CH_2 - Chapter 2 CHAPTER 2 - Describing Motion: Kinematics...

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Chapter 2 Page 1 CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d / t ; 15 km/h = (75 km)/ t , which gives t = 5.0 h . 2. We find the average speed from average speed = d / t = (280 km)/(3.2 h) = 88 km/h . 3. We find the distance traveled from average speed = d / t ; (110 km/h)/(3600 s/h) = d /(2.0 s), which gives d = 6.1 × 10 –2 km = 61 m . 4. We find the average velocity from æ = ( x 2 x 1 )/( t 2 t 1 ) = (– 4.2 cm – 3.4 cm)/(6.1 s – 3.0 s) = – 2.5 cm/s (toward – x ) . 5. We find the average velocity from æ x 2 x 1 )/( t 2 t 1 ) = (8.5 cm – 3.4 cm)/[4.5 s – (– 2.0 s)] = 0.78 cm/s (toward + x ) . Because we do not know the total distance traveled, we cannot calculate the average speed. 6. ( a ) We find the elapsed time before the speed change from speed = d 1 / t 1 ; 65 mi/h = (130 mi)/ t 1 , which gives t 1 = 2.0 h. Thus the time at the lower speed is t 2 = T t 1 = 3.33 h – 2.0 h = 1.33 h. We find the distance traveled at the lower speed from speed = d 2 / t 2 ; 55 mi/h = d 2 /(1.33 h), which gives d 2 = 73 mi. The total distance traveled is D d 1 + d 2 = 130 mi + 73 mi = 203 mi . ( b ) We find the average speed from average speed = d / t = (203 mi)/(3.33 h) = 61 mi/h . Note that the average speed is not ! (65 mi/h + 55 mi/h). The two speeds were not maintained for equal times. 7. Because there is no elapsed time when the light arrives, the sound travels one mile in 5 seconds. We find the speed of sound from speed = d / t = (1 mi)(1610 m/1 mi)/(5 s) 300 m/s . 8. ( a ) We find the average speed from average speed = d / t = 8(0.25 mi)(1.61 × 10 3 m/mi)/(12.5 min)(60 s/min) = 4.29 m/s . ( b ) Because the person finishes at the starting point, there is no displacement; thus the average velocity is æ x / t 0 . 9. ( a ) We find the average speed from average speed = d / t = (160 m + 80 m)/(17.0 s + 6.8 s) = 10.1 m/s . ( b ) The displacement away from the trainer is 160 m – 80 m = 80 m; thus the average velocity is æ x / t = (80 m)/(17.0 s + 6.8 s) = + 3.4 m/s, away from trainer . 10. Because the two locomotives are traveling with equal speeds in opposite directions, each locomotive will travel half the distance, 4.25 km. We find the elapsed time from speed 1 d 1 / t ; (95 km/h)/(60 min/h) = (4.25 km)/ t , which gives t = 2.7 min .
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Chapter 2 Page 2 11. ( a ) We find the instantaneous velocity from the slope of the straight line from t = 0 to t = 10.0 s: v 10 = x / t = (2.8 m – 0)/(10.0 s – 0) = 0.28 m/s . ( b ) We find the instantaneous velocity from the slope of a tangent to the line at t = 30.0 s: v 30 x / t = (22 m – 10 m)/(35 s – 25 s) = 1.2 m/s . ( c ) The velocity is constant for the first 17 s (a straight line), so the velocity is the same as the velocity at t = 10 s: æ 0 5 0.28 m/s . ( d ) For the average velocity we have æ 25 30 x / t = (16 m – 8 m)/(30.0 s – 25.0 s) = 1.6 m/s . ( e ) For the average velocity we have æ 40 50 x / t = (10 m – 20 m)/(50.0 s – 40.0 s) = – 1.0 m/s . 12. ( a ) Constant velocity is indicated by a straight line, which occurs from t = 0 to 17 s . ( b ) The maximum velocity is when the slope is greatest: t = 28 s . ( c ) Zero velocity is indicated by a zero slope. The tangent is horizontal at t = 38 s .
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This note was uploaded on 04/08/2008 for the course PHY 53 taught by Professor Mueller during the Fall '08 term at Duke.

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CH_2 - Chapter 2 CHAPTER 2 - Describing Motion: Kinematics...

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