Vector quantities are defined as physical quantities that have both magnitude and direction. Velocity, acceleration, displacement, and force all fit this definition. The other choices contain scalar quantities: speed, distance, work, power, kinetic energy, electric charge, temperature, mass, time, and electric potential are all described by magnitude only.

The magnitude of the sum of two vectors is equal to the sum of their individual magnitudes only in the specific case where the vectors are parallel and point in the same direction (an angle of $$0^\circ$$ between them). In all other cases, due to the triangle inequality, the magnitude of the resultant vector will be less than the sum of the individual magnitudes.

The vector has a negative x-component and a positive y-component, placing it in the second quadrant. The reference angle relative to the negative x-axis can be found using the arctangent: $$\alpha = \arctan{\left|\frac{v_y}{v_x}\right|} = \arctan{\left|\frac{4.0}{-3.0}\right|} \approx 53.1^\circ$$. The angle measured counterclockwise from the positive x-axis is $$180^\circ - \alpha = 180^\circ - 53.1^\circ = 126.9^\circ$$. Distractor A is only the reference angle. The other distractors correspond to angles in the third and fourth quadrants.

The displacement vector is the change in position, calculated as $$\Delta\vec{r} = \vec{r}_f - \vec{r}_i$$. Subtracting the components: $$x: (-4) - (2) = -6$$. $$y: (6) - (-3) = 9$$. $$z: (2) - (5) = -3$$. Therefore, $$\Delta\vec{r} = (-6\hat{i} + 9\hat{j} - 3\hat{k})$$ m. Distractor A represents the sum $$\vec{r}_i + \vec{r}_f$$. Distractor B represents the vector $$\vec{r}_i - \vec{r}_f$$. Distractor D contains calculation errors.

First, resolve each displacement vector into its x (east) and y (north) components. For the first displacement: $$d_{1x} = 6.0 \cos(30^\circ) \approx 5.20$$ m, $$d_{1y} = 6.0 \sin(30^\circ) = 3.00$$ m. For the second displacement: $$d_{2x} = -4.0 \sin(60^\circ) \approx -3.46$$ m, $$d_{2y} = 4.0 \cos(60^\circ) = 2.00$$ m. The resultant components are $$R_x = 5.20 - 3.46 = 1.74$$ m and $$R_y = 3.00 + 2.00 = 5.00$$ m. The magnitude of the resultant displacement is found using the Pythagorean theorem: $$|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(1.74)^2 + (5.00)^2} \approx 5.3$$ m. Distractor D is the scalar sum of the magnitudes.

A unit vector is a vector with a magnitude of 1. It is found by dividing the vector by its own magnitude. First, find the magnitude of $$\vec{r}$$: $$|\vec{r}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$$ m. Then, divide the vector by this magnitude: $$\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{3\hat{i} - 4\hat{j}}{5} = 0.6\hat{i} - 0.8\hat{j}$$. This resulting vector is dimensionless.

Displacement is a vector quantity representing the change in position from the start point to the end point. Since the athlete finishes at the same point they started, the displacement vector is zero. Distance is a scalar quantity representing the total path length covered. For one lap around a circular track, the distance is the circumference, $$2\pi R$$.

Velocity is the time derivative of the position vector, $$\vec{v}(t) = \frac{d\vec{r}}{dt}$$. We must differentiate each component of $$\vec{r}(t)$$ with respect to time. $$\frac{d}{dt}(3t^2 - 4t) = 6t - 4$$. $$\frac{d}{dt}(2t^3) = 6t^2$$. $$\frac{d}{dt}(-5t) = -5$$. Combining these gives the correct velocity vector. Distractor B is the integral of the position vector. Distractor C is the acceleration vector (the derivative of velocity). Distractor D omits the k-component and does not perform the differentiation.

The magnitude of a vector with components $$F_x$$ and $$F_y$$ is given by the Pythagorean theorem: $$|\vec{F}| = \sqrt{F_x^2 + F_y^2}$$. Substituting the given components: $$|\vec{F}| = \sqrt{(12)^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$ N. Distractor A is the scalar subtraction of the component magnitudes. Distractor D is the scalar addition of the component magnitudes.

The magnitude of the resultant of two vectors depends on the angle between them. The maximum possible magnitude occurs when the vectors are parallel (angle $$0^\circ$$), resulting in a magnitude of $$8 + 6 = 14$$ N. The minimum possible magnitude occurs when the vectors are anti-parallel (angle $$180^\circ$$), resulting in a magnitude of $$|8 - 6| = 2$$ N. All other possible magnitudes lie between these two extremes. Therefore, a magnitude of 1 N is not possible.