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02_InstSolManual_PDF - MOTION ALONG A STRAIGHT LINE 2 8 A 9...

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2-1 M OTION ALONG A S TRAIGHT L INE 2 Answers to Multiple-Choice Problems 1. C, D 2. C 3. C, D 4. A, B 5. D 6. C 7. A, D 8. A 9. B 10. C 11. A 12. A, C, D 13. B, D 14. D 15. C Solutions to Problems 2.1. Set Up: Let the direction be to the right in the figure. Solve: (a) The lengths of the segments determine the distance of each point from O : and (b) The displacement is the sign of indicates its direction. The distance is always positive. (i) A to B : Distance is 50 cm. (ii) B to C : Distance is 30 cm. (iii) C to D : Distance is 20 cm. (iv) A to D : Reflect: When the motion is always in the same direction during the interval the magnitude of the displacement and the distance traveled are the same. In (iv) the ant travels to the right and then to the left and the magnitude of the dis- placement is less than the distance traveled. 2.2. Set Up: From the graph the position at each time t is: and Solve: (a) The displacement is (i) (ii) (iii) (iv) (b) (i) (ii) (iii) zero (stays at 2.3. Set Up: Let the direction be to the right. Solve: Average velocity is Reflect: The average speed is greater than the magnitude of the average velocity. 2.4. Set Up: The distance traveled is 2650 miles and the displacement is 1000 miles. Solve: (a) Average speed 5 distance time 5 2650 miles 2016 h 5 1.3 mi / h time 5 12 weeks 1 168 h / week 2 5 2016 h Average speed 5 distance time 5 4.0 m 1 1.0 m 1 1.0 m 3.0 s 5 2.0 m / s v av- x 5 D x D t 5 x C 2 x A D t 5 1 6.0 m 2 2.0 m 3.0 s 5 1.3 m / s. x C 5 6.0 m. x B 5 7.0 m, x A 5 2.0 m, 1 x x 5 6.0 m) 1.0 m 1 1.0 m 5 2.0 m; 3.0 m 1 1.0 m 5 4.0 m; D x 5 x 4 2 x 2 5 0. x 3 2 x 2 5 2 1.0 m; D x 5 D x 5 x 10 2 x 3 5 1 7.0 m; D x 5 x 10 2 x 1 5 1 5.0 m; D x . x 10 5 6.0 m. x 8 5 6.0 m, x 4 5 0, x 3 5 2 1.0 m, x 2 5 0, x 1 5 1.0 m, x t Distance 5 2 1 AB 2 5 100 cm. D x 5 x D 2 x A 5 0. D x 5 x D 2 x C 5 2 5 cm 2 15 cm 5 2 20 cm. D x 5 x C 2 x B 5 1 15 cm 2 45 cm 5 2 30 cm. D x 5 x B 2 x A 5 1 45 cm 2 1 2 5 cm 2 5 1 50 cm. D x D x ; x D 5 2 5 cm. x C 5 1 15 cm, x B 5 1 45 cm, x A 5 2 5 cm, 1 x
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(b) (c) The hiking time is 2.5. Set Up: Solve: (a) A to B : B to C : A to C : (b) The velocity is always in the same direction so the distance traveled is equal to the displacement in each case, and the average speed is the same as the magnitude of the average velocity. Reflect: The average speed is different for different time intervals. 2.6. Set Up: Solve: (a) A to B : B to C : A to C : (b) For A to B and for B to C the distance traveled equals the magnitude of the displacement and the average speed equals the magnitude of the average velocity. For A to C the displacement is zero. Thus, the average velocity is zero but the distance traveled is not zero so the average speed is not zero. For the motion A to B and for B to C the veloc- ity is always in the same direction but during A to C the motion changes direction. 2.7. Set Up: The positions at time t are: Solve: (a) The distance is (b) (i) (ii) (iii) (iv) (v) Reflect: In successive 1 s time intervals the boulder travels greater distances and the average velocity for the inter- vals increases from one interval to the next.
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