Quiz 2 Solutions - schroeder(cs39963 Quiz 2 Rate of Change...

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schroeder (cs39963) – Quiz 2: Rate of Change and Kinematics – Balasubramanya – (1401310) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate lim x 1 (3 x 2 + 4 x + 1) . 1. limit doesn’t exist 2. limit = 7 3. limit = 6 4. limit = 8 correct 5. limit = 10 Explanation: By properties of limits, the limit can be taken term by term in f ( x ) = 3 x 2 + 4 x + 1 . Consequently, lim x 1 (3 x 2 + 4 x + 1) = 8 . 002 10.0 points Evaluate lim x 3 (2 x 2 - 5 x - 1) . 1. limit = 3 2. limit = 2 correct 3. limit = 4 4. limit doesn’t exist 5. limit = - 10 Explanation: Since a polynomial function is continuous, lim x 3 (2 x 2 - 5 x - 1) = (2 x 2 - 5 x - 1) vextendsingle vextendsingle vextendsingle x =3 = 2 . 003 (part 1 of 2) 10.0 points Treat the x axis as pointing in the direction of initial motion. While John is traveling along a straight interstate highway, he notices that the mile marker reads 259 km. John travels until he reaches the 146 km marker and then retraces his path to the 166 km marker. What is John’s resultant displacement from the 259 km marker? Correct answer: - 93 km. Explanation: Let : s 0 = 259 km and s f = 166 km . Δ s = s f - s 0 = 166 km - 259 km = - 93 km . 004 (part 2 of 2) 10.0 points How far has he traveled? Correct answer: 133 km. Explanation: Let : s 2 = 146 km . The distance traveled is given by d = | s 2 - s 0 | + | s f - s 2 | = | 146 km - 259 km | + | 166 km - 146 km | = 133 km .
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schroeder (cs39963) – Quiz 2: Rate of Change and Kinematics – Balasubramanya – (1401310) 2 005 10.0 points An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel 1530 m / s in seawater. How deep is the water directly below the vessel if the time delay of the echo to the ocean floor and back is 4 s? Correct answer: 3060 m. Explanation: Let : v = 1530 m / s and Δ t = 4 s . The sound takes 2 s to reach the ocean floor (and 2 s to return), so d = v t = (1530 m / s) (2 s) = 3060 m . 006 10.0 points Ann is driving down a street at 52 km / h. Suddenly a child runs into the street. If it takes Ann 0 . 767 s to react and apply the brakes, how far will she have moved before she begins to slow down? Correct answer: 11 . 0789 m. Explanation: Let : v = 52 km / h and t = 0 . 767 s . d = v t = (52 km / h) (0 . 767 s) 1 h 3600 s 1000 m 1 km = 11 . 0789 m . 007 10.0 points The velocity of the transverse waves produced by an earthquake is 4 . 14 km / s, while that of the longitudinal waves is 7 . 4934 km / s. A seis- mograph records the arrival of the transverse waves 49 . 8 s after that of the longitudinal waves. How far away was the earthquake? Correct answer: 460 . 705 km. Explanation: Let : v t = 4 . 14 km / s , v = 7 . 4934 km / s , and t = 49 . 8 s . If the distance to the earthquake is d , then the time lag is Δ t = d v t - d v = d ( v - v t ) v v t d = Δ t v v t v - v t = (49 . 8 s) (7 . 4934 km / s) (4 . 14 km / s) 4 . 14 km / s - 7 . 4934 km / s = 460 . 705 km . Note: The longitudinal wave travels faster than the transverse waves. In fact, the ratio of the longitudinal velocity to the transverse wave velocity should be about 3 .
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