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Test #1-solutions

# Test #1-solutions - Version 030/AABDC Test#1...

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Version 030/AABDC – Test #1 – Antoniewicz – (57380) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Vector A has components A x = - 2 , A y = 6 . 4 , A z = 4 . 7 , while vector B has components B x = 6 . 7 , B y = - 7 . 6 , B z = 6 . 9 . What is the angle θ AB between these vec- tors? (Answer between 0 and 180 .) 1. 143.745 2. 115.127 3. 116.363 4. 103.497 5. 104.967 6. 122.173 7. 111.897 8. 107.157 9. 109.423 10. 138.004 Correct answer: 107 . 157 . Explanation: A = A 2 x + A 2 y + A 2 z = ( - 2) 2 + 6 . 4 2 + 4 . 7 2 = 8 . 18841 and B = B 2 x + B 2 y + B 2 z = 6 . 7 2 + ( - 7 . 6) 2 + 6 . 9 2 = 12 . 2581 , so using A · B = A x B x + A y B y + A z B z = ( - 2) 6 . 7 + 6 . 4 ( - 7 . 6) + 4 . 7 (6 . 9) = - 29 . 61 , cos θ AB = A · B A B = - 29 . 61 (8 . 18841) (12 . 2581) = - 0 . 294997 θ AB = arccos( - 0 . 294997) = 107 . 157 . Two vectors can define a plane. When these two vectors are plotted in this plane, we have A B 107 . 157 002 10.0 points A car traveling at a constant speed of 32 . 6 m / s passes a trooper hidden behind a billboard. One second later the trooper starts the car with a constant acceleration of 3 . 24 m / s 2 . How long after the trooper starts the chase does he overtake the speeding car? 1. 25.6838 2. 26.4721 3. 25.1425 4. 23.018 5. 21.0782 6. 28.6634 7. 23.5233 8. 39.3158 9. 31.9697 10. 25.9804 Correct answer: 21 . 0782 s. Explanation: Let : v = 32 . 6 m / s , a = 3 . 24 m / s 2 , a s = 0 , v 0 p = 0 , and Δ t = 1 .

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Version 030/AABDC – Test #1 – Antoniewicz – (57380) 2 It is convenient to choose the origin at the position of the billboard and take t = 0 as the time the trooper begins moving. The speeder travels with constant speed, so if the trooper takes t s to overtake the car, the car has traveled ( t + 1) s. x = v i t + 1 2 a t 2 . The trooper overtakes the car when x T = x C 1 2 a t 2 = v ( t + 1) a t 2 - 2 v t - 2 v = 0 , which has a positive solution of t = 2 v + 4 v 2 + 8 a v 2 a . Since 4 v 2 + 8 a v = 4 (32 . 6 m / s) 2 + 8 (3 . 24 m / s 2 ) (32 . 6 m / s) = 5096 . 03 m 2 / s 2 , then t = 2 (32 . 6 m / s) + 5096 . 03 m 2 / s 2 2 (3 . 24 m / s 2 ) = 21 . 0782 s . 003 (part 1 of 3) 4.0 points A motorboat heads due east at 11 . 8 m/s across a river that flows toward the south at a speed of 4 . 3 m/s.
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Test #1-solutions - Version 030/AABDC Test#1...

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