Parametric Functions Modeling Planar Motion

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AP Precalculus › Parametric Functions Modeling Planar Motion

Questions 1 - 10
1

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

$(0,0)$

$(-12,12)$

$(12,12)$

$(0,24)$

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.

2

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

$(0,-2)$

$(2,0)$

$(1.13,-1.65)$

$(0,2)$

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.

3

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

$\sqrt{41}\text{ m/s}$

$41\text{ m/s}$

$4\text{ m/s}$

$3\text{ m/s}$

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.

4

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

$\frac{10\pi}{8}$ m/s

$\frac{5\pi}{4}$ m/s

$\frac{5\pi}{2}$ m/s

$\frac{25\pi}{8}$ 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.

5

A football is kicked from ground with $V_0=30\text{ m/s}$: $x(t)=30\cos\theta,t$, $y(t)=30\sin\theta,t-4.9t^2$. How does increasing $\theta$ affect trajectory?

Lower peak, longer horizontal range

Higher peak, longer horizontal range

Higher peak, shorter horizontal range

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.

6

A pendulum bob of length $2.0$ m swings with $\theta(t)=20^\circ\cos(\pi t)$; $x(t)=2\sin(\theta(t))$, $y(t)=-2\cos(\theta(t))$ (meters). What is its position at $t=1$ s?​

$(0.684\text{ m},\ -1.879\text{ m})$

$(0.684\text{ m},\ 1.879\text{ m})$

$(-0.684\text{ m},\ -1.879\text{ m})$

$(0\text{ m},\ -2.0\text{ m})$

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.

7

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

$(24,1)$

$(24,3)$

$(6,3)$

$(6,1)$

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.

8

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

$(15,33)$

$(0,33)$

$(0,18)$

$(15,18)$

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.

9

A baseball is hit with $V_0=30\text{ m/s}$ from $(0,1)$; $x(t)=V_0\cos\theta,t$, $y(t)=1+V_0\sin\theta,t-4.9t^2$. How does increasing $\theta$ affect the trajectory?

No change; only $V_0$ matters

Higher peak and longer range always

Lower peak and longer range

Higher peak and shorter range

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.

10

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

$3\text{ m/s}$

$12\text{ m/s}$

$24\text{ m/s}$

$6\text{ m/s}$

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.

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