{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math2930_2011fa_Week03_PS03Sol_Zaytsev(1)

math2930_2011fa_Week03_PS03Sol_Zaytsev(1) - 2.3 ‘2 3 23...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2.3 ‘2, 3,. 23 23.(a) Measure the positive direction of motion downward. Based on Newton’s second law, the equation of motion is given by dv _ {—0.75v+mg, 0 <1: < 10 m.— _ dt —1212+mg, t> 10 Note that gravity acts in the positive direction, and the drag force is resistive. During the first ten seconds of fall, the initial value problem is dv/dt = —v / 7.5 + 32, with initial velocity 11(0) : 0 ft/s. This differential equation is separable and linear, with solution 005) 2 240(1 — (ft/7'5). Hence 11(10): 176.7 ft/s. (b) Integrating the velocity, with 330%) = 0, the distance fallen is given by $05) = 2401: + 1800 64/” - 1800. Hence 2:00) 2 1074.5 ft. (c) For computational purposes, reset time to t = 0. For the, remainder of the motion, the initial value problem is dv/dt = —32'u/ 15 + 32, with specified initial velocity 71(0) = 176.7 ft/s. The solution is given by 71(t’)~.: 15 + 161.7(3—32 V15. As 75 —-> oo, 110$) —> ’UL : l5 ft/s. (d) Integrating the velocity, with 56(0) 2 1074.5, the distance fallen after the parachute is open is given by 33(2‘.) 2 15t — 75.86—32’5/15 + 1150.3. To find the duration of the second part of the motion, estimate the root of the transcendental equation 15T _ 75867327715 + 1150.3 2 5000. The result is T = 256.6 s. (6) Velocity "ae’elscity 180 ,2»: 150 MEI 120' l I l. ma 1, 3 a “ , a 'w 513 mu 11.50 200 2513 5 if? 1.. n 29.(a) Let 2: represent the height above the earth’s surface. The equation of motion is given by 772% = —-G&% , in which G is the universal gravitational constant. The symbols M and R are the mass and radius of the earth, respectively. By the chain rule, ~ 2710121 — —G———Mm d3: .— (R + x)2 ' This equation is separable, with min = —G’M (R + :13) ‘2dx. Integrating both sides, and invoking the initial condition 12(0) 2 V2912, the solution is v2 = 2GM(R + :c)‘1 + 29R — ZGM/R. From elementary physics, it follows that g = GM / R2. Therefore 71(95): @(R/VR + 90). (Note that g = 78, 545 Dill/1112.) (b) We now consider dac/dt = #29 (R/VR +32). This equation is also separable, With \/ R + a: d 2 M29 Rdt. By definition of the variable :17, the initial condition is 93(0) = 0. Integrating both sides, we obtain use) = (-3th + git/2))?“ — R. Setting the distance a:(T) + R = 240, 000, and solving for T, the duration of such a flight would be T m 49 hours. _ ' 72%; 7' “if: 5142-4"? W fl‘nflé F5) 2135» b’ék ,. ‘ r‘ , ”I/ < 1 '1 1" /‘2fi “3 l1 5'» “f7 [ft 1) w § L/ 7%“ J" / J if 00;!ng 2‘ C 42/91 6‘": S 6’ 6f ,4 ’5" 25¢»: 0v) (7x95) C) Farm fla f 3y; 4 wéé'bvt/ m/(i CARL/2 3 6 e a“ ' 3: . N '1 MM 0 . z u > * [3&fmb4 £054 flfl /S (64;& )zlz D J( .3.» '3( / '- , ' m ' V . 54162.52" a lam/41¢ 5 ‘ . ,. r1 (9». fer [771“? Arena-(7 / 71- j 0 < 3 r / 1%{L¢,.¢,1 /,_ / » .. r - L 4 E hm fl r16. C '2 CG! SC: > V. [1 f; '(~.' 'é"X 7/, M. C f/A'Cfi"? (51 74 j “p L/ o N I {ZVL‘L- "I . (.9 . d liq Chi/C; [x > fl“ e 6La #2097“ fiflé) 9%”: “/5 w / z M " ‘1346'220‘1 a! f!) fleas/ring m , x5 Yap/£7 5.; 5.4 7439.“ g féé EXP" / [EW/VS ,AW find/(PEGIA/r'yf «xx/”j : V10 fl»; f/Zifi/ A CM L My “ ....._ {1 -.L , , , 5' 9-bit- um 0/ eye.” flat/£9 ' an ri'flc 2 ,1" “ 14;; /€ ¢~ 1 fn‘j 7/163“ Am) M2 2‘ ,r u 1 ' ' ,- '_,_— " " 14:1 , 1 /, (2‘) W/LCV} L1 ’- Li / 9” l 4 f x,. * ' ’7’! £72,: '6 bwL/q $11 (:5 6-"er LZ m, Z L T C ’ " WJ/é' 0/ 4/491 -"Lc7’t/V\ M (f/ , a / ~ 5'» \ F} mw a M/,él,«6/ZYVMM ( 922 mi 4 rs x; $60M 7 >29ng€ . [$3315. ,7; 001’! Zia (f “’5'” 2% £1 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern