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p02_012

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

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12. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and work with distances in centimeters and times in seconds. (a) We plug into the given equation for x for t = 2 . 00 s and t = 3 . 00 s and obtain x 2 = 21 . 75 cm and x 3 = 50 . 25 cm, respectively. The average velocity during the time interval 2 . 00 t 3 . 00 s is v avg = x t = 50 . 25 cm 21 . 75 cm 3 . 00 s 2 . 00 s which yields v avg = 28 . 5 cm / s. (b) The instantaneous velocity is v = dx dt = 4 . 5 t 2 , which yields v = (4 . 5)(2 . 00) 2 = 18 . 0 cm / s at time t = 2 . 00 s. (c) At t = 3 . 00 s, the instantaneous velocity is v = (4 . 5)(3 . 00) 2 = 40 . 5 cm/s. (d) At t = 2 . 50 s, the instantaneous velocity is v = (4 . 5)(2 . 50) 2 = 28 . 1 cm/s. (e) Let t m stand for the moment when the particle is midway between
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Unformatted text preview: x 2 and x 3 (that is, when the particle is at x m = ( x 2 + x 3 ) / 2 = 36 cm). Therefore, x m = 9 . 75 + 1 . 5 t 3 m = ⇒ t m = 2 . 596 in seconds. Thus, the instantaneous speed at this time is v = 4 . 5(2 . 596) 2 = 30 . 3 cm/s. (f) The answer to part (a) is given by the slope of the straight line between t = 2 and t = 3 in this x-vs-t plot. The an-swers to parts (b), (c), (d) and (e) correspond to the slopes of tangent lines (not shown but easily imag-ined) to the curve at the appropriate points. (cm) x (a) 20 40 60 2 3 t...
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