Quest HW 2

# Quest HW 2 - roberts(eer474 Quest HW 2 seckin(56425 This...

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roberts (eer474) – Quest HW 2 – seckin – (56425) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A tank holds 100 gallons of water, which drains from the bottom of the tank in 10 minutes. The values in the table t (min) 0 2 4 6 8 10 V (gal) 100 72 41 12 4 0 show the volume, V ( t ), of water remaining in the tank (in gallons) after t minutes. If P is the point (2 , V (2)) on the graph of V as a function of time t , find the slope of the secant line PQ when Q = (6 , V (6)). 1. slope = 13 2. slope = 15 correct 3. slope = 14 4. slope = 15 5. slope = 17 6. slope = 14 7. slope = 17 8. slope = 13 Explanation: When P = (2 , V (2)) , Q = (6 , V (6)) the slope of the secant line PQ is given by rise run = V (6) V (2) 6 2 . From the table of values, therefore, we see that slope = 12 72 6 2 = 15 . 002 (part 1 of 5) 10.0 points At which point on the graph P Q R S T U (i) is the slope greatest ( i.e. , most positive)? 1. T 2. R 3. Q 4. S 5. P correct 6. U Explanation: By inspection the point is P . 003 (part 2 of 5) 10.0 points (ii) is the slope smallest ( i.e. , most negative)? 1. R correct 2. S 3. Q 4. T

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roberts (eer474) – Quest HW 2 – seckin – (56425) 2 5. P 6. U Explanation: By inspection the point is R. 004 (part 3 of 5) 10.0 points (iii) does the slope change from positive to negative? 1. R 2. S 3. Q correct 4. T 5. P 6. U Explanation: By inspection the point is Q. 005 (part 4 of 5) 10.0 points (iv) does the slope change from negative to positive? 1. S 2. T correct 3. R 4. U 5. P 6. Q Explanation: By inspection the point is T . 006 (part 5 of 5) 10.0 points (v) 1. U 2. P 3. T 4. R 5. S correct 6. Q Explanation: By inspection the point is S . keywords: slope, graph, change of slope 007 (part 1 of 2) 10.0 points After t seconds the displacement, s ( t ), of a particle moving rightwards along the x -axis is given (in feet) by s ( t ) = 7 t 2 2 t + 3 . (i) Determine the average velocity of the particle over the time interval [1 , 2]. 1. average vel. = 16 ft/sec 2. average vel. = 19 ft/sec correct 3. average vel. = 17 ft/sec 4. average vel. = 18 ft/sec 5. average vel. = 20 ft/sec Explanation: The average velocity over a time interval [ a, b ] is given by dist travelled time taken = s ( b ) s ( a ) b a . For the time interval [1 , 2], therefore, ave. vel. = s (2) s (1) 2 1 ft/sec . Now s (2) = 7 × 4 2 × 2 + 3 = 27 feet ,
roberts (eer474) – Quest HW 2 – seckin – (56425) 3 while s (1) = 7 2 + 3 = 8 feet . Consequently, average vel. = 27 8 = 19 ft/sec . 008 (part 2 of 2) 10.0 points (ii) By determining the average velocity successively over the time intervals [1 , 1 . 1] , [1 , 1 . 01] , [1 , 1 . 001] , [1 , 1 . 0001] , estimate the instantaneous velocity, v (1), of the particle at time t = 1. 1. v (1) = 15 ft/sec 2. v (1) = 13 ft/sec 3. v (1) = 12 ft/sec correct 4. v (1) = 11 ft/sec 5. v (1) = 14 ft/sec Explanation: After calculation, s (1 . 1) s (1) 0 . 1 = 12 . 7 s (1 . 01) s (1) 0 . 01 = 12 . 07 s (1 . 001) s (1) 0 . 001 = 12 . 007 s (1 . 0001) s (1) 0 . 0001 = 12 . 0007 .

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