problem02_31 solution

University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.31: a) At t = 3 s the graph is horizontal and the acceleration is 0. From t = 5 s to t = 9 s, the acceleration is constant (from the graph) and equal to 2 s 4 s m 20 s m 45 s m 3 . 6 = - . From t = 9 s to t = 13 s the acceleration is constant and equal to . s m 2 . 11 2 s 4 s m 45 0 - = - b) In the first five seconds, the area under the graph is the area of the rectangle, (20 m) (5 s) = 100 m.
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Unformatted text preview: Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s + 20 m/s)(4 s) = 130 m (compare to Eq. (2.14)), and so the total distance in the first 9 s is 230 m. Between t = 9 s and t = 13 s, the area under the triangle is m 90 s) 4 )( s m 45 )( 2 1 ( = , and so the total distance in the first 13 s is 320 m....
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• Binary relation, Equals sign, Total Distance

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