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 RA.
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 RAB. Finally, we move 2 meters in the negative x direction, represented as RBC. Our goal is to find the magnitude and direction of the final position vector RC.
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 RA:
xA = 6.2 * cos(25°) = 5.62
yA = 6.2 * sin(25°) = -2.62 (negative because it is below the x-axis)
For the displacement vector RAB:
xAB = 9.9 * cos(78°) = 2.06
yAB = 9.9 * sin(78°) = 9.68
For the displacement vector RBC (which only affects the x-component):
xBC = -2
yBC = 0
Now, we can compile these components into a table to sum them up:
Vector | x Component | y Component |
---|---|---|
RA | 5.62 | -2.62 |
RAB | 2.06 | 9.68 |
RBC | -2 | 0 |
Adding the x-components:
xC = 5.62 + 2.06 - 2 = 5.68
Adding the y-components:
yC = -2.62 + 9.68 + 0 = 7.06
Now that we have the components of the final position vector RC, we can calculate its magnitude using the Pythagorean theorem:
|RC| = √(xC2 + yC2) = √(5.682 + 7.062) = 9.06
To find the direction of the vector, we use the inverse tangent function:
θC = tan-1(yC / xC) = tan-1(7.06 / 5.68) = 51.2°
Thus, the final position vector RC has a magnitude of 9.06 units and is directed at an angle of 51.2 degrees above the positive x-axis.