Ch_2Word - Chapter 2 CHAPTER 2 Describing Motion Kinematics...

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Chapter  2 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 Page 1
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Chapter  2 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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