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AP Precalculus Quiz

AP Precalculus Quiz: Parametric Functions Modeling Planar Motion

Practice Parametric Functions Modeling Planar Motion in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 19

0 of 19 answered

A Ferris wheel has radius 12 m12\text{ m}12 m, center (0,12)(0,12)(0,12), and angular speed ω=π/10 rad/s\omega=\pi/10\text{ rad/s}ω=π/10 rad/s: x(t)=12cos⁡(ωt)x(t)=12\cos(\omega t)x(t)=12cos(ωt), y(t)=12+12sin⁡(ωt)y(t)=12+12\sin(\omega t)y(t)=12+12sin(ωt). Find position at t=5t=5t=5.

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What this quiz covers

This quiz focuses on Parametric Functions Modeling Planar Motion, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

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 Ferris wheel has radius 12 m12\text{ m}12 m, center (0,12)(0,12)(0,12), and angular speed ω=π/10 rad/s\omega=\pi/10\text{ rad/s}ω=π/10 rad/s: x(t)=12cos⁡(ωt)x(t)=12\cos(\omega t)x(t)=12cos(ωt), y(t)=12+12sin⁡(ωt)y(t)=12+12\sin(\omega t)y(t)=12+12sin(ωt). Find position at t=5t=5t=5.

  1. (0,24)(0,24)(0,24) (correct answer)
  2. (12,12)(12,12)(12,12)
  3. (0,0)(0,0)(0,0)
  4. (−12,12)(-12,12)(−12,12)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the Ferris wheel motion is described by x(t) = 12cos(ωt) and y(t) = 12 + 12sin(ωt), where ω = π/10 rad/s represents the angular velocity. Choice A is correct because at t = 5, we have ωt = π/2, giving x(5) = 12cos(π/2) = 0 and y(5) = 12 + 12sin(π/2) = 12 + 12 = 24, placing the rider at the top of the wheel. Choice B is incorrect because it represents the center of the Ferris wheel, not a position on its circumference. To help students: Use unit circle knowledge to evaluate trigonometric functions at key angles. Visualize circular motion and connect parametric equations to physical positions on the circle.

Question 2

A pendulum of length 2 m2\text{ m}2 m swings with θ(t)=0.6cos⁡(πt)\theta(t)=0.6\cos(\pi t)θ(t)=0.6cos(πt) (radians): x(t)=2sin⁡θ(t)x(t)=2\sin\theta(t)x(t)=2sinθ(t), y(t)=−2cos⁡θ(t)y(t)=-2\cos\theta(t)y(t)=−2cosθ(t). What is position at t=0t=0t=0?

  1. (0,−2)(0,-2)(0,−2)
  2. (2,0)(2,0)(2,0)
  3. (1.13,−1.65)(1.13,-1.65)(1.13,−1.65) (correct answer)
  4. (0,2)(0,2)(0,2)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the pendulum motion is described by x(t) = 2sin(θ(t)) and y(t) = -2cos(θ(t)), where θ(t) = 0.6cos(πt) gives the angular displacement. Choice C is correct because at t = 0, θ(0) = 0.6cos(0) = 0.6 radians, so x(0) = 2sin(0.6) ≈ 1.13 m and y(0) = -2cos(0.6) ≈ -1.65 m. Choice A is incorrect because it assumes θ(0) = 0, failing to evaluate the given angular function. To help students: Practice composite function evaluation, first finding θ(t) then substituting into position equations. Visualize pendulum motion and understand that y is negative because the pendulum hangs downward from its pivot.

Question 3

A drone’s position is x(t)=5t+2x(t)=5t+2x(t)=5t+2, y(t)=4t−1y(t)=4t-1y(t)=4t−1 (meters). Based on these, what is its speed at t=3t=3t=3?

  1. 3 m/s3\text{ m/s}3 m/s
  2. 4 m/s4\text{ m/s}4 m/s
  3. 41 m/s\sqrt{41}\text{ m/s}41​ m/s (correct answer)
  4. 41 m/s41\text{ m/s}41 m/s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the drone follows a linear path with x(t) = 5t + 2 and y(t) = 4t - 1, indicating constant velocity in both directions. Choice C is correct because speed is the magnitude of velocity vector, found by taking derivatives: dx/dt = 5 and dy/dt = 4, giving speed = √(5² + 4²) = √41 m/s, which is constant for all times. Choice B is incorrect because it represents only the vertical component of velocity, ignoring the horizontal motion. To help students: Emphasize that speed in parametric motion requires finding both velocity components and using the Pythagorean theorem. Practice distinguishing between velocity components and total speed, reinforcing vector magnitude calculations.

Question 4

A Ferris wheel is modeled by x(t)=10cos⁡(π8t)x(t)=10\cos\left(\frac{\pi}{8}t\right)x(t)=10cos(8π​t), y(t)=10+10sin⁡(π8t)y(t)=10+10\sin\left(\frac{\pi}{8}t\right)y(t)=10+10sin(8π​t) (meters). Based on these, calculate the speed at t=4t=4t=4 s.​

  1. 5π2\frac{5\pi}{2}25π​ m/s
  2. 5π4\frac{5\pi}{4}45π​ m/s (correct answer)
  3. 10π8\frac{10\pi}{8}810π​ m/s
  4. 25π8\frac{25\pi}{8}825π​ m/s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on calculating speed in circular motion. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the Ferris wheel motion is described by x(t) = 10cos(πt/8) and y(t) = 10 + 10sin(πt/8), representing uniform circular motion with radius 10 m. Choice B is correct because the velocity components are vx = -10(π/8)sin(πt/8) and vy = 10(π/8)cos(πt/8), and the speed is constant at |v| = 10(π/8) = 5π/4 m/s for uniform circular motion. Choice A is incorrect because it doubles the correct speed value. To help students: Remember that for uniform circular motion, speed equals radius times angular velocity (v = rω). Practice differentiating trigonometric parametric equations and finding the magnitude of velocity vectors.

Question 5

A football is kicked from ground with V0=30 m/sV_0=30\text{ m/s}V0​=30 m/s: x(t)=30cos⁡θ tx(t)=30\cos\theta\,tx(t)=30cosθt, y(t)=30sin⁡θ t−4.9t2y(t)=30\sin\theta\,t-4.9t^2y(t)=30sinθt−4.9t2. How does increasing θ\thetaθ affect trajectory?

  1. Higher peak, shorter horizontal range (correct answer)
  2. Lower peak, longer horizontal range
  3. Higher peak, longer horizontal range
  4. No change in peak or range

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the football's trajectory depends on launch angle θ, with x(t) = 30cos(θ)t and y(t) = 30sin(θ)t - 4.9t². Choice A is correct because increasing θ (up to 90°) increases the vertical component 30sin(θ) while decreasing the horizontal component 30cos(θ), resulting in higher maximum height but shorter horizontal range. Choice C is incorrect because it fails to recognize that horizontal range decreases as angle increases beyond 45°. To help students: Analyze how trigonometric functions change with angle, noting that sin increases while cos decreases from 0° to 90°. Discuss that maximum range occurs at 45° for projectile motion.

Question 6

A pendulum bob of length 2.02.02.0 m swings with θ(t)=20∘cos⁡(πt)\theta(t)=20^\circ\cos(\pi t)θ(t)=20∘cos(πt); x(t)=2sin⁡(θ(t))x(t)=2\sin(\theta(t))x(t)=2sin(θ(t)), y(t)=−2cos⁡(θ(t))y(t)=-2\cos(\theta(t))y(t)=−2cos(θ(t)) (meters). What is its position at t=1t=1t=1 s?​

  1. (0 m, −2.0 m)(0\text{ m},\ -2.0\text{ m})(0 m, −2.0 m)
  2. (0.684 m, −1.879 m)(0.684\text{ m},\ -1.879\text{ m})(0.684 m, −1.879 m)
  3. (−0.684 m, −1.879 m)(-0.684\text{ m},\ -1.879\text{ m})(−0.684 m, −1.879 m) (correct answer)
  4. (0.684 m, 1.879 m)(0.684\text{ m},\ 1.879\text{ m})(0.684 m, 1.879 m)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on pendulum motion with angular displacement. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the pendulum bob's position is given by x(t) = 2sin(θ(t)) and y(t) = -2cos(θ(t)), where θ(t) = 20°cos(πt) describes the angular displacement. Choice C is correct because at t = 1 s, θ(1) = 20°cos(π) = -20°, giving x(1) = 2sin(-20°) ≈ -0.684 m and y(1) = -2cos(-20°) ≈ -1.879 m. Choice B is incorrect because it has the wrong sign for the x-coordinate, failing to account for the negative angle. To help students: Carefully track the sign of angles when evaluating trigonometric functions. Visualize pendulum motion and understand that negative angles correspond to displacement in the opposite direction.

Question 7

A roller coaster is modeled by x(t)=4tx(t)=4tx(t)=4t and y(t)=3+2sin⁡(πt/4)y(t)=3+2\sin(\pi t/4)y(t)=3+2sin(πt/4) (meters). What is the position at t=6t=6t=6?

  1. (24,1)(24,1)(24,1) (correct answer)
  2. (24,3)(24,3)(24,3)
  3. (6,1)(6,1)(6,1)
  4. (6,3)(6,3)(6,3)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the roller coaster moves with x(t) = 4t (constant horizontal speed) and y(t) = 3 + 2sin(πt/4) (oscillating vertical motion). Choice A is correct because at t = 6, x(6) = 4(6) = 24 m and y(6) = 3 + 2sin(3π/2) = 3 + 2(-1) = 1 m, giving position (24, 1). Choice B is incorrect because it fails to evaluate the sine function correctly, possibly assuming sin(3π/2) = 0 instead of -1. To help students: Practice evaluating trigonometric functions at multiples of π/2. Visualize how the vertical component oscillates while horizontal motion continues steadily.

Question 8

A Ferris wheel has radius 15 m15\text{ m}15 m, center at (0,18)(0,18)(0,18), and angular speed ω=0.2 rad/s\omega=0.2\text{ rad/s}ω=0.2 rad/s: x(t)=15cos⁡(0.2t)x(t)=15\cos(0.2t)x(t)=15cos(0.2t), y(t)=18+15sin⁡(0.2t)y(t)=18+15\sin(0.2t)y(t)=18+15sin(0.2t). Find its position at t=0t=0t=0.

  1. (15,18)(15,18)(15,18) (correct answer)
  2. (0,33)(0,33)(0,33)
  3. (15,33)(15,33)(15,33)
  4. (0,18)(0,18)(0,18)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on circular motion and initial position. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the Ferris wheel follows x(t) = 15cos(0.2t) and y(t) = 18 + 15sin(0.2t), where the center is at (0,18) and radius is 15 m. Choice A is correct because at t = 0, x(0) = 15cos(0) = 15 and y(0) = 18 + 15sin(0) = 18, giving position (15,18). Choice C (15,33) is incorrect because it adds the radius to the center height instead of using the sine function. To help students: Visualize circular motion starting from the rightmost point when using standard trigonometric functions. Remember that cos(0) = 1 and sin(0) = 0.

Question 9

A baseball is hit with V0=30 m/sV_0=30\text{ m/s}V0​=30 m/s from (0,1)(0,1)(0,1); x(t)=V0cos⁡θ tx(t)=V_0\cos\theta\,tx(t)=V0​cosθt, y(t)=1+V0sin⁡θ t−4.9t2y(t)=1+V_0\sin\theta\,t-4.9t^2y(t)=1+V0​sinθt−4.9t2. How does increasing θ\thetaθ affect the trajectory?

  1. Higher peak and shorter range (correct answer)
  2. Lower peak and longer range
  3. Higher peak and longer range always
  4. No change; only V0V_0V0​ matters

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on how launch angle affects projectile trajectory. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the baseball follows standard projectile motion with x(t) = V₀cos(θ)t and y(t) = 1 + V₀sin(θ)t - 4.9t². Choice A is correct because increasing θ (up to 90°) increases the vertical component V₀sin(θ) leading to higher peak, but decreases the horizontal component V₀cos(θ) leading to shorter range. Choice C is incorrect because while higher angles give higher peaks, they don't always give longer range (maximum range occurs at 45° for level ground). To help students: Visualize how velocity components change with angle. Remember that projectile motion involves trade-offs between height and distance.

Question 10

A rider moves on a circular track: x(t)=12cos⁡(0.5t)x(t)=12\cos(0.5t)x(t)=12cos(0.5t), y(t)=12sin⁡(0.5t)y(t)=12\sin(0.5t)y(t)=12sin(0.5t) (meters). Based on these, calculate the speed at t=4 st=4\text{ s}t=4 s.

  1. 6 m/s6\text{ m/s}6 m/s (correct answer)
  2. 24 m/s24\text{ m/s}24 m/s
  3. 12 m/s12\text{ m/s}12 m/s
  4. 3 m/s3\text{ m/s}3 m/s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on calculating speed from parametric equations. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the rider follows circular motion with x(t) = 12cos(0.5t) and y(t) = 12sin(0.5t), representing a circle of radius 12 m. Choice A is correct because speed = √[(dx/dt)² + (dy/dt)²], where dx/dt = -6sin(0.5t) and dy/dt = 6cos(0.5t), giving speed = √[36sin²(0.5t) + 36cos²(0.5t)] = 6 m/s at any time. Choice C (12 m/s) is incorrect because it uses the radius instead of calculating the derivative. To help students: Remember that speed in circular motion equals radius times angular velocity (12 × 0.5 = 6). Practice differentiating parametric equations and using the Pythagorean identity.

Question 11

A soccer ball is kicked from (0,0)(0,0)(0,0) with V0=18 m/sV_0=18\text{ m/s}V0​=18 m/s at 60∘60^\circ60∘: x(t)=9tx(t)=9tx(t)=9t, y(t)=15.59t−4.9t2y(t)=15.59t-4.9t^2y(t)=15.59t−4.9t2. When does it reach x=27 mx=27\text{ m}x=27 m?

  1. t=2 st=2\text{ s}t=2 s
  2. t=3 st=3\text{ s}t=3 s (correct answer)
  3. t=4 st=4\text{ s}t=4 s
  4. t=27 st=27\text{ s}t=27 s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the soccer ball follows x(t) = 9t and y(t) = 15.59t - 4.9t², where the velocity components come from 18cos(60°) = 9 and 18sin(60°) ≈ 15.59. Choice B is correct because when x = 27 m, we solve 9t = 27 to get t = 3 seconds, which is when the ball reaches the specified horizontal position. Choice D is incorrect because it confuses the x-coordinate value (27 m) with the time value, showing a common error in parametric problems. To help students: Clearly distinguish between position coordinates and time values. Practice solving for time when given specific position requirements in either x or y.

Question 12

A soccer ball follows x(t)=18tx(t)=18tx(t)=18t, y(t)=1.5+12t−4.9t2y(t)=1.5+12t-4.9t^2y(t)=1.5+12t−4.9t2 (meters, ttt in seconds). Determine when it reaches x=54 mx=54\text{ m}x=54 m.

  1. t=2 st=2\text{ s}t=2 s
  2. t=54 st=54\text{ s}t=54 s
  3. t=3 st=3\text{ s}t=3 s (correct answer)
  4. t=6 st=6\text{ s}t=6 s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on finding when an object reaches a specific horizontal position. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the soccer ball follows x(t) = 18t and y(t) = 1.5 + 12t - 4.9t², representing horizontal and vertical motion respectively. Choice C is correct because when x = 54, we solve 18t = 54, giving t = 3 seconds. Choice B (t = 54 s) is incorrect because it confuses the position value with the time value. To help students: Emphasize that parametric equations require solving for the parameter t first, then using that value in other calculations. Practice identifying which equation to use based on what information is given.

Question 13

A pendulum bob of length 2 m2\text{ m}2 m swings with θ(t)=30∘cos⁡(2t)\theta(t)=30^\circ\cos(2t)θ(t)=30∘cos(2t): x(t)=2sin⁡θ(t)x(t)=2\sin\theta(t)x(t)=2sinθ(t), y(t)=−2cos⁡θ(t)y(t)=-2\cos\theta(t)y(t)=−2cosθ(t). What is the bob’s position at t=0t=0t=0?

  1. (0,−2)(0,-2)(0,−2)
  2. (1.00,−1.73)(1.00,-1.73)(1.00,−1.73) (correct answer)
  3. (2,0)(2,0)(2,0)
  4. (1.73,−1.00)(1.73,-1.00)(1.73,−1.00)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on pendulum motion with angular displacement. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the pendulum bob follows x(t) = 2sin(θ(t)) and y(t) = -2cos(θ(t)), where θ(t) = 30°cos(2t) and the length is 2 m. Choice B is correct because at t = 0, θ(0) = 30°cos(0) = 30° = π/6 radians, giving x(0) = 2sin(π/6) = 1.00 and y(0) = -2cos(π/6) = -1.73. Choice A (0,-2) is incorrect because it assumes θ = 0° instead of θ = 30°. To help students: Convert degrees to radians when using trigonometric functions. Visualize pendulum motion with the bob hanging down from the pivot point.

Question 14

A roller coaster car follows x(t)=10tx(t)=10tx(t)=10t, y(t)=5sin⁡(π4t)+12y(t)=5\sin\left(\frac{\pi}{4}t\right)+12y(t)=5sin(4π​t)+12 (meters). Determine the time when it first reaches x=50 mx=50\text{ m}x=50 m.

  1. t=4 st=4\text{ s}t=4 s
  2. t=5 st=5\text{ s}t=5 s (correct answer)
  3. t=50 st=50\text{ s}t=50 s
  4. t=10 st=10\text{ s}t=10 s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on finding when an object reaches a specific position. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the roller coaster follows x(t) = 10t and y(t) = 5sin(π/4·t) + 12, combining linear horizontal motion with sinusoidal vertical motion. Choice B is correct because when x = 50, we solve 10t = 50, giving t = 5 seconds. Choice C (t = 50 s) is incorrect because it confuses the position value with the time value. To help students: Focus on which parametric equation contains the given information. Remember that for linear motion, solving for time is straightforward division.

Question 15

A ball is launched from (0,2)(0,2)(0,2) with V0=25 m/sV_0=25\text{ m/s}V0​=25 m/s at 30∘30^\circ30∘; x(t)=21.65tx(t)=21.65tx(t)=21.65t, y(t)=2+12.5t−4.9t2y(t)=2+12.5t-4.9t^2y(t)=2+12.5t−4.9t2. Using y(t)y(t)y(t), find the maximum height.

  1. 10.0 m10.0\text{ m}10.0 m
  2. 18.0 m18.0\text{ m}18.0 m (correct answer)
  3. 34.9 m34.9\text{ m}34.9 m
  4. 2.0 m2.0\text{ m}2.0 m

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on finding maximum height of a projectile launched from an elevated position. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the ball follows y(t) = 2 + 12.5t - 4.9t², starting from height 2 m with initial vertical velocity component 12.5 m/s. Choice B is correct because maximum height occurs when dy/dt = 12.5 - 9.8t = 0, giving t = 1.276 seconds, and y(1.276) = 2 + 12.5(1.276) - 4.9(1.276)² ≈ 18.0 m. Choice C (34.9 m) is incorrect because it might result from calculation errors or using wrong initial conditions. To help students: Remember to include the initial height when calculating maximum height. Verify that the answer is reasonable given the initial velocity.

Question 16

A drone’s path is x(t)=2+5tx(t)=2+5tx(t)=2+5t, y(t)=10+3ty(t)=10+3ty(t)=10+3t (meters). Starting at t=0t=0t=0, what is its position at t=6 st=6\text{ s}t=6 s?

  1. (32,28)(32,28)(32,28) (correct answer)
  2. (12,13)(12,13)(12,13)
  3. (30,18)(30,18)(30,18)
  4. (8,16)(8,16)(8,16)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on linear motion and position calculation. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the drone follows a linear path with x(t) = 2 + 5t and y(t) = 10 + 3t, starting from position (2,10) at t = 0. Choice A is correct because at t = 6, x(6) = 2 + 5(6) = 32 and y(6) = 10 + 3(6) = 28, giving position (32,28). Choice C (30,18) is incorrect because it appears to use incorrect arithmetic or wrong time value. To help students: Practice substituting time values carefully into parametric equations. Verify answers by checking that the motion follows a straight line with constant velocity.

Question 17

A Ferris wheel has radius 121212 m, center at (0,12)(0,12)(0,12) m, and rotates at ω=π/10\omega=\pi/10ω=π/10 rad/s; x(t)=12cos⁡(ωt)x(t)=12\cos(\omega t)x(t)=12cos(ωt), y(t)=12+12sin⁡(ωt)y(t)=12+12\sin(\omega t)y(t)=12+12sin(ωt). What is its position at t=5t=5t=5 s?

  1. (0 m, 24 m)(0\text{ m},\ 24\text{ m})(0 m, 24 m) (correct answer)
  2. (12 m, 12 m)(12\text{ m},\ 12\text{ m})(12 m, 12 m)
  3. (0 m, 12 m)(0\text{ m},\ 12\text{ m})(0 m, 12 m)
  4. (−12 m, 12 m)(-12\text{ m},\ 12\text{ m})(−12 m, 12 m)

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on circular motion of a Ferris wheel. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the Ferris wheel has parametric equations x(t) = 12cos(ωt) and y(t) = 12 + 12sin(ωt), where ω = π/10 rad/s, representing circular motion with center at (0,12). Choice A is correct because at t = 5 s, ωt = π/2, giving x(5) = 12cos(π/2) = 0 m and y(5) = 12 + 12sin(π/2) = 24 m, placing the rider at the top. Choice B is incorrect because it represents a position at a different angle on the wheel. To help students: Visualize circular motion using unit circle concepts, noting that π/2 radians corresponds to the top position. Practice evaluating trigonometric functions at key angles like π/2, π, and 3π/2.

Question 18

A car drives in a circle of radius 50 m50\text{ m}50 m centered at the origin: x(t)=50cos⁡(0.2t)x(t)=50\cos(0.2t)x(t)=50cos(0.2t), y(t)=50sin⁡(0.2t)y(t)=50\sin(0.2t)y(t)=50sin(0.2t). What is its speed at t=7t=7t=7?

  1. 0.2 m/s0.2\text{ m/s}0.2 m/s
  2. 10 m/s10\text{ m/s}10 m/s (correct answer)
  3. 50 m/s50\text{ m/s}50 m/s
  4. 50 m/s\sqrt{50}\text{ m/s}50​ m/s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the car follows circular motion with x(t) = 50cos(0.2t) and y(t) = 50sin(0.2t), where the coefficient 0.2 represents angular velocity in rad/s. Choice B is correct because for circular motion, speed equals radius times angular velocity: v = rω = 50 × 0.2 = 10 m/s, which remains constant throughout the motion. Choice D is incorrect because it might result from incorrectly calculating velocity components at a specific time without recognizing the constant speed property. To help students: Emphasize that uniform circular motion has constant speed despite changing velocity direction. Practice relating linear speed to angular velocity using v = rω.

Question 19

A skateboarder starts at (0,0)(0,0)(0,0) with x(t)=3tx(t)=3tx(t)=3t and y(t)=2t−0.5t2y(t)=2t-0.5t^2y(t)=2t−0.5t2 (meters). When does x=24 mx=24\text{ m}x=24 m occur?

  1. t=6 st=6\text{ s}t=6 s
  2. t=8 st=8\text{ s}t=8 s (correct answer)
  3. t=12 st=12\text{ s}t=12 s
  4. t=24 st=24\text{ s}t=24 s

Explanation: This question tests understanding of parametric functions in modeling planar motion, focusing on interpreting and solving real-world problems. Parametric functions allow us to describe an object's position over time in two dimensions using separate equations for x(t) and y(t). In this scenario, the skateboarder follows a path described by x(t) = 3t and y(t) = 2t - 0.5t², representing horizontal motion at constant velocity and vertical motion with deceleration. Choice B is correct because when x = 24 m, we solve 3t = 24 to get t = 8 seconds, which is when the skateboarder reaches the specified horizontal position. Choice C is incorrect because it might result from confusing the x-coordinate value with the time value or making an arithmetic error. To help students: Emphasize that parametric equations allow us to find when specific positions are reached by solving one equation for t. Practice problems where students must determine times for specific x or y values, reinforcing the independence of parametric components.