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p02_003

Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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3. We use Eq. 2-2 and Eq. 2-3. During a time t c when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with ∆ x = v t c . (a) During the first part of the motion, the displacement is ∆ x 1 = 40 km and the time interval is t 1 = (40 km) (30 km / h) = 1 . 33 h . During the second part the displacement is ∆ x 2 = 40 km and the time interval is t 2 = (40 km) (60 km / h) = 0 . 67 h . Both displacements are in the same direction, so the total displacement is ∆ x = ∆ x 1 + ∆ x 2 = 40 km + 40 km = 80 km. The total time for the trip is t = t 1 + t 2 = 2 . 00 h. Consequently, the average velocity is v avg = (80 km) (2 . 0 h)
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Unformatted text preview: 0 h) = 40 km / h . (b) In this example, the numerical result For the average speed is the same as the average velocity 40 km / h. (c) In the interest oF saving space, we briefly describe the graph (with kilometers and hours understood): two contiguous line segments, the frst having a slope oF 30 and connecting the origin to ( t 1 , x 1 ) = (1 . 33 , 40) and the second having a slope oF 60 and connecting ( t 1 , x 1 ) to ( t, x ) = (2 . 00 , 80). The average velocity, From the graphical point oF view, is the slope oF a line drawn From the origin to ( t, x )....
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