AP Calculus BC : 9.6 Solving Motion Problems Using Parametric and Vector- Valued Functions- Exam Style questions with Answer- MCQ

Question

For time t> 0 , the position of an object moving in the xy-plane is given by the parametric equations x(t) =tcos(t/2) and \(y(t)=\sqrt{t^{2}+2t}\). What is the speed of the object at time t= 1 ?
A 1.155
B 1.319
C 1.339
D 1.810

Answer/Explanation

 

Question

A particle moves on a plane curve so that at any time t > 0 its x-coordinate is\( t^3- t \) and its y-coordinate is (3 ). The acceleration vector of the particle at t =1 is

(A) (0,1 )                                                            B) (2,3 )                                                 (C) (2,6)                                               (D) (6,12)                                       (E) ( 6, 24) 

Answer/Explanation

Ans:E

 

Question

The position of a particle moving in the xy-plane is given by the parametric equations\( x(t)=\frac{6t}{t+1}and y(t)=\frac{-8}{t^{2}+4}\).  What is the slope of the line tangent to the path of the particle at the point where t=2?
(A) \(\frac{1}{2}\)
(B)\(\frac{2}{3}\)
(C)\(\frac{3}{4}\)
(D) \(\frac{4}{3}\)

Answer/Explanation

 

Question

What is the speed of a particle whose motion is defined by \(y=-2t^{2}+5t\) and x = 6t when t = 1?
(A) 9
(B)\(\sqrt{37}\)
(C) \(\sqrt{57}\)
(D) \(\sqrt{117}\)

Answer/Explanation

Ans:(B)

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