Motion in Two or Three Dimensions - AP Physics C Mechanics

A particle undergoes uniform circular motion in the xy-plane. At $t = 0$ , the particle is at position $(R, 0)$ and has velocity $(0, v_0)$ where $v_0 > 0$ . At what time will the particle first return to a position where its x-coordinate equals $\frac{R}{2}$ ?
Answer: $\frac{\pi R}{3v_0}$ For uniform circular motion, $\omega = \frac{v_0}{R}$ . The position is $x(t) = R\cos(\omega t)$ and $y(t) = R\sin(\omega t)$ since at $t = 0$ , $x = R$ and $y = 0$ . We need $x(t) = \frac{R}{2}$ , so $R\cos(\omega t) = \frac{R}{2}$ , giving $\cos(\omega t) = \frac{1}{2}$ . The first positive solution is $\omega t = \frac{\pi}{3}$ , so $t = \frac{\pi}{3\omega} = \frac{\pi R}{3v_0}$ . Choice B corresponds to $\omega t = \frac{\pi}{6}$ which gives $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \neq \frac{1}{2}$ . Choice C corresponds to $\omega t = \frac{2\pi}{3}$ which gives $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ , the wrong sign. Choice D corresponds to $\omega t = \frac{\pi}{2}$ which gives $\cos(\frac{\pi}{2}) = 0$ .
A particle has an initial velocity of $\vec{v}_0 = (4\hat{i} + 1\hat{j})$ m/s and is subject to a constant acceleration of $\vec{a} = (2\hat{i} + 2\hat{j})$ m/s $^2$ . What is the speed of the particle after 2.0 seconds?
Answer: $\sqrt{89}$ m/s First, find the velocity vector at $t = 2.0$ s using $\vec{v}(t) = \vec{v}_0 + \vec{a}t$ . $\vec{v}(2) = (4\hat{i} + 1\hat{j}) + (2\hat{i} + 2\hat{j})(2) = (4\hat{i} + 1\hat{j}) + (4\hat{i} + 4\hat{j}) = (8\hat{i} + 5\hat{j})$ m/s. The speed is the magnitude of the velocity vector: $|\vec{v}(2)| = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89}$ m/s.
Based on the scenario, how does the velocity vector change over time for a projectile under uniform gravity with no air resistance?
Answer: $v_x$ constant; $v_y$ decreases linearly 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.
A particle's position is given by the vector $\vec{r}(t) = (2t^3 - 4t)\hat{i} + (6 - 3t^2)\hat{j}$ , where $t$ is in seconds and $\vec{r}$ is in meters. What is the acceleration vector of the particle at $t = 2$ s?
Answer: $(24\hat{i} - 6\hat{j})$ m/s $^2$ To find the acceleration vector, we must take the second derivative of the position vector with respect to time. First, find the velocity vector: $\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t^2 - 4)\hat{i} - 6t\hat{j}$ . Then, find the acceleration vector: $\vec{a}(t) = \frac{d\vec{v}}{dt} = 12t\hat{i} - 6\hat{j}$ . Finally, substitute $t = 2$ s into the acceleration function: $\vec{a}(2) = 12(2)\hat{i} - 6\hat{j} = (24\hat{i} - 6\hat{j})$ m/s $^2$ .
A projectile is launched from level ground with initial speed $v_0$ at an angle $\theta$ above the horizontal. In terms of $v_0$ , $\theta$ , and $g$ , what is the radius of curvature of its path at the apex of its trajectory?
Answer: $\frac{v_0^2 \cos^2\theta}{g}$ At the apex of the trajectory, the velocity is purely horizontal, with magnitude $v = v_0 \cos\theta$ . The acceleration is $g$ , directed vertically downward. At this specific point, the acceleration is perpendicular to the velocity, so it acts entirely as centripetal acceleration, $a_c = g$ . The formula for centripetal acceleration is $a_c = v^2/R$ , where R is the radius of curvature. Therefore, $g = \frac{(v_0 \cos\theta)^2}{R}$ . Solving for R gives $R = \frac{v_0^2 \cos^2\theta}{g}$ .
A ball is thrown with an initial velocity $\vec{v}_i$ at an angle $\theta$ above the horizontal. It lands at the same height from which it was thrown after a time $T$ . Neglecting air resistance, what is the change in the ball's velocity vector, $\Delta\vec{v} = \vec{v}_f - \vec{v}_i$ , during this time?
Answer: A vector of magnitude $gT$ directed vertically downward. The change in velocity is given by $\Delta\vec{v} = \vec{a} \Delta t$ . In projectile motion, the acceleration is constant, $\vec{a} = -g\hat{j}$ (where $\hat{j}$ is the unit vector in the upward vertical direction). The time interval is $\Delta t = T$ . Therefore, $\Delta\vec{v} = (-g\hat{j}) T = -gT\hat{j}$ . This is a vector with magnitude $gT$ directed vertically downward.
A rock is thrown from a cliff of height $h$ with an initial velocity $\vec{v}_0 = v_{0x}\hat{i} + v_{0y}\hat{j}$ . What is the velocity vector of the rock just before it hits the ground below? Neglect air resistance, and let the origin be the base of the cliff.
Answer: $\vec{v}_f = v_{0x}\hat{i} - \sqrt{v_{0y}^2 + 2gh}\hat{j}$ The horizontal component of velocity, $v_x$ , remains constant, so $v_{fx} = v_{0x}$ . The final vertical component, $v_{fy}$ , can be found using the kinematic equation $v_{fy}^2 = v_{0y}^2 + 2a_y \Delta y$ . Here, $a_y = -g$ and the vertical displacement is $\Delta y = -h$ . So, $v_{fy}^2 = v_{0y}^2 + 2(-g)(-h) = v_{0y}^2 + 2gh$ . Since the rock is moving downward just before impact, we take the negative root: $v_{fy} = -\sqrt{v_{0y}^2 + 2gh}$ . Combining the components gives the final velocity vector.
A particle's position is described by the vector function $\vec{r}(t) = (t^3)\hat{i} + (2t^2)\hat{j}$ , where $t$ is in seconds and position is in meters. What is the average velocity vector of the particle from $t = 0$ s to $t = 2$ s?
Answer: $(4\hat{i} + 4\hat{j})$ 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.
A particle moves along a curved path in the xy-plane. At a certain instant, its velocity vector is $\vec{v} = 3\hat{i} + 4\hat{j}$ m/s and its acceleration vector is $\vec{a} = 2\hat{i} - 1\hat{j}$ m/s $^2$ . At this instant, which of the following is true?
Answer: The particle is speeding up and its path is curving. To determine if the particle is speeding up or slowing down, we evaluate the dot product of the velocity and acceleration vectors. $\vec{v} \cdot \vec{a} = (3)(2) + (4)(-1) = 6 - 4 = 2$ . Since the dot product is positive, the component of acceleration parallel to velocity is in the same direction as velocity, so the particle is speeding up. For the path to be straight, the acceleration vector must be parallel to the velocity vector. Since $\vec{a}$ is not a scalar multiple of $\vec{v}$ , the path is not straight; it is curving.
The position of a particle moving in three-dimensional space is given by $\vec{r}(t) = (2t^2)\hat{i} + (\cos(\pi t))\hat{j} + (3t)\hat{k}$ in SI units. What is the speed of the particle at $t = 1$ s?
Answer: $5$ m/s 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.
Two projectiles are launched simultaneously from the same point. Projectile A is launched at angle $30°$ with initial speed $v_A$ , and projectile B is launched at angle $60°$ with initial speed $v_B$ . If both projectiles have the same range, what is the ratio $\frac{v_A}{v_B}$ ?
Answer: $1$ The range formula is $R = \frac{v^2\sin(2\theta)}{g}$ . For projectile A: $R_A = \frac{v_A^2\sin(60°)}{g} = \frac{v_A^2\frac{\sqrt{3}}{2}}{g}$ . For projectile B: $R_B = \frac{v_B^2\sin(120°)}{g} = \frac{v_B^2\frac{\sqrt{3}}{2}}{g}$ since $\sin(120°) = \sin(60°) = \frac{\sqrt{3}}{2}$ . Setting $R_A = R_B$ : $\frac{v_A^2\frac{\sqrt{3}}{2}}{g} = \frac{v_B^2\frac{\sqrt{3}}{2}}{g}$ , which simplifies to $v_A^2 = v_B^2$ , so $v_A = v_B$ and $\frac{v_A}{v_B} = 1$ . This makes physical sense because complementary angles ( $30°$ and $60°$ are complementary since $30° + 60° = 90°$ ) give the same range for the same initial speed. Choice B comes from incorrectly using $\frac{\sin(60°)}{\sin(30°)} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$ . Choice C results from the reciprocal calculation with a factor error. Choice D is the reciprocal of choice B.
A projectile is fired from the edge of a cliff at angle $\theta$ above the horizontal with initial speed $v_0$ . The projectile lands at a horizontal distance $d$ from the launch point and at a vertical distance $h$ below the launch point. If the projectile were instead fired horizontally with the same initial speed from the same point, at what horizontal distance from the launch point would it land?
Answer: $v_0\sqrt{\frac{2h}{g}}$ For horizontal launch, the time of flight is determined by the vertical motion: $h = \frac{1}{2}gt^2$ , so $t = \sqrt{\frac{2h}{g}}$ . The horizontal distance is then $d_{horizontal} = v_0 t = v_0\sqrt{\frac{2h}{g}}$ . This is independent of the original trajectory parameters $\theta$ and $d$ because horizontal launch only depends on the initial horizontal speed and the fall height. Choice A incorrectly tries to relate it to the original range through a cosine factor. Choice B attempts a complex relationship involving the original angle and trajectory, which is unnecessary for horizontal launch. Choice D over-complicates the relationship and includes terms from the angled launch that don't affect the horizontal launch scenario.
Using the information given, determine the object's acceleration at this point for a car on a $50\,\text{m}$ radius track at $12\,\text{m/s}$ .
Answer: $2.88\,\text{m/s}^2$ inward 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.
A particle moves in the xy-plane such that its x-coordinate is given by $x(t) = R\cos(\omega t)$ and its y-coordinate is given by $y(t) = R\sin(\omega t)$ , where $R$ and $\omega$ are positive constants. Which statement correctly describes the magnitude of the particle's acceleration vector?
Answer: It is constant and directed toward the origin. 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.
Using the information given, determine the object's acceleration at this point for a block sliding down a frictionless $30^\circ$ incline.
Answer: $4.9\,\text{m/s}^2$ down the incline This question tests AP Physics C kinematics, specifically motion on inclined planes in two dimensions. The motion requires decomposing gravitational acceleration into components parallel and perpendicular to the incline surface. In this scenario, a block slides down a frictionless 30° incline, experiencing only the component of gravity parallel to the surface. Choice B is correct because a_parallel = g·sin(30°) = 9.8 × 0.5 = 4.9 m/s² down the incline. Choice A incorrectly uses the full gravitational acceleration, C and D show wrong magnitudes or directions. To help students: Practice decomposing weight into components using free body diagrams on inclines. Emphasize that on frictionless surfaces, only the parallel component of weight causes acceleration. Watch for: using cosine instead of sine, or confusing the direction of acceleration with the normal force direction.
A particle's velocity is given by $\vec{v}(t) = (5)\hat{i} + (4t-2)\hat{j}$ m/s. The particle starts at position $\vec{r}(0) = 10\hat{i}$ m. At what time $t > 0$ is its acceleration vector perpendicular to its position vector?
Answer: $t = 1.0$ s First, find the acceleration vector: $\vec{a}(t) = \frac{d\vec{v}}{dt} = 4\hat{j}$ m/s $^2$ . Next, find the position vector by integrating velocity: $\vec{r}(t) = \int \vec{v}(t) dt = (5t)\hat{i} + (2t^2-2t)\hat{j} + \vec{C}$ . Using the initial condition $\vec{r}(0) = 10\hat{i}$ , we find $\vec{C} = 10\hat{i}$ . So, $\vec{r}(t) = (5t+10)\hat{i} + (2t^2-2t)\hat{j}$ . Two vectors are perpendicular if their dot product is zero: $\vec{a} \cdot \vec{r} = 0$ . $(4\hat{j}) \cdot ((5t+10)\hat{i} + (2t^2-2t)\hat{j}) = 4(2t^2-2t) = 0$ . This gives $8t(t-1) = 0$ . The solutions are $t=0$ and $t=1$ . The question asks for $t>0$ , so the answer is $t=1$ s.
A particle starts at the origin (0,0) at $t=0$ . Its velocity is given by $\vec{v}(t) = (4t)\hat{i} + (3)\hat{j}$ m/s. What is the magnitude of the particle's displacement from the origin at $t=2$ s?
Answer: 10 m 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.
Two projectiles are launched from the same point on level ground with the same initial speed. Projectile A is launched at an angle of 30° above the horizontal, and Projectile B is launched at 60° above the horizontal. Neglecting air resistance, which of the following statements correctly compares their time of flight and horizontal range?
Answer: Projectile B has a longer time of flight, and both projectiles have the same range. Time of flight is given by $T = \frac{2v_0 \sin\theta}{g}$ . Since $\sin(60°) > \sin(30°)$ , Projectile B has a longer time of flight. Horizontal range is given by $R = \frac{v_0^2 \sin(2\theta)}{g}$ . For complementary angles like 30° and 60°, $\sin(2 \cdot 30°) = \sin(60°)$ and $\sin(2 \cdot 60°) = \sin(120°) = \sin(60°)$ . Thus, their ranges are identical.
An object is dropped from rest from a height $H$ and simultaneously another object is launched horizontally with an initial speed $v_0$ from the same height $H$ . Which object reaches the horizontal ground first, assuming negligible air resistance?
Answer: Both objects reach the ground at the same time. The horizontal and vertical components of motion are independent. For both objects, the initial vertical velocity is zero, and they both fall the same vertical distance $H$ under the same acceleration due to gravity, $g$ . The time to fall depends only on the vertical motion, which is identical for both. Therefore, they reach the ground at the same time.
A particle's position is given by $\vec{r}(t) = (t^2 - 8t)\hat{i} + (4t)\hat{j}$ . At what time $t>0$ is the particle's speed at a minimum?
Answer: $t=4$ 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.

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