2001 AB

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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:t6 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 ) ...
View Full Document

## This document was uploaded on 05/03/2012.

Ask a homework question - tutors are online