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Chapter 3 16

# Chapter 3 16 - 166 CHAPTER 3 DIFFERENTIATION RULES(c The...

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Unformatted text preview: 166 CHAPTER 3 DIFFERENTIATION RULES (c) The particle is at rest when 12(15): 0. 3132 — 2415+ 36 — 0 —> 3(t 2)(t 6) — 0 -> t — 2 s or 6 s. (d) The particle is moving in the positive direction when v(t) > 0. 3(t # 2)(t — 6) > 0 <:> 0 S t < 2 or t > 6. (e) Since the particle is moving in the positive direction and in the (f) t = 8, s = 32 negative direction. we need to calculate the distance traveled in the intervals [0, 2]. [2, 6]. and [6, 8] separately. If(2) - f(0)| : |32 — 0| = 32- |f(6) - f(2)| = |0 - 32I = 32- If(8) - f(6)| = I32 - 0| = 32 The total distance is 32 + 32 + 32 : 96 ft. 4. (a) s:f(t) =t4-4t+1 => v(t) =f’(t) :4t374 (b) v(3) = 4(3)3 ; 4 2 104 ft/s (c) It is atrestwhen v(t) : 4(t3 — 1): 4(t — 1)(t2 + t + 1): 0 <:> t : 1 s. (d) It moves in the positive direction when 4(153 — 1) > 0 4:) t > 1. (e) Distance in positive direction 2 |f(8) - f(1)| : |4065 — (—2)| : 4067 ft Distance in negative direction : |f(1) — f(0)| 2 I72 — 1| : 3 ft Total distance traveled : 4067 —I— 3 = 4070 ft 6) t=& s : 4065 [—0— —i + //— -> S t , (t2 + 1)(1) ‘ t(2t) _ 1 — t2 5. (a) s = t2 + 1 ”(t) — 3 (t) — (t2 + 1)2 (t2 + 1)? 1—m2_1e92:§D,3 “971(3) ‘ (32 + 1)? 102 100 25 ft/S (c)Itisatrestwhenv:0 4:) 1-t2:0 <:> tzls [t#~1sincet20]. (d) It moves in the positive direction when 1) > 0 <=> 1 e t2 > 0 4:) t2 < 1 4i) 0 S t < 1. (e) Distance in positive direction : [5(1) , s(0)| = I; — 0| : % ft Distance in negative direction = [5(8) 7 s(1)l : )é - é) : % ft ‘ — l A). _ E Total distance traveled 7 2 + 130 A 65 ft (f) .. U .0 r4 ~ II II NI~:-‘ > S rat—- 6. (a) s : \/E (312 — 35t + 90) z 3155/2 — 35W2 + 90W2 => 15 1 2 #1 2 15 —1/2 2 _ i _ # w) : s’(t) : L255)” — —135t/ +45t / - —2 t (t 7t+6) ——2\/E(t 1)(t 6) ...
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