Unformatted text preview: 8 C. 61. } v k m } 5 coasting distance } (0.86)( k 30.84) }5 0.97 k < 27.343 s ( t ) 5 } v k m } (1 2 e 2 ( k / m ) t ) s ( t ) 5 0.97(1 2 e 2 (27.343/30.84) t ) s ( t ) 5 0.97(1 2 e 2 0.8866 t ) A graph of the model is shown superimposed on a graph of the data. [0, 3] by [0,1] Chapter 7 Applications of Definite Integrals ■ Section 7.1 Integral as Net Change (pp. 363–374) Exploration 1 Revisiting Example 2 1. s ( t ) 5 E 1 t 2 2 } ( t 1 8 1) 2 } 2 dt 5 } t 3 3 } 1 } t 1 8 1 } 1 C s (0) 5 } 3 3 } 1 } 1 8 1 } 1 C 5 9 ⇒ C 5 1 Thus, s ( t ) 5 } t 3 3 } 1 } t 1 8 1 } 1 1. 2. s (1) 5 } 1 3 3 } 1 } 1 1 8 1 } 1 1 5 } 1 3 6 } . This is the same as the answer we found in Example 2a. 3. s (5) 5 } 5 3 3 } 1 } 5 1 8 1 } 1 1 5 44. This is the same answer we found in Example 2b. 288 Section 7.1...
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 Spring '08
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