AP Calculus BC 9.6 Solving Motion Problems - Exam Style Questions - MCQs - New Syllabus
Question
A particle has velocity \(\langle \sin(2t),\, 2+\cos(3t)\rangle\) for \( t\ge 0 \). What is the total distance traveled from \( t=0 \) to \( t=1 \)?
(A) \(1.655\)
(B) \(2.226\)
(C) \(2.432\)
(D) \(5.259\)
(B) \(2.226\)
(C) \(2.432\)
(D) \(5.259\)
▶️ Answer/Explanation
Correct answer: (B) \(2.226\)
Speed: \[ \|\mathbf v(t)\|=\sqrt{\sin^{2}(2t)+\big(2+\cos(3t)\big)^{2}}. \] Distance: \[ \int_{0}^{1}\!\sqrt{\sin^{2}(2t)+\big(2+\cos(3t)\big)^{2}}\,dt \;\approx\; 2.226. \]
Question
A particle moves in the \(xy\)-plane with parametric equations \(x(t)=t^2-t^3\) and \(y(t)=9-6t\). What is the speed of the particle at time \(t=2\)?
(A) \(\tfrac{3}{4}\)
(B) \(5\)
(C) \(10\)
(D) \(14\)
(B) \(5\)
(C) \(10\)
(D) \(14\)
▶️ Answer/Explanation
Correct answer: (C) \(10\)
\(x'(t)=2t-3\times t^2\Rightarrow x'(2)=4-12=-8.\) \(y'(t)=-6\Rightarrow y'(2)=-6.\)
Speed \(=\sqrt{\,[x'(2)]^2+[y'(2)]^2\,}=\sqrt{(-8)^2+(-6)^2}=\sqrt{64+36}=10.\)