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140Bquiz12asols - 1 MATH 140B QUIZ#12 Solutions 1 Take time...

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1 MATH 140B QUIZ#12 Solutions Dec 10, 2007 1. Take time zero to be when the breaks are applied. We are given a = - 68 . 5 , v 0 = 102 . 7 and s 0 = 0 . Therefore, v ( t ) = - 68 . 5 t + 102 . 7 and s ( t ) = - 34 . 25 t 2 + 102 . 7 t + 0 . We want t when v = 0 . This gives t = 102 . 7 / 68 . 5 = 1 . 499 . Finally, s (1 . 499) = - 34 . 25(1 . 499) 2 +102 . 7(1 . 499) = 76 . 98 77 . Therefore, C) is the answer. 2. We know that Δ x = π - 0 4 = π 2 . The cut-points are x 0 = 0 , x 1 = π/ 4 , x 2 = π/ 2 , x 3 = 3 π/ 4 and x 4 = π. Therefore, π 4 bracketleftbigg cos( π 4 ) + cos( π 2 ) + ( 3 π 4 ) + ( π 2 ) bracketrightbigg = π 4 parenleftbigg 1 2 + 0 - 1 2 - 1 parenrightbigg = - π 4 . Therefore, B) is the answer. 3. The work done is given by the definite integral integraldisplay 11 2 (71 x - 10 . 3 x 2 + 0 . 986 x 3 ) dx = 35 x 2 - 3 . 433 x 3 + 0 . 2465 x 4 bracketrightbigg 11 2 = 3274 . 2 - 116 . 48 = 3157 . 72 Therefore, E) is the answer. 4. Expanding the integrand gives:
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