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

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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 – fnd all choices beFore answering. 001 10.0 points Vector ± A has components A x = - 2 ,A y =6 . 4 z =4 . 7 , while vector ± B has components B x . 7 ,B y = - 7 . 6 z . 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 . 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 defne a plane. When these two vectors are plotted in this plane, we have A B 107 . 157 002 10.0 points AcartravelingataconstantspeedoF32 . 6m / 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 . / s , a =3 . 24 m / s 2 , a s =0 , v 0 p , 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 =0a s the time the trooper begins moving. The speeder travels with constant speed, so if the trooper takes t stoovertakethecar ,thecar has traveled ( t +1)s. x = v i t + 1 2 at 2 . The trooper overtakes the car when x T = x C 1 2 2 = v ( t +1) 2 - 2 vt - 2 v =0 , which has a positive solution of t = 2 v + 4 v 2 +8 av 2 a . Since 4 v 2 =4(32 . 6m / s) 2 +8(3 . 24 m / s 2 )(32 . / s) =5096 . 03 m 2 / s 2 , then t = 2(32 . / s) + ± 5096 . 03 m 2 / s 2 2(3 . 24 m / s 2 ) = 21 . 0782 s . 003 (part 1 of 3) 4.0 points Am o t o r b o a th e a d sd u ee a s ta t1 1 . 8m / s across a river that Fows toward the south at a speed of 4 .
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This note was uploaded on 03/08/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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

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