Final answer:
The magnitude of vector A+B is approximately 6.2 units and the direction is approximately -68.2°.
Step-by-step explanation:
To find the magnitude and direction of the sum of vectors A and B, we can use vector addition. First, let's break down vector A into its x and y components.
The x component of A, Ax, is equal to A * cos(45°). Since vector A makes an angle of 45° with the positive x-axis, Ax is equal to 8 * cos(45°) = 5.66 units. The y component of A, Ay, is equal to A * sin(45°).
Since vector A makes an angle of 45° with the positive x-axis, Ay is equal to 8 * sin(45°) = 5.66 units.
Now let's break down vector B into its x and y components. Since vector B is along the negative x-axis, its x component, Bx, is equal to -8 units. Since vector B is perpendicular to the y-axis, its y component, By, is equal to 0 units.
To find the sum of vectors A and B, we can add their respective components. The x component of the sum, Rx, is equal to Ax + Bx = 5.66 - 8
= -2.34 units
The y component of the sum, Ry, is equal to Ay + By = 5.66 + 0
= 5.66 units
The magnitude of the sum, R, is equal to the square root of (Rx^2 + Ry^2).
R = sqrt((-2.34)^2 + (5.66)^2)
≈ 6.2 units
The direction of the sum, theta, can be found using the inverse tangent function. theta = tan^-1(Ry/Rx) = tan^-1(5.66/-2.34) ≈ -68.2°.
Therefore, the magnitude of A+B is approximately 6.2 units and the direction is approximately -68.2°.