V6 55 55 18 6 6 0 at dt 215 1 415 115 v0 2 at

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Unformatted text preview: me t = 12 because v(12) v(0)  12 ¨0 a(t ) dt  0 . (c) The absolute maximum velocity is 115 ft/sec at t = 6. The absolute maximum must occur at t = 6 or at an endpoint. v(6)  55  55 18 ¨6 6 ¨0 a(t )dt 2(15) 1 (4)(15)  115  v(0) 2 a(t ) dt  0 so v(18)  v(6) £ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ 4:¦ ¤ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¥ ¦ 1:t6 1 : absolute maximum velocity 1 : identifies t  6 and t  18 as candidates or indicates that v increases, decreases, then increases 1 : eliminates t  18 (d) The car’s velocity is never equal to 0. The absolute minimum occurs at t = 16 where v(16)  115 16 ¨6 a(t ) dt  115 105  10  0 . £ 1 : answer ¦ 2:¦ ¤ ¦ 1 : reason ¦ ¥ Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 4 AP® CALCULUS AB 2001 SCORING GUIDELINES Question 4 Let h be a function defined for all x L 0 such that h(4)  by h a(x ) ...
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This document was uploaded on 05/03/2012.

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