, the sum V1 + V2 has a magnitude of approximately 5.81 units and points at an angle of approximately -28.0° with respect to the +x axis.