Kinematics · quiz worked solutions

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

Gravity in Depth · question 1

Back to the question · AP Topic 2.6

Question: What is the attraction force between Drew and Jake given that their masses are 2 kg each, G = 1 m 3 / ( kg ⋅ s 2 ) and the distance between them is 2 meters?

Answer: 1 N.

1. Use F = Gm₁m₂/r² with the deliberately supplied G = 1. Why: This is the hypothetical constant in the problem, not measured universal G.

2. Calculate (1)(2)(2)/2² = 1 N. Why: Both masses enter the numerator and the separation is squared.

Gravity in Depth · question 2

Back to the question · AP Topic 2.6

Question: What would be the force between two objects given that the mass of both objects is 3 kg , and that the distance between them is 9 meters?

Answer: Force magnitude G/9 in the stated numerical units.

1. Insert the two 3 kg masses and 9 m separation into F = Gm₁m₂/r². Why: The gravitational interaction depends on both masses.

2. Simplify G(3)(3)/9² = G/9. Why: This expression is a magnitude; the actual force points toward the other object.

Gravity in Depth · question 3

Back to the question · AP Topic 2.6

Question: Two objects have a gravitational attraction of F → a when they are 1 m apart. If the masses remain constant, what will the new gravitational force be if the radius between the objects is increased by 3 meters?

Answer: F_b = F_a/16.

1. The distance increases by 3 m, from 1 m to 4 m. Why: An increase by 3 is not a new distance of 3.

2. Use F_b/F_a = (1/4)² = 1/16. Why: Fixed masses leave only inverse-square distance scaling.

Gravity in Depth · question 4

Back to the question · AP Topic 2.6

Question: Two objects initially have equal mass and attract with a force of 4 N at separation 2 m. Their separation increases to 3 m while one mass stays fixed. By what percentage must the other mass increase to keep the force at 4 N?

Answer: 125% increase.

1. Keep F constant: m_new/m_old = (3/2)² = 2.25. Why: Increasing one mass compensates for the larger squared separation.

2. Calculate the percentage increase (2.25 − 1)100% = 125%. Why: 225% is the final mass relative to the original, not the increase.

Gravity in Depth · question 5

Back to the question · Extra exploration

Question: A person of mass m₁ is anywhere inside an otherwise empty spherical shell whose mass m₂ is distributed uniformly over the shell. Ignoring all other bodies, what gravitational force does the shell exert on the person?

Answer: 0 N.

1. Use the uniform spherical-shell theorem. Why: All shell elements contribute, rather than one point mass at the center.

2. Inside the empty shell, their gravitational fields cancel everywhere. Why: This enrichment result requires spherical symmetry; outside matter and a nonuniform shell would change it.

FFR - Kepler’s Laws · question 1

Back to the question · Extra exploration

Question: An ellipse is given by the equation x 2 4 + y 2 16 = 25 . What is the width of the ellipse along its x-axis?

Answer: Width 20 units.

1. Divide the equation by 25 to get x²/100 + y²/400 = 1. Why: This puts the ellipse into standard semiaxis form.

2. The x semiaxis is √100 = 10, so full x width is 20. Why: The historical keyed 10 is the semiaxis, not the width; use 20 for the stated question.

FFR - Kepler’s Laws · question 2

Back to the question · Extra exploration

Question: Where are the foci located on an ellipse with the following equation? x 2 16 + y 2 64 = 1

Answer: Foci at (0, ±6.928).

1. The larger denominator gives the y semimajor axis a = 8, while b = 4. Why: The ellipse is longer vertically.

2. Find c = √(a² − b²) = √48 = 6.928. Why: Place the two foci symmetrically on the major axis.

FFR - Kepler’s Laws · question 3

Back to the question · AP Topic 6.6

Question: A negligible-mass planet has orbital period 50 Earth years around the Sun. Find its orbital semimajor axis in au.

Answer: 13.57 au.

1. For a negligible-mass body around the Sun, use T² ≈ a³ with T in years and a in au. Why: The Earth-normalized form uses the same central mass.

2. Find a = 50^(2/3) ≈ 13.57 au. Why: This gives semimajor axis, not necessarily the instantaneous orbital distance.

FFR - Kepler’s Laws · question 4

Back to the question · Extra exploration

Question: A planet has the shown orbit around a star. At which of the marked points would the planet be moving the fastest?

Answer: Point D.

1. Identify the closest marked point to the star. Why: A central attractive force speeds the body on its inward part of the orbit.

2. Choose periapsis, D. Why: Conservation of orbital energy or equal-areas reasoning puts the largest speed nearest the star.

FFR - Kepler’s Laws · question 5

Back to the question · AP Topic 6.6

Question: A negligible-mass planet orbiting the Sun has semimajor axis 45 au. Find its orbital period in Earth years.

Answer: 302 years.

1. Use T² ≈ a³ for the same solar central mass. Why: The supplied semimajor axis is 45 au.

2. Calculate T = 45^(3/2) ≈ 301.87 years. Why: The period grows faster than the radius in this relation.

Projectile Motion · question 1

Back to the question · AP Topic 1.5

Question: If a ball is thrown as shown, what with the magnitude of its velocity be when it returns to its original y-value?

Answer: 20 m/s.

1. Under gravity alone, vₓ is constant. Why: There is no horizontal acceleration in the ideal projectile model.

2. At the original height vᵧ reverses sign but keeps its magnitude, so the total speed returns to the shown launch speed 20 m/s. Why: Equal component magnitudes give equal speed; the velocity direction is different.

Projectile Motion · question 2

Back to the question · AP Topic 1.5

Question: This is the same ball as in question 1. How long does the ball take to reach its initial y-value?

Answer: 3.34 s.

1. Resolve the shown 20 m/s launch at 55°: vᵧ₀ = 20 sin55° = 16.383 m/s. Why: The flight time back to the same height depends on vertical motion.

2. Use 0 = vᵧ₀t − ½gt², giving t = 2vᵧ₀/9.81 ≈ 3.340 s. Why: Discard t = 0 because it is the launch instant.

Projectile Motion · question 3

Back to the question · AP Topic 1.5

Question: A 50 g ball is launched at 20 m/s and 55° above horizontal, as shown. A constant 10 N force acts horizontally to the left throughout flight. Find its horizontal landing coordinate when it returns to launch height. Use g = 9.81 m/s²; take right as +x.

Answer: About 1077 m left of launch.

1. Find t = 2(20 sin55°)/9.81 ≈ 3.340 s. Why: The extra force is horizontal, so vertical flight time stays the same.

2. Use aₓ = −10/0.050 = −200 m/s² and vₓ₀ = 20 cos55° ≈ 11.472 m/s. Why: The small mass gives a large leftward acceleration.

3. Calculate x = vₓ₀t + ½aₓt² ≈ −1077.29 m. Why: The negative coordinate means the ball lands far to the left in this idealized constant-force model.

Projectile Motion · question 4

Back to the question · AP Topic 1.5

Question: With gravity as the only force and no air resistance, a projectile launched upward has the same speed when it returns to its launch height: vₓ is unchanged and vᵧ has the opposite sign.

Answer: True.

1. Keep the gravity-only model and compare the two equal-height events. Why: There is no change in horizontal velocity.

2. The vertical speed has equal magnitude on ascent and descent, with opposite sign. Why: Thus speed is the same although the velocity vector is not.

Projectile Motion · question 5

Back to the question · AP Topic 1.5

Question: In the following diagram, a ball is shot over a hill. What is the smallest initial angle needed to make it over the hill, assuming it peaks atop the hill.

Answer: 23.05° under the stated peak-at-the-hill assumption.

1. Set the peak height to 50 m: (80 sinθ)²/(2g) = 50. Why: At the peak the vertical velocity is zero.

2. Find θ = arcsin(√(2·9.81·50)/80) ≈ 23.05°. Why: This is the smallest angle reaching that height; without hill-distance data it does not independently guarantee clearing an arbitrary hill profile.

Projectile Motion · question 1

Back to the question · AP Topic 1.5

Question: What is the horizontal acceleration typically assumed in projectile motion?

Answer: 0 m/s².

1. Use the ideal projectile assumption: only gravity acts. Why: Near Earth’s surface gravity is vertical.

2. The horizontal net force is zero, so aₓ = 0. Why: Horizontal velocity remains constant in this model.

Projectile Motion · question 2

Back to the question · AP Topic 1.5

Question: If a projectile is launched with the shown trajectory, how far away will it land? The landing spot is at the same elevation as the launch point.

Answer: 362 m.

1. Read v₀ = 60 m/s and θ = 40° and find t = 2v₀ sinθ/g. Why: The launch and landing heights are equal.

2. Find R = v₀ cosθ·t = v₀² sin80°/9.81 ≈ 361.40 m. Why: The nearest listed value is 362 m.

Projectile Motion · question 3

Back to the question · AP Topic 1.5

Question: For a fixed launch speed, which angle maximizes the range over flat ground at the launch elevation? Neglect air resistance.

Answer: 45°.

1. Write R = v₀² sin2θ/g for equal launch and landing heights. Why: The speed and g are fixed.

2. Maximize sin2θ at 2θ = 90°, hence θ = 45°. Why: This conclusion needs the stated flat-ground, no-drag conditions.

Projectile Motion · question 4

Back to the question · AP Topic 1.5

Question: Three ballistas are arranged along a castle’s wall. If they each try to hit a target 100 m away from the base of the wall by launching the bolt horizontally, what speed will the middle ballista need to fire its bolt?

Answer: 49.5 m/s.

1. Read the middle ballista’s height 20 m; t = √(2·20/9.81) ≈ 2.019 s. Why: A horizontal launch starts with zero vertical velocity.

2. Find vₓ = 100/t ≈ 49.52 m/s. Why: The same flight time applies to the independent horizontal motion.

Projectile Motion · question 5

Back to the question · AP Topic 1.5

Question: Match each situation to the point on the projectile’s path it best applies to.

Answer: Targets 1–4: launch Δx = 0; peak vᵧ = 0; ascent vᵧ > 0; descent vᵧ < 0.

1. Choose the origin at launch with upward positive. Why: That fixes the coordinate and velocity signs.

2. Match rising, top and falling points using the slope of the path. Why: The top has zero vertical velocity, but still nonzero horizontal velocity and downward acceleration.

Variables of Motion · question 1

Back to the question · AP Topic 1.2

Question: An object moves in the +x direction: 4 m in 3 s, then 8 m in 10 s, waits 5 s, then moves 9 m in 4 s. What is its average velocity?

Answer: 0.95i m/s.

1. Add displacement 4 + 8 + 9 = 21 m and elapsed time 3 + 10 + 5 + 4 = 22 s. Why: The wait counts in elapsed time even though it adds no displacement.

2. Find average velocity 21/22 ≈ 0.9545 m/s along +x. Why: Average velocity is total displacement divided by total time.

Variables of Motion · question 2

Back to the question · AP Topic 1.2

Question: If a car drives at 10 m / s , then accelerates to 15 m / s in 5 s , then accelerates again to 20 m / s in 10 s , drives at that speed for another 10 s , and then decelerates to 0 m / s in 30 s , what is the car’s average acceleration in the direction of motion??

Answer: −0.18 m/s².

1. Use total velocity change 0 − 10 = −10 m/s. Why: Average acceleration depends on endpoints.

2. Use total stated time 5 + 10 + 10 + 30 = 55 s, giving −10/55 ≈ −0.1818 m/s². Why: Do not average the accelerations of unequal-duration stages directly.

Variables of Motion · question 3

Back to the question · AP Topic 1.3

Question: A car starts from rest and then accelerates at a constant acceleration of 5 m / s 2 . What is the car’s speed after 40 s ?

Answer: 200 m/s.

1. Start with v_f = v_i + at. Why: The acceleration is constant.

2. Calculate 0 + (5)(40) = 200 m/s. Why: The units m/s² times s become m/s.

Variables of Motion · question 4

Back to the question · AP Topic 1.3

Question: A 50 kg skydiver falls for 18 s before deploying their parachute. If he started with a vertical velocity of 0 m / s , what is the vertical velocity just prior to deploying the parachute? Neglect the air resistance.

Answer: −176.6 m/s with upward positive.

1. Use aᵧ = −9.81 m/s² and vᵧ₀ = 0. Why: The problem explicitly neglects air resistance.

2. Find vᵧ = 0 − (9.81)(18) = −176.58 m/s. Why: Mass cancels from free fall in this model; real air drag would matter.

Variables of Motion · question 5

Back to the question · AP Topic 1.2

Question: A man travels a distance of 4km over 5 hours. What is his average speed?

Answer: 0.22 m/s.

1. Convert 4 km = 4000 m and 5 h = 18,000 s. Why: Use compatible SI units before dividing.

2. Find average speed 4000/18000 ≈ 0.222 m/s. Why: Speed uses path distance, so no direction is attached.

Variables of Motion · question 1

Back to the question · AP Topic 1.2

Question: In the position–time graph, what is the speed at 14 s? The vertical axis is position in metres.

Answer: 5 m/s.

1. At 14 s use the position–time segment from (13 s, 10 m) to (15 s, 0 m). Why: Velocity is the slope of the local segment, not the position value.

2. Find v = (0 − 10)/(15 − 13) = −5 m/s; speed = 5 m/s. Why: Speed is the nonnegative magnitude of velocity; the printed position-axis unit is repaired to metres by the prompt.

Variables of Motion · question 2

Back to the question · AP Topic 1.2

Question: For the following velocity versus time graph, what is the acceleration in the direction of motion at 6 s ?

Answer: 0.25 m/s².

1. At 6 s use the velocity–time segment from (5 s, 1 m/s) to (9 s, 2 m/s). Why: Acceleration is the graph’s slope.

2. Find a = (2 − 1)/(9 − 5) = 0.25 m/s². Why: This is positive because velocity increases on that segment.

Variables of Motion · question 3

Back to the question · AP Topic 1.2

Question: Draw the position–time graph of a one-dimensional particle that starts at 0 m with velocity 5 m/s, changes to −2 m/s at 3 s, continues at that velocity for another 10 s, then remains stopped for 2 s.

Answer: Option 2: rise to 15 m, fall to −5 m, then stay flat.

1. From 0–3 s draw slope +5 m/s, ending at 15 m. Why: Position accumulates the velocity times the interval.

2. From 3–13 s draw slope −2 m/s, ending at −5 m; from 13–15 s draw a horizontal line. Why: A reversal changes slope sign, while stopping makes slope zero.

Variables of Motion · question 4

Back to the question · AP Topic 1.2

Question: Which of the following position versus time graphs show a particle that is accelerating?

Answer: Option 4, the curved position graph.

1. Compare graph slopes at successive times. Why: The position–time slope is velocity.

2. Select the curve whose slope changes. Why: A straight line has constant velocity and zero acceleration, even if its slope is nonzero.

Variables of Motion · question 5

Back to the question · AP Topic 1.2

Question: Match the labels with the position versus time graph.

Answer: Targets 1–3: stopped, reverses, accelerates.

1. Assign stopped to a horizontal position segment and reverses to a peak where slope changes sign. Why: Zero slope means zero velocity; a sign change means opposite direction.

2. Assign accelerates to the curved part whose slope changes. Why: Curvature shows changing velocity rather than merely changing position.

Variables of Motion · question 1

Back to the question · AP Topic 1.3

Question: If a ball is thrown upwards at 10 m / s , how long does it take to fall to 1 m below where it was initially thrown?

Answer: 2.13 s.

1. Use −1 = 10t − ½(9.81)t² with upward positive. Why: The destination is one metre below the launch height.

2. Solve t = [10 + √(100 + 19.62)]/9.81 ≈ 2.134 s. Why: The other root is negative and lies before launch.

Variables of Motion · question 2

Back to the question · AP Topic 1.3

Question: If a car is accelerating at 4 m / s 2 , and it starts at rest, how long does it take to go 60 m ?

Answer: 5.48 s.

1. Use 60 = ½(4)t² because v₀ = 0. Why: Constant acceleration lets us use the displacement equation.

2. Find t = √30 ≈ 5.477 s. Why: The physical elapsed time is the positive root.

Variables of Motion · question 3

Back to the question · AP Topic 1.3

Question: If two cars are racing each other, and they have the velocities as shown, and car A has an acceleration of 2 m / s 2 , and car B has an acceleration of 5 m / s 2 , how long will it take for car B to pass car A?

Answer: 4.73 s.

1. Set x_A = 10 + 25t + t² and x_B = 20t + 2.5t² using the diagram’s 10 m lead. Why: Both cars need positions in the same coordinate frame.

2. Solve 1.5t² − 5t − 10 = 0, giving t = (5 + √85)/3 ≈ 4.740 s. Why: The nearest old option is 4.73 s; the negative root is not a future passing time.

Variables of Motion · question 4

Back to the question · AP Topic 1.3

Question: If a skydiver jumps out of a plane that is flying straight up at 30 m / s , how long does it take for the skydiver to fall 500 m ? Neglect air resistance.

Answer: 13.6 s.

1. Set upward positive and use −500 = 30t − ½(9.81)t². Why: The jumper initially shares the plane’s upward velocity.

2. Find t = [30 + √(30² + 2·9.81·500)]/9.81 ≈ 13.607 s. Why: The upward first part adds time before the downward fall.

Variables of Motion · question 5

Back to the question · AP Topic 1.3

Question: Two people are running 5 km . The first person runs at 7 m / s for the first 2 km , but then decelerates constantly at 0.5 m / s 2 for the next 7 seconds, then stays at that new speed for the rest of the race. The second person runs at 4.5 m / s for the entire race. If they started at the same time, who finishes first?

Answer: The second runner.

1. Runner 1 takes 2000/7 = 285.714 s, then slows to 3.5 m/s in 7 s while covering ((7 + 3.5)/2) × 7 = 36.75 m. Why: A constant-acceleration stage uses average endpoint speed.

2. The remaining 2963.25 m takes 846.643 s, so total ≈ 1139.36 s; runner 2 takes 5000/4.5 ≈ 1111.11 s. Why: The second runner finishes about 28.25 s sooner despite the slower opening speed.

Model references

These original worked steps use the stated question models. Physics references: Projectile motion, Orbits. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.