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Chapter 4 136 - 398 CHAPTER 4 APPLICATIONS OF...

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Unformatted text preview: 398 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 75. a(t) 2 k. the initial velocity is 30 mi/h = 30 - %% = 44 fI/s. and the final velocity (after 5 seconds) is 50mi/h=5 50- 333%=%ft/:.Sov(t)=kt—l—Candv(0)=44 => 0:44. Thus v(t)=l~ct—I—44 => 11(5) 4 5k + 44. But v(5)— so 5k + 44 4 23—0 4 5k 4 838 _> k 4 f: m 5.87 ft/s2. 76. a(t) : —16 :> 1.2(t) : —16t + 120 where 110 is the car‘s speed (in ft/s) when the brakes were applied. The car stops when —16t+ ’00— — 0 4:} t— — 1—16110. Now 3(t) = %(—16)t2 + vot : 48152 + wit. The car travels 200 ft in the time that it takes to stop. so 5(116710) — 200 —> 200 — 8(1167102) + 110(1i 6120) : i713 => 113 = 32 ~ 200 4 6400 :> 120 4 80 ft/s (54E mi/h). 77. Let the acceleration be a(t) : k km/hz. We have 71(0) 2 100 km/ h and we can take the initial position 5(0) to be 0. We want the time tf for which v(t) : 0 to satisfy 5(t) < 0.08 km. In general. 71’ (t) : a(t) : k, so 11(t) : kt + C. where C 2 11(0) : 100. Now s'(t) : 11(t) = kt + 100. so 5(t) : ékt2 + 10015 + D, where D : 5(0) 2 0. Thus. 5(t) = §kt2 + 100t. Since 12(tf) = 0. we have ktf + 100 : 0 0r tf = —100/k, so 2 8(tf) — ;k( 1—00) I 100< 100) 10.000 (i 4 l) = —m The condition 5(tf) must satisfy 18 k k 2k k [c —O':00 < 0.08 :> —% > k [k is negative] => k < —62.500 km/h2. or equivalently. k < —3—61385 3 —4.82 m/s2 . 78. (a) For 0 g t S 3 we have a(t) 4 60t 4> v(t) 4 30152 I C 11(0) 0 C 11(t) 30t2. so 3(t) : 10113 i C 5(0) 0 C s 3(t) 1013. Note thatv(3) 4 270 and 5(3) 4 270. F0r3 < t g 17: a(t) : —g : 732 ft/s v(t) — 32(t 3) l C 12(3) — 270 — C «0(t) 4 —32(t 3) l 270 5(t) 16(t 3)2 : 270(1 3) + 0 => 3(3) 4 270 = 0 => 5(t) : 416(154 3)2 + 270(t — 3) + 270. Note that 71(17) 2 —178 and 3(17) : 914. For 17 < t g 22: The velocity increases linearly from 4178 ft/s to 418 ft/s during this period. so Av —18 — (—178)_ 1_60 ——:———— =32 Thum():32(t417)417s => At 22417 5 3(1) 4 16(t 4 17)2 4 178(t 4 17) + 914 and 5(2) : 424 ft. Fort > 22: 11(t) : —18 => 3(t) : —18(t - 22) + C. But 8(22) : 424 : C => 5(t) = —18(t 4 22) + 424. Therefore. until the rocket lands. we have 30t2 if 0 g t g 3 432(143)+270 if3<tgi7 v(t) : . 32(t 4 17) 4 178 if 17 < t g 22 418 if t > 22 and 1013 if 0 g t g 3 —16(t 4 3)2 + 270(t 4 3) + 270 if 3 < t g 17 16(t 4 17)2 4 178 (t 4 17) + 914 if 17 < t g 22 418(t 4 22) + 424 if t > 22 ...
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