AP Physics C Mechanics Quiz: Motion In Two Or Three Dimensions
20 questions · exam conditions
0:00
Motion In Two Or Three DimensionsQuestion 1 of 20

A stone is thrown horizontally with a speed of 10 m/s from the top of a cliff 80 m high. How far from the base of the cliff does the stone land? Use g≈10g \approx 10 m/s2^2.

20 m
40 m
80 m
160 m
← Back to quizzes

AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Motion In Two Or Three Dimensions

Practice Motion In Two Or Three Dimensions in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Motion In Two Or Three Dimensions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A stone is thrown horizontally with a speed of 10 m/s from the top of a cliff 80 m high. How far from the base of the cliff does the stone land? Use g≈10g \approx 10 m/s2^2.

  1. 20 m
  2. 40 m (correct answer)
  3. 80 m
  4. 160 m
Explanation: The motion is analyzed in two independent components. The vertical motion determines the time of flight. Using Δy=v0yt+12ayt2\Delta y = v_{0y}t + \frac{1}{2}a_y t^2, with Δy=−80\Delta y = -80 m, v0y=0v_{0y} = 0, and ay=−10a_y = -10 m/s2^2: −80=0−5t2-80 = 0 - 5t^2, which gives t2=16t^2 = 16 and t=4t = 4 s. The horizontal distance (range) is determined by the horizontal motion: Δx=vxt=(10 m/s)(4 s)=40\Delta x = v_x t = (10 \text{ m/s})(4 \text{ s}) = 40 m.

Question 2

A particle moves in the xy-plane such that its x-coordinate is given by x(t)=Rcos⁡(ωt)x(t) = R\cos(\omega t) and its y-coordinate is given by y(t)=Rsin⁡(ωt)y(t) = R\sin(\omega t), where RR and ω\omega are positive constants. Which statement correctly describes the magnitude of the particle's acceleration vector?

  1. It is zero because the particle's speed is constant.
  2. It increases with time as the particle covers more distance.
  3. It is constant and directed toward the origin. (correct answer)
  4. It is constant and directed away from the origin.
Explanation: This is uniform circular motion. The position vector is r⃗(t)=Rcos⁡(ωt)i^+Rsin⁡(ωt)j^\vec{r}(t) = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}. Differentiating twice gives the acceleration vector: v⃗(t)=−Rωsin⁡(ωt)i^+Rωcos⁡(ωt)j^\vec{v}(t) = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j} and a⃗(t)=−Rω2cos⁡(ωt)i^−Rω2sin⁡(ωt)j^\vec{a}(t) = -R\omega^2\cos(\omega t)\hat{i} - R\omega^2\sin(\omega t)\hat{j}. This can be written as a⃗(t)=−ω2r⃗(t)\vec{a}(t) = -\omega^2 \vec{r}(t). The magnitude of the acceleration is ∣a⃗∣=ω2∣r⃗∣=ω2R|\vec{a}| = \omega^2 |\vec{r}| = \omega^2 R, which is constant. The direction is opposite to the position vector r⃗\vec{r}, meaning it is always directed toward the origin.

Question 3

The position of a particle moving in three-dimensional space is given by r⃗(t)=(2t2)i^+(cos⁡(πt))j^+(3t)k^\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=1t = 1 s?

  1. (4i^+3k^)(4\hat{i} + 3\hat{k}) m/s
  2. 14\sqrt{14} m/s
  3. 55 m/s (correct answer)
  4. 16+π4\sqrt{16 + \pi^4} m/s
Explanation: First, find the velocity vector by taking the derivative of the position vector: v⃗(t)=dr⃗dt=(4t)i^−(πsin⁡(πt))j^+(3)k^\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=1t = 1 s: v⃗(1)=(4(1))i^−(πsin⁡(π))j^+3k^=4i^−0j^+3k^=(4i^+3k^)\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: ∣v⃗(1)∣=42+32=16+9=25=5|\vec{v}(1)| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 m/s.

Question 4

A particle's position is described by the vector function r⃗(t)=(t3)i^+(2t2)j^\vec{r}(t) = (t^3)\hat{i} + (2t^2)\hat{j}, where tt is in seconds and position is in meters. What is the average velocity vector of the particle from t=0t = 0 s to t=2t = 2 s?

  1. (8i^+8j^)(8\hat{i} + 8\hat{j}) m/s
  2. (6i^+4j^)(6\hat{i} + 4\hat{j}) m/s
  3. (3i^+4j^)(3\hat{i} + 4\hat{j}) m/s
  4. (4i^+4j^)(4\hat{i} + 4\hat{j}) m/s (correct answer)
Explanation: Average velocity is defined as the displacement divided by the time interval, v⃗avg=Δr⃗Δt\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t}. First, find the displacement vector: Δr⃗=r⃗(2)−r⃗(0)\Delta\vec{r} = \vec{r}(2) - \vec{r}(0). r⃗(2)=(23)i^+(2(2)2)j^=8i^+8j^\vec{r}(2) = (2^3)\hat{i} + (2(2)^2)\hat{j} = 8\hat{i} + 8\hat{j}. r⃗(0)=0i^+0j^\vec{r}(0) = 0\hat{i} + 0\hat{j}. So, Δr⃗=8i^+8j^\Delta\vec{r} = 8\hat{i} + 8\hat{j}. The time interval is Δt=2−0=2\Delta t = 2 - 0 = 2 s. Therefore, v⃗avg=8i^+8j^2=(4i^+4j^)\vec{v}_{avg} = \frac{8\hat{i} + 8\hat{j}}{2} = (4\hat{i} + 4\hat{j}) m/s.

Question 5

A particle starts from the origin with an initial velocity of v⃗0=5i^\vec{v}_0 = 5\hat{i} m/s. It experiences a time-varying acceleration given by a⃗(t)=6ti^−4j^\vec{a}(t) = 6t\hat{i} - 4\hat{j} m/s2^2. What is the velocity vector v⃗(t)\vec{v}(t) of the particle at t=3t = 3 s?

  1. (27i^−12j^)(27\hat{i} - 12\hat{j}) m/s
  2. (32i^−12j^)(32\hat{i} - 12\hat{j}) m/s (correct answer)
  3. (18i^−4j^)(18\hat{i} - 4\hat{j}) m/s
  4. (59i^−12j^)(59\hat{i} - 12\hat{j}) m/s
Explanation: The velocity vector is found by integrating the acceleration vector with respect to time: v⃗(t)=∫a⃗(t)dt+C⃗\vec{v}(t) = \int \vec{a}(t) dt + \vec{C}. v⃗(t)=∫(6ti^−4j^)dt=3t2i^−4tj^+C⃗\vec{v}(t) = \int (6t\hat{i} - 4\hat{j}) dt = 3t^2\hat{i} - 4t\hat{j} + \vec{C}. The constant of integration C⃗\vec{C} is the initial velocity v⃗(0)=v⃗0=5i^\vec{v}(0) = \vec{v}_0 = 5\hat{i}. So, v⃗(t)=(3t2+5)i^−4tj^\vec{v}(t) = (3t^2 + 5)\hat{i} - 4t\hat{j}. Evaluating at t=3t = 3 s gives v⃗(3)=(3(3)2+5)i^−4(3)j^=(27+5)i^−12j^=(32i^−12j^)\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.

Question 6

A particle's position is given by r⃗(t)=(t2−8t)i^+(4t)j^\vec{r}(t) = (t^2 - 8t)\hat{i} + (4t)\hat{j}. At what time t>0t>0 is the particle's speed at a minimum?

  1. t=2t=2 s
  2. t=4t=4 s (correct answer)
  3. t=8t=8 s
  4. The speed is never at a minimum for t>0t>0.
Explanation: First, find the velocity vector: v⃗(t)=dr⃗dt=(2t−8)i^+4j^\vec{v}(t) = \frac{d\vec{r}}{dt} = (2t - 8)\hat{i} + 4\hat{j}. The speed ss is the magnitude of the velocity: s(t)=∣v⃗(t)∣=(2t−8)2+42s(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+16f(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: dfdt=2(2t−8)(2)=0\frac{df}{dt} = 2(2t - 8)(2) = 0, which solves to 2t−8=02t - 8 = 0, or t=4t = 4 s.

Question 7

A particle's motion is described by the parametric equations x(t)=4tx(t) = 4t and y(t)=20t−5t2y(t) = 20t - 5t^2, where xx and yy are in meters and tt is in seconds. Which equation describes the trajectory of the particle in the xy-plane?

  1. y=5x−516x2y = 5x - \frac{5}{16}x^2 (correct answer)
  2. y=5x−54x2y = 5x - \frac{5}{4}x^2
  3. y=20−10ty = 20 - 10t
  4. y=80t−20t2y = 80t - 20t^2
Explanation: To find the trajectory equation y(x)y(x), we must eliminate the parameter tt from the two equations. From the equation for x, we can solve for tt: t=x/4t = x/4. Now, substitute this expression for tt into the equation for y: y=20(x/4)−5(x/4)2=5x−5(x2/16)=5x−516x2y = 20(x/4) - 5(x/4)^2 = 5x - 5(x^2/16) = 5x - \frac{5}{16}x^2.

Question 8

A particle starts at the origin (0,0) at t=0t=0. Its velocity is given by v⃗(t)=(4t)i^+(3)j^\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=2t=2 s?

  1. 73\sqrt{73} m
  2. 10 m (correct answer)
  3. 14 m
  4. (8i^+6j^)(8\hat{i} + 6\hat{j}) m
Explanation: To find the displacement, we must integrate the velocity vector to get the position vector r⃗(t)\vec{r}(t). r⃗(t)=∫v⃗(t)dt=∫((4t)i^+3j^)dt=(2t2)i^+(3t)j^+C⃗\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, r⃗(0)=0\vec{r}(0) = 0, so C⃗=0\vec{C} = 0. At t=2t=2 s, the position (and displacement from origin) is r⃗(2)=(2(2)2)i^+(3(2))j^=8i^+6j^\vec{r}(2) = (2(2)^2)\hat{i} + (3(2))\hat{j} = 8\hat{i} + 6\hat{j}. The magnitude of this displacement is ∣r⃗(2)∣=82+62=64+36=100=10|\vec{r}(2)| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 m.

Question 9

A particle's position is given by the vector r⃗(t)=(2t3−4t)i^+(6−3t2)j^\vec{r}(t) = (2t^3 - 4t)\hat{i} + (6 - 3t^2)\hat{j}, where tt is in seconds and r⃗\vec{r} is in meters. What is the acceleration vector of the particle at t=2t = 2 s?

  1. (8i^−6j^)(8\hat{i} - 6\hat{j}) m/s2^2
  2. (44i^−12j^)(44\hat{i} - 12\hat{j}) m/s2^2
  3. (24i^−6j^)(24\hat{i} - 6\hat{j}) m/s2^2 (correct answer)
  4. (12ti^−6j^)(12t\hat{i} - 6\hat{j}) m/s2^2
Explanation: To find the acceleration vector, we must take the second derivative of the position vector with respect to time. First, find the velocity vector: v⃗(t)=dr⃗dt=(6t2−4)i^−6tj^\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t^2 - 4)\hat{i} - 6t\hat{j}. Then, find the acceleration vector: a⃗(t)=dv⃗dt=12ti^−6j^\vec{a}(t) = \frac{d\vec{v}}{dt} = 12t\hat{i} - 6\hat{j}. Finally, substitute t=2t = 2 s into the acceleration function: a⃗(2)=12(2)i^−6j^=(24i^−6j^)\vec{a}(2) = 12(2)\hat{i} - 6\hat{j} = (24\hat{i} - 6\hat{j}) m/s2^2.

Question 10

A projectile is fired from ground level with an initial velocity of 50 m/s at an angle of 37° above the horizontal. Assuming negligible air resistance and using g≈10g \approx 10 m/s2^2, what is the maximum vertical height reached by the projectile? (sin 37° ≈ 0.6, cos 37° ≈ 0.8)

  1. 125 m
  2. 80 m
  3. 90 m
  4. 45 m (correct answer)
Explanation: First, find the initial vertical component of velocity: v0y=v0sin⁡θ=(50 m/s)sin⁡(37°)≈50(0.6)=30v_{0y} = v_0 \sin\theta = (50 \text{ m/s}) \sin(37°) \approx 50(0.6) = 30 m/s. At the maximum height, the vertical velocity is zero. Using the kinematic equation vy2=v0y2+2ayΔyv_y^2 = v_{0y}^2 + 2a_y \Delta y, we have 02=(30)2+2(−10)H0^2 = (30)^2 + 2(-10) H. Solving for H gives 20H=90020H = 900, so H=45H = 45 m.

Question 11

An object is dropped from rest from a height HH and simultaneously another object is launched horizontally with an initial speed v0v_0 from the same height HH. Which object reaches the horizontal ground first, assuming negligible air resistance?

  1. The object dropped from rest reaches the ground first because it travels a shorter distance.
  2. The object launched horizontally reaches the ground first because it has a greater initial speed.
  3. Both objects reach the ground at the same time. (correct answer)
  4. The answer depends on the value of the initial horizontal speed v0v_0.
Explanation: 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 HH under the same acceleration due to gravity, gg. The time to fall depends only on the vertical motion, which is identical for both. Therefore, they reach the ground at the same time.

Question 12

A particle has an initial velocity of v⃗0=(4i^+1j^)\vec{v}_0 = (4\hat{i} + 1\hat{j}) m/s and is subject to a constant acceleration of a⃗=(2i^+2j^)\vec{a} = (2\hat{i} + 2\hat{j}) m/s2^2. What is the speed of the particle after 2.0 seconds?

  1. 89\sqrt{89} m/s (correct answer)
  2. 1313 m/s
  3. 17\sqrt{17} m/s
  4. 65\sqrt{65} m/s
Explanation: First, find the velocity vector at t=2.0t = 2.0 s using v⃗(t)=v⃗0+a⃗t\vec{v}(t) = \vec{v}_0 + \vec{a}t. v⃗(2)=(4i^+1j^)+(2i^+2j^)(2)=(4i^+1j^)+(4i^+4j^)=(8i^+5j^)\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: ∣v⃗(2)∣=82+52=64+25=89|\vec{v}(2)| = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89} m/s.

Question 13

A particle's velocity is given by v⃗(t)=(5)i^+(4t−2)j^\vec{v}(t) = (5)\hat{i} + (4t-2)\hat{j} m/s. The particle starts at position r⃗(0)=10i^\vec{r}(0) = 10\hat{i} m. At what time t>0t > 0 is its acceleration vector perpendicular to its position vector?

  1. t=0.5t = 0.5 s
  2. t=1.0t = 1.0 s (correct answer)
  3. t=2.0t = 2.0 s
  4. t=0t = 0 s
Explanation: First, find the acceleration vector: a⃗(t)=dv⃗dt=4j^\vec{a}(t) = \frac{d\vec{v}}{dt} = 4\hat{j} m/s2^2. Next, find the position vector by integrating velocity: r⃗(t)=∫v⃗(t)dt=(5t)i^+(2t2−2t)j^+C⃗\vec{r}(t) = \int \vec{v}(t) dt = (5t)\hat{i} + (2t^2-2t)\hat{j} + \vec{C}. Using the initial condition r⃗(0)=10i^\vec{r}(0) = 10\hat{i}, we find C⃗=10i^\vec{C} = 10\hat{i}. So, r⃗(t)=(5t+10)i^+(2t2−2t)j^\vec{r}(t) = (5t+10)\hat{i} + (2t^2-2t)\hat{j}. Two vectors are perpendicular if their dot product is zero: a⃗⋅r⃗=0\vec{a} \cdot \vec{r} = 0. (4j^)⋅((5t+10)i^+(2t2−2t)j^)=4(2t2−2t)=0(4\hat{j}) \cdot ((5t+10)\hat{i} + (2t^2-2t)\hat{j}) = 4(2t^2-2t) = 0. This gives 8t(t−1)=08t(t-1) = 0. The solutions are t=0t=0 and t=1t=1. The question asks for t>0t>0, so the answer is t=1t=1 s.

Question 14

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?

  1. Projectile A has a longer time of flight, and both projectiles have the same range.
  2. Projectile B has a longer time of flight, and Projectile B has a longer range.
  3. Both projectiles have the same time of flight and the same range.
  4. Projectile B has a longer time of flight, and both projectiles have the same range. (correct answer)
Explanation: Time of flight is given by T=2v0sin⁡θgT = \frac{2v_0 \sin\theta}{g}. Since sin⁡(60°)>sin⁡(30°)\sin(60°) > \sin(30°), Projectile B has a longer time of flight. Horizontal range is given by R=v02sin⁡(2θ)gR = \frac{v_0^2 \sin(2\theta)}{g}. For complementary angles like 30° and 60°, sin⁡(2⋅30°)=sin⁡(60°)\sin(2 \cdot 30°) = \sin(60°) and sin⁡(2⋅60°)=sin⁡(120°)=sin⁡(60°)\sin(2 \cdot 60°) = \sin(120°) = \sin(60°). Thus, their ranges are identical.

Question 15

A ball is thrown with an initial velocity v⃗i\vec{v}_i at an angle θ\theta above the horizontal. It lands at the same height from which it was thrown after a time TT. Neglecting air resistance, what is the change in the ball's velocity vector, Δv⃗=v⃗f−v⃗i\Delta\vec{v} = \vec{v}_f - \vec{v}_i, during this time?

  1. Zero, because the final speed is the same as the initial speed.
  2. A vector of magnitude 2visin⁡θ2 v_i \sin\theta directed vertically upward.
  3. A vector of magnitude 2vicos⁡θ2 v_i \cos\theta directed horizontally.
  4. A vector of magnitude gTgT directed vertically downward. (correct answer)
Explanation: The change in velocity is given by Δv⃗=a⃗Δt\Delta\vec{v} = \vec{a} \Delta t. In projectile motion, the acceleration is constant, a⃗=−gj^\vec{a} = -g\hat{j} (where j^\hat{j} is the unit vector in the upward vertical direction). The time interval is Δt=T\Delta t = T. Therefore, Δv⃗=(−gj^)T=−gTj^\Delta\vec{v} = (-g\hat{j}) T = -gT\hat{j}. This is a vector with magnitude gTgT directed vertically downward.

Question 16

A rock is thrown from a cliff of height hh with an initial velocity v⃗0=v0xi^+v0yj^\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.

  1. v⃗f=v0xi^−v0y2+2ghj^\vec{v}_f = v_{0x}\hat{i} - \sqrt{v_{0y}^2 + 2gh}\hat{j} (correct answer)
  2. v⃗f=v0xi^+v0y2+2ghj^\vec{v}_f = v_{0x}\hat{i} + \sqrt{v_{0y}^2 + 2gh}\hat{j}
  3. v⃗f=0i^−v0y2+2ghj^\vec{v}_f = 0\hat{i} - \sqrt{v_{0y}^2 + 2gh}\hat{j}
  4. v⃗f=v0xi^−v0y2−2ghj^\vec{v}_f = v_{0x}\hat{i} - \sqrt{v_{0y}^2 - 2gh}\hat{j}
Explanation: The horizontal component of velocity, vxv_x, remains constant, so vfx=v0xv_{fx} = v_{0x}. The final vertical component, vfyv_{fy}, can be found using the kinematic equation vfy2=v0y2+2ayΔyv_{fy}^2 = v_{0y}^2 + 2a_y \Delta y. Here, ay=−ga_y = -g and the vertical displacement is Δy=−h\Delta y = -h. So, vfy2=v0y2+2(−g)(−h)=v0y2+2ghv_{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: vfy=−v0y2+2ghv_{fy} = -\sqrt{v_{0y}^2 + 2gh}. Combining the components gives the final velocity vector.

Question 17

A particle moves along a curved path in the xy-plane. At a certain instant, its velocity vector is v⃗=3i^+4j^\vec{v} = 3\hat{i} + 4\hat{j} m/s and its acceleration vector is a⃗=2i^−1j^\vec{a} = 2\hat{i} - 1\hat{j} m/s2^2. At this instant, which of the following is true?

  1. The particle is slowing down and its path is curving.
  2. The particle is speeding up and moving in a straight line.
  3. The particle is moving at a constant speed and its path is curving.
  4. The particle is speeding up and its path is curving. (correct answer)
Explanation: To determine if the particle is speeding up or slowing down, we evaluate the dot product of the velocity and acceleration vectors. v⃗⋅a⃗=(3)(2)+(4)(−1)=6−4=2\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 a⃗\vec{a} is not a scalar multiple of v⃗\vec{v}, the path is not straight; it is curving.

Question 18

A projectile is launched from level ground with initial speed v0v_0 at an angle θ\theta above the horizontal. In terms of v0v_0, θ\theta, and gg, what is the radius of curvature of its path at the apex of its trajectory?

  1. v02cos⁡2θg\frac{v_0^2 \cos^2\theta}{g} (correct answer)
  2. v02sin⁡2θg\frac{v_0^2 \sin^2\theta}{g}
  3. v02sin⁡(2θ)g\frac{v_0^2 \sin(2\theta)}{g}
  4. Zero, because the vertical velocity is zero at the apex.
Explanation: At the apex of the trajectory, the velocity is purely horizontal, with magnitude v=v0cos⁡θv = v_0 \cos\theta. The acceleration is gg, directed vertically downward. At this specific point, the acceleration is perpendicular to the velocity, so it acts entirely as centripetal acceleration, ac=ga_c = g. The formula for centripetal acceleration is ac=v2/Ra_c = v^2/R, where R is the radius of curvature. Therefore, g=(v0cos⁡θ)2Rg = \frac{(v_0 \cos\theta)^2}{R}. Solving for R gives R=v02cos⁡2θgR = \frac{v_0^2 \cos^2\theta}{g}.

Question 19

A projectile is launched from a flat, horizontal surface. At the highest point of its trajectory, which of the following statements is true? Neglect air resistance.

  1. The velocity vector is horizontal, and the acceleration vector is directed vertically downward. (correct answer)
  2. Both the velocity and acceleration vectors are zero.
  3. The velocity vector is zero, but the acceleration vector is non-zero and directed vertically downward.
  4. The velocity vector is horizontal, and the acceleration vector is zero.
Explanation: 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.

Question 20

Based on the scenario, how does the velocity vector change over time for uniform circular motion?

  1. Magnitude constant; direction rotates; Δv⃗\Delta\vec v points inward (correct answer)
  2. Magnitude increases; direction constant; Δv⃗\Delta\vec v is tangent
  3. Magnitude constant; direction constant; Δv⃗=0⃗\Delta\vec v=\vec 0
  4. Magnitude decreases; direction rotates; Δv⃗\Delta\vec v points outward
Explanation: This question tests AP Physics C kinematics, specifically motion in two or three dimensions for uniform circular motion velocity vectors. The motion requires understanding how velocity vectors behave in circular motion - constant magnitude but continuously changing direction. In this scenario, we analyze how the velocity vector evolves during uniform circular motion. Choice A is correct because it accurately describes that velocity magnitude stays constant while direction rotates, and the change in velocity (Δv) points inward toward the center, creating centripetal acceleration. Choice C is incorrect because it suggests the velocity vector doesn't change at all, which would mean no acceleration and thus no circular motion. To help students: Use vector diagrams showing velocity at different points around the circle. Demonstrate how subtracting consecutive velocity vectors yields an inward-pointing Δv. Watch for: confusion between speed (scalar) and velocity (vector) and misunderstanding how vector subtraction works.