p02_027

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

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27. The constant acceleration stated in the problem permits the use of the equations in Table 2-1. (a) We solve v = v 0 + at for the time: t = v v 0 a = 1 10 ( 3 . 0 × 10 8 m / s ) 9 . 8m / s 2 =3 . 1 × 10 6 s which is equivalent to 1 . 2months
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Unformatted text preview: (b) We evaluate x = x + v t + 1 2 at 2 , with x = 0. The result is x = 1 2 ³ 9 . 8 m / s 2 ´ ( 3 . 1 × 10 6 s ) 2 = 4 . 7 × 10 13 m ....
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