FINAL.pdf - Version 004 young(toy68 FINAL villafuerte...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 004 – young (toy68) – FINAL – villafuerte altu – (53780) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points After t seconds the displacement, s ( t ), of a particle moving rightwards along the t -axis is given (in feet) by s ( t ) = 1 4 t 3 . Find the average velocity of the particle over the time interval [1 , 1 + h ]. 1. avg. vel. = 3 4 + 3 4 h + 1 4 h 2 ft/sec correct 2. avg. vel. = 1 4 + 3 4 h + 1 4 h 2 ft/sec 3. avg. vel. = 1 4 + 3 4 h + 1 4 h 3 ft/sec 4. avg. vel. = 1 4 + 1 4 h + 3 4 h 3 ft/sec 5. avg. vel. = 3 4 + 1 4 h + 3 4 h 3 ft/sec 6. avg. vel. = 3 4 + 1 4 h + 1 4 h 2 ft/sec Explanation: The average velocity of the particle over the time interval [1 , 1 + h ] is given by dist travelled time taken = s (1 + h ) s (1) (1 + h ) 1 . Now s (1 + h ) = 1 4 (1 + h ) 3 = 1 4 (1 + 3 h + 3 h 2 + h 3 ) , while s (1) = 1 4 . Thus s (1 + h ) s (1) (1 + h ) 1 = 1 4 braceleftBig 3 h + 3 h 2 + h 3 h bracerightBig . Consequently, over the interval [1 , 1 + h ] the particle has average vel. = 3 4 + 3 4 h + 1 4 h 2 ft/sec . 002 10.0points Shown are the graphs of distance versus time for three runners A, B, and C who run a 100 -m race and finish in tie. Which of the follow- ing statements about the runners is false ? 2 4 6 8 A B C 100 t [seconds] s [meters] 1. At t = 7, runner B has a lower velocity than runner A. correct 2. Runner B runs as a constant speed throughout the race. 3. At t = 1, runner A has a higher velocity than B. 4. Runner A gradually slows down through- out the race. 5. Runner C gradually speeds up throughout the race. Explanation: Statement (A) is false. At t = 7, the graph associated to runner B is steeper than the graph associated to runner A. Thus runner B has a higher velocity at t = 7 than does runner A.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Version 004 – young (toy68) – FINAL – villafuerte altu – (53780) 2 keywords: velocity, position function 003 10.0points Determine if lim x 0 2 x 2 + x x exists, and if it does, find its value. 1. limit = 2 2. limit = 1 3. limit does not exist 4. limit = 1 2 correct 5. limit = 1 2 6. limit = 2 7. limit = 1 Explanation: After rationalization we see that 2 x 2 + x = (2 x ) (2 + x ) 2 x + 2 + x = 2 x 2 x + 2 + x . Thus 2 x 2 + x x = 2 2 x + 2 + x for all x negationslash = 0. But lim x 0 ( 2 x + 2 + x ) = 2 2 . Consequently, by Properties of Limits, limit = 2 2 2 = 1 2 . 004 10.0points Find all values of x at which the function f defined by f ( x ) = x 2 2 x 15 x 2 9 x + 20 , x negationslash = 5, 9 , x = 5, is continuous, expressing your answer in in- terval notation. 1. ( −∞ , 5) ( 5 , 4) ( 4 , ) 2. ( −∞ , 4) ( 4 , ) 3. ( −∞ , 4) (4 , 5) (5 , ) correct 4. ( −∞ , 5) ( 5 , 4) (4 , ) 5. ( −∞ , 4) ( 4 , 5) (5 , ) Explanation: After factorization f becomes f ( x ) = ( x 5)( x + 3) ( x 4)( x 5) = x + 3 x 4 for values of x negationslash = 5. But lim x 5 f ( x ) = lim x 5 x + 3 x 4 = 8 negationslash = f (5) , so f is defined but not continuous at x = 5.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern