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# ch02 - Chapter 2 1 The speed(assumed constant is v =(90...

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21 Chapter 2 1. The speed (assumed constant) is v = (90 km/h)(1000 m/km) (3600 s/h) = 25 m/s. Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s) 13 m. 2. (a) Using the fact that time = distance/velocity while the velocity is constant, we find avg 73.2 m 73.2 m 2.85 m 1.22 m/s 73.2 m 73.2 m 1.71 m/s. v (b) Using the fact that distance = vt while the velocity v is constant, we find v avg m / s)(60 s) m / s)(60 s) s m / s. ( . ( . . 122 305 120 214 (c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer. 3. Since the trip consists of two parts, let the displacements during first and second parts of the motion be x 1 and x 2, and the corresponding time intervals be t 1 and t 2 , respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is x = x 1 + x 2 , and the total time for the trip is t = t 1 + t 2 . Using the definition of average velocity given in Eq. 2-2, we have 1 2 avg 1 2 . x x x v t t t       To find the average speed, we note that during a time t if the velocity remains a positive constant, then the speed is equal to the magnitude of velocity, and the distance is equal to the magnitude of displacement, with | | d x v t     .

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CHAPTER 2 22 (a) During the first part of the motion, the displacement is x 1 = 40 km and the time interval is t 1 40 133 ( . km) (30 km / h) h. Similarly, during the second part the displacement is x 2 = 40 km and the time interval is t 2 40 0 67 ( . km) (60 km / h) h. The total displacement is x = x 1 + x 2 = 40 km + 40 km = 80 km, and the total time elapsed is t = t 1 + t 2 = 2.00 h. Consequently, the average velocity is avg (80 km) 40 km/h. (2.0 h) x v t (b) In this case, the average speed is the same as the magnitude of the average velocity: avg 40 km/h. s (c) The graph of the entire trip is shown below; it consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to ( t 1 , x 1 ) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting ( t 1 , x 1 ) to ( t, x ) = (2.00 h, 80 km). 4. Average speed, as opposed to average velocity, relates to the total distance, as
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ch02 - Chapter 2 1 The speed(assumed constant is v =(90...

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