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 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.

This problem involves analyzing the motion of a particle using parametric equations. To find the velocity at t = 3, differentiate $r(t) = \langle 5t, t^2 - 4 \rangle$ to obtain $v(t) = \langle 5, 2t \rangle$. Evaluate at t = 3: $v(3) = \langle 5, 6 \rangle$. This vector gives the instantaneous rate of change in position. A tempting distractor is C, $\sqrt{61}$, which is the magnitude of $\langle 5, 6 \rangle$ 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.

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.

This problem involves analyzing the motion of a boat using parametric equations. To find the displacement, subtract the position at $t=1$ from $t=3$: $r(3) = \langle 3,9 \rangle$ and $r(1)=\langle 1,1 \rangle$, so displacement is $\langle 2,8 \rangle$. 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 $\sqrt{68}$, 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.

This problem involves analyzing the motion of a car using parametric equations. To find the velocity, differentiate: $ dx/dt = 3t^2 $ and $ dy/dt = 4t $. At t=1, this is $ \langle 3,4 \rangle $. The components show the rates in x and y directions. A tempting distractor might be 5, which is the speed $ \sqrt{9+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.

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.