In this problem, we are tasked with determining the final position vector after a series of movements starting from an initial position. The initial position is given as a vector with a magnitude of 6.2 units at an angle of 25 degrees below the x-axis. This means that the vector points into the fourth quadrant of the coordinate system. To visualize this, we can represent the initial position vector as R_A.

Next, we travel a distance of 9.9 units at an angle of 78 degrees above the positive x-axis, which we will label as R_AB. Finally, we move 2 meters in the negative x direction, represented as R_BC. Our goal is to find the magnitude and direction of the final position vector R_C.

To solve this, we first break down each vector into its x and y components. The components of a vector can be calculated using the following equations:

x = r * cos(θ)

y = r * sin(θ)

For the initial position vector R_A:

x_A = 6.2 * cos(25°) = 5.62

y_A = 6.2 * sin(25°) = -2.62 (negative because it is below the x-axis)

For the displacement vector R_AB:

x_AB = 9.9 * cos(78°) = 2.06

y_AB = 9.9 * sin(78°) = 9.68

For the displacement vector R_BC (which only affects the x-component):

x_BC = -2

y_BC = 0

Now, we can compile these components into a table to sum them up:

Adding the x-components:

x_C = 5.62 + 2.06 - 2 = 5.68

Adding the y-components:

y_C = -2.62 + 9.68 + 0 = 7.06

Now that we have the components of the final position vector R_C, we can calculate its magnitude using the Pythagorean theorem:

|R_C| = √(x_C² + y_C²) = √(5.68² + 7.06²) = 9.06

To find the direction of the vector, we use the inverse tangent function:

θ_C = tan^-1(y_C / x_C) = tan^-1(7.06 / 5.68) = 51.2°

Thus, the final position vector R_C has a magnitude of 9.06 units and is directed at an angle of 51.2 degrees above the positive x-axis.