140Bquiz12asols - 1 MATH 140B QUIZ#12 Solutions Dec 10,...

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Unformatted text preview: 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, v0 = 102.7 and s0 = 0. Therefore, v(t) = -68.5t + 102.7 and s(t) = -34.25t2 + 102.7t + 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. -0 4 2. We know that x = x4 = . Therefore, = . The cut-points are x0 = 0, x1 = /4, x2 = /2, x3 = 3/4 and 2 1 1 +0- -1 2 2 - . 4 3 cos( ) + cos( ) + ( ) + ( ) = 4 4 2 4 2 4 Therefore, B) is the answer. 3. The work done is given by the definite integral 11 2 = 11 (71x - 10.3x2 + 0.986x3 ) dx = 35x2 - 3.433x3 + 0.2465x4 Therefore, E) is the answer. 2 = 3274.2 - 116.48 = 3157.72 4. Expanding the integrand gives: 3 1 1 x- x 2 3 dx = 0 1 x -2+ 2 x 2 1 dx = x - 2x - x 3 3 1 1 = (9 - 6 - 3 ) - ( 3 - 2 - 1) = 16 . 3 0 Therefore, A) is the answer. 5. I(t) = 3.389e0.1049t dt = 32.307e0.1049t + C. But I(0) = 0, so C = -32.307. Therefore, I(t) = 32.307e0.1049t - 32.307. Now, we evalute to get I(27) = 516 and I(34) = 1111. Subtracting to the number in the last 7 weeks, gives 595. Therefore, D) is the answer. ...
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This note was uploaded on 03/30/2008 for the course MATH 140B taught by Professor Fabbri,marc during the Fall '07 term at Pennsylvania State University, University Park.

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