M408L_HW04

# M408L_HW04 - hyun(hh7953 – HW04 – gogolev –(57440 1...

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Unformatted text preview: hyun (hh7953) – HW04 – gogolev – (57440) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points A Calculus student begins walking in a straight line from the RLM building towards the PCL Library. After t minutes his velocity v = v ( t ) is given (in multiples of 10 yards per minute) by the function whose graph is 2 4 6 8 10- 2 2 4 6- 2 t v How far is the student from the RLM build- ing at time t = 4? 1. dist = 160 yards 2. dist = 130 yards 3. dist = 140 yards correct 4. dist = 150 yards 5. dist = 180 yards Explanation: The student is walking towards the PCL Library whenever v ( t ) > 0 and towards RLM whenever v ( t ) < 0. The distance he has walked is given by 10 times the area between the graph of v and the t axis since 1 unit is equivalent to 10 yards. Now at t = 4 this area is 14 units, so the student has walked a distance of 140 yards. 002 (part 2 of 4) 10.0 points How far is the student from the RLM build- ing at time t = 6? 1. dist = 240 yards 2. dist = 230 yards correct 3. dist = 250 yards 4. dist = 270 yards 5. dist = 220 yards Explanation: The distance the student has walked is given by the area between the graph of v and the t axis. Now at t = 6 this area is 23 units, so the student has walked a distance of 230 yards. 003 (part 3 of 4) 10.0 points What is the total distance walked by the student from time t = 0 to t = 10? 1. dist = 290 yards correct 2. dist = 300 yards 3. dist = 280 yards 4. dist = 270 yards 5. dist = 320 yards Explanation: The distance the student has walked is given by the area between the graph of v and the t axis. Now at t = 10 this area is 29 units (area is always positive remember), so the student has walked a distance of 290 yards. 004 (part 4 of 4) 10.0 points How far is the student from the RLM build- ing when he turns back? 1. dist = 250 yards 2. dist = 280 yards hyun (hh7953) – HW04 – gogolev – (57440) 2 3. dist = 230 yards 4. dist = 260 yards correct 5. dist = 300 yards Explanation: The student will turn back when his veloc- ity changes from postive to negative, i.e. , at t = 7. At this time he is 260 yards from the RLM building. keywords: distance, time, graph, velocity, area 005 10.0 points The graph of f is shown in the figure 2 4 6 8 2 4 6 If F is an anti-derivative of f and integraldisplay 8 3 f ( x ) dx = 23 , find the value of F (8)- F (2). 1. F (8)- F (2) = 29 2. F (8)- F (2) = 28 correct 3. F (8)- F (2) = 32 4. F (8)- F (2) = 31 5. F (8)- F (2) = 30 Explanation: We already know that the area under the graph on the interval 3 ≤ x ≤ 8 is equal to 23, alternatively, by the Fundamental Theorem of Calculus we can say that F (8)- F (3) = 23 ....
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## This note was uploaded on 01/19/2010 for the course M 57440 taught by Professor Gogolev during the Fall '09 term at University of Texas.

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M408L_HW04 - hyun(hh7953 – HW04 – gogolev –(57440 1...

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