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Unformatted text preview: Version 076 – Quiz 1 – Coker – (56625) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This is the first and easiest quiz, covering only the material in 3 chapters, Chs. 2, 3 and 4. 001 10.0 points A particle moves in a straight line with veloc ity v ( t ) = 3 t 2 − 6 t . If it initially starts moving from 0 (where x = 0), then its position x ( t ) is equal to 1. x ( t ) = t 3 − 3 t 2 correct 2. x ( t ) = 6 t − 6 3. x ( t ) = 6 t 4. x ( t ) = t 3 − 3 t 2 + 1 5. x ( t ) = 6 t − 3 t 2 Explanation: v ( t ) = 3 t 2 − 6 t d s ( t ) dt = 3 t 2 − 6 t integraldisplay d s ( t ) dt dt = integraldisplay ( 3 t 2 − 6 t ) dt s ( t ) = t 3 − 3 t 2 + C x (0) = 0, so 0 = C and s ( t ) = t 3 − 3 t 2 . 002 10.0 points Consider the vector −→ M = (+ b, + c ) = + b ˆ ı + c ˆ . − d − b − a − c c a b d − d − b − a − c c a b d D H B G C A E F x y Which vector is a valid representation of −→ M ? 1. E only. 2. C and G , but no others. 3. E and H , but no others. 4. F only. 5. G only. 6. F , C , D , and G , but no others. 7. E , A , H , and B , but no others. 8. A and B , but no others. correct 9. F and D , but no others. 10. B only. Explanation: Vectors A and B , but no others, have an xdisplacement of + b and ydisplacement of + c . The vector −→ M plotted with its tail at the origin is Version 076 – Quiz 1 – Coker – (56625) 2 − d − b − a − c c a b d − d − b − a − c c a b d M x y Its length and direction are the same as vectors A and B , but no others. 003 10.0 points Vector vector A has a magnitude of 8 and points in the positive xdirection. Vector vector B has a magnitude of 21 and makes an angle of 27 ◦ with the positive xaxis. What is the magnitude of vector A − vector B ? 1. 16.0801 2. 17.6379 3. 9.86934 4. 17.0749 5. 9.68872 6. 9.45618 7. 13.5208 8. 14.3395 9. 11.4659 10. 13.8608 Correct answer: 14 . 3395. Explanation: A A B − B A − B A − B θ θ The component of vector vector B along the xaxis is B x = B cos θ = 21 cos 27 ◦ = 18 . 7111 and the component along the yaxis is B y = B sin θ = 21 sin 27 ◦ = 9 . 5338 . Vector vector A points in the xdirection, so it has no component along yaxis, so ( vector A − vector B ) x = A x − B x = 8 − 18 . 7111 = − 10 . 7111 , ( vector A − vector B ) y = A y − B y = 0 − 9 . 5338 = − 9 . 5338 , and bardbl vector A − vector B bardbl = radicalBig ( − 10 . 7111) 2 + ( − 9 . 5338) 2 = 14 . 3395 . 004 10.0 points A target lies flat on the ground 9 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m / s 2 . Air resistance is negligible....
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This note was uploaded on 11/17/2010 for the course PHY 56630 taught by Professor Coker during the Spring '10 term at University of Texas at Austin.
 Spring '10
 COKER

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