This is uniform circular motion. The position vector is $$\vec{r}(t) = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}$$. Differentiating twice gives the acceleration vector: $$\vec{v}(t) = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j}$$ and $$\vec{a}(t) = -R\omega^2\cos(\omega t)\hat{i} - R\omega^2\sin(\omega t)\hat{j}$$. This can be written as $$\vec{a}(t) = -\omega^2 \vec{r}(t)$$. The magnitude of the acceleration is $$|\vec{a}| = \omega^2 |\vec{r}| = \omega^2 R$$, which is constant. The direction is opposite to the position vector $$\vec{r}$$, meaning it is always directed toward the origin.

First, find the velocity vector by taking the derivative of the position vector: $$\vec{v}(t) = \frac{d\vec{r}}{dt} = (4t)\hat{i} - (\pi\sin(\pi t))\hat{j} + (3)\hat{k}$$. Next, evaluate the velocity vector at $$t = 1$$ s: $$\vec{v}(1) = (4(1))\hat{i} - (\pi\sin(\pi))\hat{j} + 3\hat{k} = 4\hat{i} - 0\hat{j} + 3\hat{k} = (4\hat{i} + 3\hat{k})$$ m/s. The speed is the magnitude of this vector: $$ |\vec{v}(1)| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $$ m/s.

Average velocity is defined as the displacement divided by the time interval, $$\vec{v}{avg} = \frac{\Delta\vec{r}}{\Delta t}$$. First, find the displacement vector: $$\Delta\vec{r} = \vec{r}(2) - \vec{r}(0)$$. $$\vec{r}(2) = (2^3)\hat{i} + (2(2)^2)\hat{j} = 8\hat{i} + 8\hat{j}$$. $$\vec{r}(0) = 0\hat{i} + 0\hat{j}$$. So, $$\Delta\vec{r} = 8\hat{i} + 8\hat{j}$$. The time interval is $$\Delta t = 2 - 0 = 2$$ s. Therefore, $$\vec{v}{avg} = \frac{8\hat{i} + 8\hat{j}}{2} = (4\hat{i} + 4\hat{j})$$ m/s.

The velocity vector is found by integrating the acceleration vector with respect to time: $$\vec{v}(t) = \int \vec{a}(t) dt + \vec{C}$$. $$\vec{v}(t) = \int(6t\hat{i} - 4\hat{j}) dt = 3t^2\hat{i} - 4t\hat{j} + \vec{C}$$. The constant of integration $$\vec{C}$$ is the initial velocity $$\vec{v}(0) = \vec{v}_0 = 5\hat{i}$$. So, $$\vec{v}(t) = (3t^2 + 5)\hat{i} - 4t\hat{j}$$. Evaluating at $$t = 3$$ s gives $$\vec{v}(3) = (3(3)^2 + 5)\hat{i} - 4(3)\hat{j} = (27 + 5)\hat{i} - 12\hat{j} = (32\hat{i} - 12\hat{j})$$ m/s.

First, find the velocity vector: $$\vec{v}(t) = \frac{d\vec{r}}{dt} = (2t - 8)\hat{i} + 4\hat{j}$$. The speed $$s$$ is the magnitude of the velocity: $$s(t) = |\vec{v}(t)| = \sqrt{(2t - 8)^2 + 4^2}$$. To find the minimum speed, we can find the minimum of the function inside the square root, $$f(t) = (2t - 8)^2 + 16$$. This is a parabola that opens upward, and its minimum occurs at the vertex. The vertex is where the derivative is zero: $$\frac{df}{dt} = 2(2t - 8)(2) = 0$$, which solves to $$2t - 8 = 0$$, or $$t = 4$$ s.

To find the trajectory equation $$y(x)$$, we must eliminate the parameter $$t$$ from the two equations. From the equation for x, we can solve for $$t$$: $$t = x/4$$. Now, substitute this expression for $$t$$ into the equation for y: $$y = 20(x/4) - 5(x/4)^2 = 5x - 5(x^2/16) = 5x - \frac{5}{16}x^2$$.

To find the displacement, we must integrate the velocity vector to get the position vector $$\vec{r}(t)$$. $$\vec{r}(t) = \int \vec{v}(t) dt = \int((4t)\hat{i} + 3\hat{j}) dt = (2t^2)\hat{i} + (3t)\hat{j} + \vec{C}$$. Since the particle starts at the origin, $$\vec{r}(0) = 0$$, so $$\vec{C} = 0$$. At $$t=2$$ s, the position (and displacement from origin) is $$\vec{r}(2) = (2(2)^2)\hat{i} + (3(2))\hat{j} = 8\hat{i} + 6\hat{j}$$. The magnitude of this displacement is $$|\vec{r}(2)| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ m.

For projectile motion, the acceleration due to gravity is always constant and directed vertically downward. At the highest point of the trajectory, the vertical component of the velocity is momentarily zero, but the horizontal component of the velocity remains constant (and non-zero, unless launched vertically). Thus, the velocity vector is purely horizontal, while the acceleration vector is directed downward.

This question tests AP Physics C kinematics, specifically understanding velocity components in projectile motion under uniform gravity. The motion in two dimensions requires recognizing that gravity acts only vertically, leaving horizontal motion unaffected. In this scenario with no air resistance, only gravity acts on the projectile, providing constant downward acceleration. Choice B is correct because vₓ remains constant (no horizontal forces) while vᵧ decreases linearly due to constant gravitational acceleration (vᵧ = v₀ᵧ - gt). Choice A reverses the components, C incorrectly suggests both components change, and D misunderstands that while speed changes, velocity components change predictably. To help students: Emphasize the independence of horizontal and vertical motion in projectile problems. Practice analyzing force diagrams to identify which components experience acceleration. Watch for: confusing speed with velocity or forgetting that only vertical motion is affected by gravity.

This question tests AP Physics C kinematics, specifically centripetal acceleration in circular motion. The motion in two dimensions for uniform circular motion requires understanding that acceleration always points toward the center of the circular path. In this scenario, a car travels on a 50 m radius track at constant speed 12 m/s, requiring centripetal acceleration a = v²/r. Choice B is correct because a = (12)²/50 = 144/50 = 2.88 m/s² directed inward toward the center. Choice A incorrectly states the direction as outward, violating Newton's laws for circular motion. Choices C and D show incorrect calculations of the acceleration magnitude. To help students: Emphasize that centripetal acceleration always points toward the center, regardless of the direction of motion. Practice identifying the radius and speed in various circular motion scenarios. Watch for: confusion between centripetal and centrifugal effects, or errors in applying the v²/r formula.