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AP Calculus BC Quiz

AP Calculus BC Quiz: Motion Problems Parametric Vector Valued Functions

Practice Motion Problems Parametric Vector Valued Functions in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A particle has position r(t)=\langle 2\cos t,\,2\sin t\rangle; which gives its velocity vector at t=π/3t=\pi/3t=π/3?

Select an answer to continue

What this quiz covers

This quiz focuses on Motion Problems Parametric Vector Valued Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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 particle has position r(t)=\langle 2\cos t,\,2\sin t\rangle; which gives its velocity vector at t=π/3t=\pi/3t=π/3?

  1. ⟨−2sin⁡(π/3), 2cos⁡(π/3)⟩\langle -2\sin(\pi/3),\,2\cos(\pi/3)\rangle⟨−2sin(π/3),2cos(π/3)⟩ (correct answer)
  2. ⟨2cos⁡(π/3), 2sin⁡(π/3)⟩\langle 2\cos(\pi/3),\,2\sin(\pi/3)\rangle⟨2cos(π/3),2sin(π/3)⟩
  3. ∣⟨−2sin⁡(π/3), 2cos⁡(π/3)⟩∣\left\lvert\langle -2\sin(\pi/3),\,2\cos(\pi/3)\rangle\right\rvert∣⟨−2sin(π/3),2cos(π/3)⟩∣
  4. ⟨2sin⁡(π/3), 2cos⁡(π/3)⟩\langle 2\sin(\pi/3),\,2\cos(\pi/3)\rangle⟨2sin(π/3),2cos(π/3)⟩
  5. ⟨−2cos⁡(π/3), −2sin⁡(π/3)⟩\langle -2\cos(\pi/3),\,-2\sin(\pi/3)\rangle⟨−2cos(π/3),−2sin(π/3)⟩

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the velocity vector, differentiate each component of r(t) = <2 cos t, 2 sin t> with respect to t, resulting in v(t) = <-2 sin t, 2 cos t>. Evaluate at t = π/3, where sin(π/3) = √3/2 and cos(π/3) = 1/2, giving v(π/3) = <-2(√3/2), 2(1/2)> = <-√3, 1>, which matches <-2 sin(π/3), 2 cos(π/3)>. Note that the expression is left in terms of sine and cosine as presented in the choices. A tempting distractor is B, <2 cos(π/3), 2 sin(π/3)>, which is actually the position vector at t = π/3, confusing position with velocity. A transferable motion-analysis strategy is to remember that velocity is the derivative of position, so always apply differentiation to parametric components before evaluation.

Question 2

A particle follows r(t)=\langle t-2,\,(t-2)^2\rangle; what is its speed at t=2t=2t=2?

  1. ⟨1,0⟩\langle 1,0\rangle⟨1,0⟩
  2. 2\sqrt{2}2​
  3. 111 (correct answer)
  4. 000
  5. ⟨0,1⟩\langle 0,1\rangle⟨0,1⟩

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the speed, differentiate: dx/dt = 1 and dy/dt = 2(t-2). At t=2, velocity is <1,0>, and speed is √(1+0)=1. This indicates motion only in the x-direction at that point. A tempting distractor might be √2, perhaps from evaluating at t=1 or t=3 where the y-component is nonzero. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.

Question 3

A particle moves as x=t2+1x=t^2+1x=t2+1 and y=4−ty=4-ty=4−t; what is its velocity at t=3t=3t=3?

  1. ⟨6,−1⟩\langle 6,-1\rangle⟨6,−1⟩ (correct answer)
  2. 37\sqrt{37}37​
  3. ⟨10,1⟩\langle 10,1\rangle⟨10,1⟩
  4. 6\sqrt{6}6​
  5. 555

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the velocity, differentiate the given x and y: dx/dt = 2t and dy/dt = -1. At t=3, this yields <6, -1>. The velocity vector shows both components of the motion at that instant. A tempting distractor might be √37, which is the speed instead of the velocity vector. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.

Question 4

A robot follows r(t)=⟨cos⁡t, sin⁡t⟩r(t)=\langle \cos t,\,\sin t\rangler(t)=⟨cost,sint⟩; what is its velocity vector at t=π/2t=\pi/2t=π/2?

  1. 111
  2. ⟨−1,0⟩\langle -1,0\rangle⟨−1,0⟩ (correct answer)
  3. ⟨0,1⟩\langle 0,1\rangle⟨0,1⟩
  4. ⟨1,0⟩\langle 1,0\rangle⟨1,0⟩
  5. 000

Explanation: This problem involves analyzing the motion of a robot using parametric equations. To find the velocity vector, differentiate the position components: dx/dt=−sin⁡tdx/dt = -\sin tdx/dt=−sint and dy/dt=cos⁡tdy/dt = \cos tdy/dt=cost. At t=π/2t=\pi/2t=π/2, this gives ⟨−1,0⟩\langle -1, 0 \rangle⟨−1,0⟩. The velocity vector indicates the instantaneous direction and rate of change at that point. A tempting distractor might be ⟨0,1⟩\langle 0,1 \rangle⟨0,1⟩, which is the position at t=π/2t=\pi/2t=π/2, confusing position with velocity. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.

Question 5

A drone has position r(t)=⟨2t−1, t3⟩r(t)=\langle 2t-1,\, t^3\rangler(t)=⟨2t−1,t3⟩; what is its speed at t=1t=1t=1?

  1. ⟨2,3⟩\langle 2,3\rangle⟨2,3⟩
  2. 555
  3. 13\sqrt{13}13​ (correct answer)
  4. ⟨1,1⟩\langle 1,1\rangle⟨1,1⟩
  5. 5\sqrt{5}5​

Explanation: This problem involves analyzing the motion of a drone using parametric equations. To find the speed, compute the velocity by differentiating: dx/dt=2dx/dt = 2dx/dt=2 and dy/dt=3t2dy/dt = 3t^2dy/dt=3t2. At t=1, velocity is ⟨2,3⟩\langle 2, 3 \rangle⟨2,3⟩, and speed is 4+9=13\sqrt{4 + 9} = \sqrt{13}4+9​=13​. This magnitude represents how fast the drone is moving regardless of direction. A tempting distractor might be 5, which is the magnitude of ⟨1,2⟩\langle 1,2 \rangle⟨1,2⟩ if someone misdifferentiates the y-component. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.

Question 6

A particle has velocity v(t)=⟨sin⁡t, cos⁡t⟩v(t)=\langle \sin t,\, \cos t\ranglev(t)=⟨sint,cost⟩; what is its speed at t=πt=\pit=π?

  1. 000
  2. ⟨0,−1⟩\langle 0,-1\rangle⟨0,−1⟩
  3. −1-1−1
  4. 111 (correct answer)
  5. ⟨0,1⟩\langle 0,1\rangle⟨0,1⟩

Explanation: When velocity is given directly, speed is simply its magnitude. At t = π\piπ, we have v(π)=⟨sin⁡π,cos⁡π⟩=⟨0,−1⟩v(\pi) = \langle \sin \pi, \cos \pi \rangle = \langle 0, -1 \ranglev(π)=⟨sinπ,cosπ⟩=⟨0,−1⟩. The speed is ∣v(π)∣=02+(−1)2=1=1|v(\pi)| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1∣v(π)∣=02+(−1)2​=1​=1. Students might think the speed is 0 because the x-component is 0 (choice A) or report -1 because of the negative y-component (choice C). Speed is always non-negative; it's the magnitude of velocity regardless of direction.

Question 7

A particle moves with r(t)=\langle t^2-1,\,3t\rangle for 0≤t≤20\le t\le 20≤t≤2; what is its speed at t=1t=1t=1?

  1. 13\sqrt{13}13​ (correct answer)
  2. 10\sqrt{10}10​
  3. 131313
  4. ⟨2,3⟩\langle 2,3\rangle⟨2,3⟩
  5. ⟨0,3⟩\langle 0,3\rangle⟨0,3⟩

Explanation: This problem involves finding the speed of a particle moving along a parametric path. To find speed at t=1, we first compute the velocity vector v(t) = r'(t) = ⟨2t, 3⟩. At t=1, we get v(1) = ⟨2, 3⟩. Speed is the magnitude of velocity, so speed = |v(1)| = √(2² + 3²) = √(4 + 9) = √13. A common error would be to report the velocity vector ⟨2, 3⟩ itself (choice D) rather than its magnitude. Remember: velocity is a vector quantity showing direction and rate of change, while speed is the scalar magnitude of velocity.

Question 8

For r(t)=\langle 3t,\,\sin t\rangle, what is the particle’s speed at t=0t=0t=0?

  1. ⟨3,cos⁡0⟩\langle 3,\cos 0\rangle⟨3,cos0⟩
  2. 10\sqrt{10}10​ (correct answer)
  3. 333
  4. 32+(cos⁡0)2\sqrt{3^2+(\cos 0)^2}32+(cos0)2​
  5. ⟨0,0⟩\langle 0,0\rangle⟨0,0⟩

Explanation: This problem requires finding the speed of a particle with mixed polynomial and trigonometric parametric equations. First, we find the velocity vector by differentiating: v(t)=r′(t)=⟨3,cos⁡t⟩v(t) = r'(t) = \langle 3, \cos t \ranglev(t)=r′(t)=⟨3,cost⟩. At t=0t=0t=0, this becomes v(0)=⟨3,cos⁡(0)⟩=⟨3,1⟩v(0) = \langle 3, \cos(0) \rangle = \langle 3, 1 \ranglev(0)=⟨3,cos(0)⟩=⟨3,1⟩. The speed is the magnitude of this velocity vector: ∣v(0)∣=32+12=9+1=10|v(0)| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}∣v(0)∣=32+12​=9+1​=10​. Choice C gives only the x-component of velocity, ignoring the y-component contribution to speed. For parametric motion, speed requires considering both components of velocity in the magnitude calculation.

Question 9

A car’s position is r(t)=\langle t^3,\;4t\rangle (meters). What is its velocity vector at t=2t=2t=2?

  1. ⟨8,  8⟩\langle 8,\;8\rangle⟨8,8⟩
  2. ⟨12,  4⟩\langle 12,\;4\rangle⟨12,4⟩ (correct answer)
  3. ⟨6,  4⟩\langle 6,\;4\rangle⟨6,4⟩
  4. ⟨16,  8⟩\langle 16,\;8\rangle⟨16,8⟩
  5. ⟨8,  4⟩\langle 8,\;4\rangle⟨8,4⟩

Explanation: This problem requires computing the velocity vector of a car at a specific time using parametric equations. Given r(t) = ⟨t³, 4t⟩, we find velocity by differentiating each component: r'(t) = ⟨3t², 4⟩. At t=2, we substitute to get r'(2) = ⟨3(2)², 4⟩ = ⟨3(4), 4⟩ = ⟨12, 4⟩. The velocity vector tells us both the speed and direction of motion at that instant. Choice C (⟨6, 4⟩) might result from incorrectly computing the derivative as ⟨3t, 4⟩ instead of ⟨3t², 4⟩, a common error when differentiating powers. Always differentiate parametric components carefully: d/dt(t^n) = nt^(n-1).

Question 10

A particle moves with r(t)=\langle t^2,\,t^3\rangle; what is its speed at t=2t=2t=2?

  1. ⟨4,12⟩\langle 4,12\rangle⟨4,12⟩
  2. 160\sqrt{160}160​ (correct answer)
  3. 160160160
  4. 42+82\sqrt{4^2+8^2}42+82​
  5. ⟨4,8⟩\langle 4,8\rangle⟨4,8⟩

Explanation: This problem involves finding speed for power function parametric equations. We differentiate to get velocity: v(t)=r′(t)=⟨2t,3t2⟩v(t) = r'(t) = \langle 2t, 3t^2 \ranglev(t)=r′(t)=⟨2t,3t2⟩. At t=2t=2t=2, this gives v(2)=⟨2(2),3(2)2⟩=⟨4,12⟩v(2) = \langle 2(2), 3(2)^2 \rangle = \langle 4, 12 \ranglev(2)=⟨2(2),3(2)2⟩=⟨4,12⟩. The speed is the magnitude: ∣v(2)∣=42+122=16+144=160|v(2)| = \sqrt{4^2 + 12^2} = \sqrt{16 + 144} = \sqrt{160}∣v(2)∣=42+122​=16+144​=160​. Choice A gives the velocity vector components, while choice C incorrectly states 160 without taking the square root. For parametric motion, speed is always the square root of the sum of squared velocity components.

Question 11

A robot’s position is r(t)=\langle t,\;t^2\rangle (meters). Which gives its velocity at t=4t=4t=4?

  1. 0˘07fr(4)\u007fr(4)0˘07fr(4)
  2. ∣0˘07fr(4)∣\left|\u007fr(4)\right|​0˘07fr(4)​
  3. 0˘07fr′(4)\u007fr'(4)0˘07fr′(4) (correct answer)
  4. ∣0˘07fr′(4)∣\left|\u007fr'(4)\right|​0˘07fr′(4)​
  5. 0˘07fr(4)−0˘07fr(0)\u007fr(4)-\u007fr(0)0˘07fr(4)−0˘07fr(0)

Explanation: This problem asks for the velocity vector of a robot following a parabolic parametric path. Given r(t) = ⟨t, t²⟩, velocity is found by differentiating each component: r'(t) = ⟨1, 2t⟩. This represents the instantaneous rate of change of position. At any time t, including t=4, the velocity vector is r'(t), not its magnitude |r'(t)|. Choice D (|r'(4)|) would give the speed √(1 + 16) = √17, losing the directional information that makes velocity a vector. For parametric motion, velocity r'(t) is a vector quantity that includes both magnitude and direction information.

Question 12

A particle moves with r(t)=\langle \cos t,\;\sin t\rangle. What is its speed at t=π/2t=\pi/2t=π/2?

  1. 000
  2. 111 (correct answer)
  3. 2\sqrt{2}2​
  4. π/2\pi/2π/2
  5. 222

Explanation: This problem involves finding the speed of a particle moving on a unit circle described parametrically. Given r(t) = ⟨cos t, sin t⟩, we find velocity: r'(t) = ⟨-sin t, cos t⟩. At t=π/2, we have r'(π/2) = ⟨-sin(π/2), cos(π/2)⟩ = ⟨-1, 0⟩. The speed is |r'(π/2)| = √((-1)² + 0²) = √1 = 1. Choice C (√2) might result from evaluating at a different angle or confusing the speed formula. For circular motion with radius R, parametric equations r(t) = ⟨R cos t, R sin t⟩ always yield constant speed |r'(t)| = R.

Question 13

For r(t)=\langle t,\,t^2\rangle, what is the displacement vector from t=1t=1t=1 to t=3t=3t=3?

  1. ⟨2, 8⟩\langle 2,\,8\rangle⟨2,8⟩ (correct answer)
  2. ⟨4, 10⟩\langle 4,\,10\rangle⟨4,10⟩
  3. ⟨2, 10⟩\langle 2,\,10\rangle⟨2,10⟩
  4. ⟨3, 9⟩\langle 3,\,9\rangle⟨3,9⟩
  5. 68\sqrt{68}68​

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the displacement vector from t = 1 to t = 3, evaluate the position at both times: r(3) = <3, 9> and r(1) = <1, 1>. Subtract the initial position from the final position to get <3 - 1, 9 - 1> = <2, 8>. This vector represents the net change in position over the interval. A tempting distractor is E, √68, which is the magnitude of <2, 8> or the distance traveled, but the question asks for the displacement vector, not its length. A transferable motion-analysis strategy is to compute displacement as the difference in position vectors for any time interval in parametric motion.

Question 14

A drone’s position is r(t)=⟨5t, t2−4⟩r(t)=\langle 5t,\,t^2-4\rangler(t)=⟨5t,t2−4⟩; what is its velocity at t=3t=3t=3?

  1. ⟨5, 6⟩\langle 5,\,6\rangle⟨5,6⟩ (correct answer)
  2. ⟨15, 5⟩\langle 15,\,5\rangle⟨15,5⟩
  3. 61\sqrt{61}61​
  4. ⟨15, 9⟩\langle 15,\,9\rangle⟨15,9⟩
  5. ⟨5, 5⟩\langle 5,\,5\rangle⟨5,5⟩

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the velocity at t = 3, differentiate r(t)=⟨5t,t2−4⟩r(t) = \langle 5t, t^2 - 4 \rangler(t)=⟨5t,t2−4⟩ to obtain v(t)=⟨5,2t⟩v(t) = \langle 5, 2t \ranglev(t)=⟨5,2t⟩. Evaluate at t = 3: v(3)=⟨5,6⟩v(3) = \langle 5, 6 \ranglev(3)=⟨5,6⟩. This vector gives the instantaneous rate of change in position. A tempting distractor is C, 61\sqrt{61}61​, which is the magnitude of ⟨5,6⟩\langle 5, 6 \rangle⟨5,6⟩ or the speed, but the question specifically asks for the velocity vector. A transferable motion-analysis strategy is to differentiate each parametric component separately to build the velocity vector for motion analysis.

Question 15

A boat’s position is r(t)=⟨t, t2⟩r(t)=\langle t,\,t^2\rangler(t)=⟨t,t2⟩; what is its displacement from t=1t=1t=1 to t=3t=3t=3?

  1. 68\sqrt{68}68​
  2. ⟨8,2⟩\langle 8,2\rangle⟨8,2⟩
  3. ⟨2,8⟩\langle 2,8\rangle⟨2,8⟩ (correct answer)
  4. ⟨3,9⟩\langle 3,9\rangle⟨3,9⟩
  5. 101010

Explanation: This problem involves analyzing the motion of a boat using parametric equations. To find the displacement, subtract the position at t=1t=1t=1 from t=3t=3t=3: r(3)=⟨3,9⟩r(3) = \langle 3,9 \rangler(3)=⟨3,9⟩ and r(1)=⟨1,1⟩r(1)=\langle 1,1 \rangler(1)=⟨1,1⟩, so displacement is ⟨2,8⟩\langle 2,8 \rangle⟨2,8⟩. This vector represents the net change in position over the interval. It differs from the total distance traveled, which would require integrating speed. A tempting distractor might be 68\sqrt{68}68​, which is the magnitude of the displacement, confusing vector with scalar distance. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.

Question 16

A car’s position is r(t)=⟨t3, 2t2⟩r(t)=\langle t^3,\,2t^2\rangler(t)=⟨t3,2t2⟩; what is its velocity at t=1t=1t=1?

  1. ⟨1,2⟩\langle 1,2\rangle⟨1,2⟩
  2. 13\sqrt{13}13​
  3. ⟨3,4⟩\langle 3,4\rangle⟨3,4⟩ (correct answer)
  4. 555
  5. ⟨4,3⟩\langle 4,3\rangle⟨4,3⟩

Explanation: This problem involves analyzing the motion of a car using parametric equations. To find the velocity, differentiate: dx/dt=3t2dx/dt = 3t^2dx/dt=3t2 and dy/dt=4tdy/dt = 4tdy/dt=4t. At t=1, this is ⟨3,4⟩\langle 3,4 \rangle⟨3,4⟩. The components show the rates in x and y directions. A tempting distractor might be 5, which is the speed 9+16=5\sqrt{9+16} = 59+16​=5, confusing vector with scalar. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.

Question 17

A particle moves with r(t)=\langle e^t,\,\ln(t+1)\rangle; what is its speed at t=0t=0t=0?

  1. ⟨1,1⟩\langle 1,1\rangle⟨1,1⟩
  2. 2\sqrt{2}2​ (correct answer)
  3. 000
  4. 111
  5. ⟨1,0⟩\langle 1,0\rangle⟨1,0⟩

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the speed, differentiate: dx/dt = e^t and dy/dt = 1/(t+1). At t=0, velocity is <1,1>, and speed is √(1+1)=√2. This gives the instantaneous rate of change in position. A tempting distractor might be <1,1>, which is the velocity vector, not its magnitude. Always remember to integrate velocity to find position or differentiate position to find velocity and speed when analyzing parametric motion.

Question 18

A car’s motion is r(t)=⟨4cos⁡t, 4sin⁡t⟩r(t)=\langle 4\cos t,\,4\sin t\rangler(t)=⟨4cost,4sint⟩; what is its speed at t=π/2t=\pi/2t=π/2?

  1. 444 (correct answer)
  2. ⟨−4, 0⟩\langle -4,\,0\rangle⟨−4,0⟩
  3. ⟨0, 4⟩\langle 0,\,4\rangle⟨0,4⟩
  4. 000
  5. 161616

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the speed at t=π/2t = \pi/2t=π/2, differentiate r(t)=⟨4cos⁡t,4sin⁡t⟩r(t) = \langle 4 \cos t, 4 \sin t \rangler(t)=⟨4cost,4sint⟩ to get v(t)=⟨−4sin⁡t,4cos⁡t⟩v(t) = \langle -4 \sin t, 4 \cos t \ranglev(t)=⟨−4sint,4cost⟩. Evaluate at t=π/2t = \pi/2t=π/2: v(π/2)=⟨−4(1),4(0)⟩=⟨−4,0⟩v(\pi/2) = \langle -4(1), 4(0) \rangle = \langle -4, 0 \ranglev(π/2)=⟨−4(1),4(0)⟩=⟨−4,0⟩. The speed is (−4)2+02=16=4\sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4(−4)2+02​=16​=4. A tempting distractor is B, ⟨−4,0⟩\langle -4, 0 \rangle⟨−4,0⟩, which is the velocity vector, but the question asks for speed, the scalar magnitude. A transferable motion-analysis strategy is to recognize that for circular motion parametrized by sine and cosine, the speed is constant and equal to the radius times the angular speed.

Question 19

A drone’s position is r(t)=\langle 2\cos t,\;2\sin t\rangle (meters). Which expression gives its velocity at time ttt?

  1. 0˘07fr(t)\u007fr(t)0˘07fr(t)
  2. ∣0˘07fr′(t)∣\left|\u007fr'(t)\right|​0˘07fr′(t)​
  3. 0˘07fr′(t)\u007fr'(t)0˘07fr′(t) (correct answer)
  4. ∣0˘07fr(t)∣\left|\u007fr(t)\right|​0˘07fr(t)​
  5. 0˘07fr(t)−0˘07fr(0)\u007fr(t)-\u007fr(0)0˘07fr(t)−0˘07fr(0)

Explanation: This problem asks for the velocity expression of a drone following a circular parametric path. For parametric motion, velocity is the derivative of the position vector: if r(t) = ⟨x(t), y(t)⟩, then velocity v(t) = r'(t) = ⟨x'(t), y'(t)⟩. Taking derivatives component-wise: r'(t) = ⟨-2sin t, 2cos t⟩. This gives us the velocity vector at any time t. Choice B (|r'(t)|) would give speed (a scalar), not velocity (a vector), missing the directional information crucial for understanding the drone's motion. For parametric motion problems, distinguish between velocity r'(t) (vector) and speed |r'(t)| (scalar magnitude).

Question 20

A particle moves with r(t)=⟨t3, t2⟩r(t)=\langle t^3,\,t^2\rangler(t)=⟨t3,t2⟩; which is its speed at t=2t=2t=2?

  1. 160\sqrt{160}160​ (correct answer)
  2. 160160160
  3. ⟨12, 4⟩\langle 12,\,4\rangle⟨12,4⟩
  4. 20\sqrt{20}20​
  5. ⟨8, 4⟩\langle 8,\,4\rangle⟨8,4⟩

Explanation: This problem involves analyzing the motion of a particle using parametric equations. To find the speed at t=2t = 2t=2, differentiate r(t)=⟨t3,t2⟩r(t) = \langle t^3, t^2 \rangler(t)=⟨t3,t2⟩ to get v(t)=⟨3t2,2t⟩v(t) = \langle 3t^2, 2t \ranglev(t)=⟨3t2,2t⟩. Evaluate at t=2t = 2t=2: v(2)=⟨3(4),4⟩=⟨12,4⟩v(2) = \langle 3(4), 4 \rangle = \langle 12, 4 \ranglev(2)=⟨3(4),4⟩=⟨12,4⟩. The speed is the magnitude 122+42=144+16=160\sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160}122+42​=144+16​=160​. A tempting distractor is B, 160, which comes from adding the squares without taking the square root, mistaking it for speed. A transferable motion-analysis strategy is to derive velocity from position derivatives and then find its magnitude to determine speed at any point.