p02_033

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

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33. The problem statement (see part (a)) indicates that a = constant, which allows us to use Table 2-1. (a) We take x 0 = 0, and solve x = v 0 t + 1 2 at 2 (Eq. 2-15) for the acceleration: a =2 ( x v 0 t ) /t 2 . Substituting x =24 . 0m, v 0 =56 . 0km / h=15 . 55 m / sand t =2 . 00 s, we Fnd a = 2(24 . 0m (15 . 55 m / s)(2 . 00 s)) (2 . 00 s) 2 = 3 . 56 m / s 2 .
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Unformatted text preview: The car is slowing down. (b) We evaluate v = v + at as follows: v = 15 . 55 m / s − ³ 3 . 56 m / s 2 ´ (2 . 00 s) = 8 . 43 m / s which is equivalent to 30 . 3 km/h....
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