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.

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.

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.

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

This problem involves analyzing the motion of a drone using parametric equations. To find the speed, compute the velocity by differentiating: $dx/dt = 2$ and $dy/dt = 3t^2$. At t=1, velocity is $\langle 2, 3 \rangle$, and speed is $\sqrt{4 + 9} = \sqrt{13}$. This magnitude represents how fast the drone is moving regardless of direction. A tempting distractor might be 5, which is the magnitude of $\langle 1,2 \rangle$ 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.

When velocity is given directly, speed is simply its magnitude. At t = $\pi$, we have $v(\pi) = \langle \sin \pi, \cos \pi \rangle = \langle 0, -1 \rangle$. The speed is $|v(\pi)| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1$. 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.

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.

This problem requires finding the speed of a particle with mixed polynomial and trigonometric parametric equations. First, we find the velocity vector by differentiating: $v(t) = r'(t) = \langle 3, \cos t \rangle$. At $t=0$, this becomes $v(0) = \langle 3, \cos(0) \rangle = \langle 3, 1 \rangle$. The speed is the magnitude of this velocity vector: $|v(0)| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$. 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.

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).

This problem involves finding speed for power function parametric equations. We differentiate to get velocity: $v(t) = r'(t) = \langle 2t, 3t^2 \rangle$. At $t=2$, this gives $v(2) = \langle 2(2), 3(2)^2 \rangle = \langle 4, 12 \rangle$. The speed is the magnitude: $|v(2)| = \sqrt{4^2 + 12^2} = \sqrt{16 + 144} = \sqrt{160}$. 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.