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RCW2 solutions

# RCW2 solutions - Recitation ClassWork 2 solutions Physics...

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Recitation ClassWork 2 solutions Physics 203 1. Cutnell & Johnson 8 th Edition, Chapter 2 Problem 68. Also, explain how you can tell from the graph (without doing any calculations) when the bus has a zero acceleration and a non-zero acceleration. REASONING The average velocity for each segment is the slope of the line for that segment. SOLUTION Taking the direction of motion as positive, we have from the graph for segments A , B , and C , 1 1 10.0 km – 40.0 km –2.0 10 km/h 1.5 h – 0.0 h 20.0 km – 10.0 km 1.0 10 km/h 2.5 h – 1.5 h 40.0 km – 20.0 km 40 km/h 3.0 h – 2.5 h A B C v v v = = × = = × = = 2. Problem 65 REASONING AND SOLUTION The average acceleration for each segment is the slope of that segment. 2 A 2 B 2 C 40 m/s 0 m/s 1.9 m/s 21 s 0 s 40 m/s 0 m/s 0 m/s 48 s 21 s 80 m/s 0 m/s 3.3 m/s 60 s 48 s a a a = = − 4 = = − 4 = = 3. Problem 6 REASONING The distance traveled by the Space Shuttle is equal to its speed multiplied by the time. The number of football fields is equal to this distance divided by the length L of one football field. SOLUTION The number of football fields is ( )( ) 3 3 7.6 10 m /s 110 10 s Number = 9.1 91.4 m x vt L L × × = = = 4. Problem 7 REASONING In a race against el-Guerrouj, Bannister would run a distance given by his average speed times the time duration of the race (see Equation 2.1). The time duration of the race would be el-Guerrouj’s winning time of 3:43.13 (223.13 s). The difference between Bannister’s distance and the length of the race is el-Guerrouj’s winning margin. SOLUTION From the table of conversion factors on the page facing the front cover, we find that one mile corresponds to 1609 m. According to Equation 2.1, Bannister’s average speed is Distance 1609 m Average speed = Elapsed time 239.4 s = Had he run against el-Guerrouj at this average speed for the 223.13-s duration of the race, he would have

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