ef157-m1-examsolution-2018 (1)

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The University of Tennessee, Knoxville *

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157

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Aerospace Engineering

Date

Oct 30, 2023

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pdf

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EF 157-Fall 2018-Exam 1-Page 3 of7 Section_ Name ______________ _ g = acceleration due to gravity= 9.81 m/s 2 = 32.2 ft/s 2 , 1 mile= 5280 ft, 1 yard= 3 ft, 1 in= 2.54 cm 3) (20 pts) Emily is traveling in a straight line on the Ped Walkway on her way to Perkins Hall. Because of the heavy pedestrian traffic during class change time, she cannot maintain a constant speed. In fact, she was initially "pushed backwards" by the large crowd (hence, an initial negative velocity). Her velocity-time ( v-t) graph is shown at right. Assume she marked the intersection of the Ped Walkway and Volunteer Blvd as the origin, and she begins her walk at s = + 15 ft. Also, assume 3 SF for all numbers. a) Construct Emily's position-time (s-t) graph. Hint: calculate her position at 0, JO, 20, and 30 sand then ''.freehand" draw the s-t graph using these points. b) Construct Emily's acceleration-time (a-t) graph. c) What is the total distance that Emily travels? ,_ ,!J l.-, - - :: .!) I op -e -::: C\. pt:: 3-(-3)¾ ~ ---=---- 20 --6 ..S · 0- /o) -~ z o - 3 w_s 5~0 ·:: ~~ ~ pS-z ~ -? ; +~ :' /.~ r;. _st.> - 26 3 s· (f-t-) tJof. t1tfo<J C ·1ri c:°'vL Emily·s s-t Graph on the Ped Walkway \J l 1 1 , UC l,(; V Cu~1 CC\ v.:.., u. {~ ..r,:- 8 \ ~, -- v ,-6 I r, ·r {"C ·i-u,v( _,,_/ ~ pc1.1. ·f' o------.c:"'--- --b- ~ =-- - , C . --20 ... l. ' j 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 4 , 3. Vl 2 • :: . 1 - > -;:o I ] -1 ·A , . /. r g;' -2 ' . I ff\ . / ~ --------- I -3 --------- -4 \ :. I 1 ,.__ l time (t) seconds ' '· 0 2 " 6 8 10 12 14 16 18 20 22 2,\ 26 28 30 32 0.. C (,_;_ ✓ ..., \ , , <- _ time (t) seconds rys""J ctcL,,. ,cv,.,,.,o , Emily's a-t Graph on the Ped Walkway 6,c.o · G,'-1 • o~ : -= ----------- ~, -u:Z.• { ~6. L/ · - o,(p I • 0 2 4 6 8 10 12 14 16 18 20 22 2,1 26 28 30 32 time (t). seconds -6)5 = - I j- ft- f- /~ ¢c- x< -::/.J ~ ..1 - J (t J fJ f3c -2<>) s :: ~

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EF 157 - Fall 2018 - Exam 1 - Page 5 of 7 Section_ Name _____ _____ __ ___ _ g = acceleration due to gravity= 9.81 m/s 2 = 32.2 ft /s 2 , 1 mile= 5280 ft, 1 yard= 3 ft, 1 in= 2.54 cm 5) (30 pts) Josh attempts to kick a football over the goalpost, which is 15 ft high and 25 ft away from him. The football leaves the ground at point A with a speed of VA= 52 ft/sat an angle of 64° from the horizontal. t 1r ((-:,-) V = 2 ft /s A A j T h t 15 ft 0 B 4.5 ft ~(;] a) The football "clears the goalpost" by a height h above O __i.:t..,.-4~-- -=====" =----u__ _______ ___J~ __.____.__ > ::::7 the horizontal crossbar. Determine h. -- 25ft- -- --- d --- - 6 b) An "enthusiastic fan" named Sam runs onto the field and catches the football at a height of 4.5 ft above the ground (point B). Determine the distance d from the goalpost where Sam needs to be in order to catch the football. c) What is the velocity of the football right before Sam catches it at point B? Give your answer in magnitude and angle (with respect to the horizontal) format. d) Sam was initially at some unknown distance away from point B so she was running to the left at a speed of 13 ft/s when she caught the football. What is the speed of the football relative to Sam just before the football is caught at point B? There is Extra Space for this problem on the next page, please use it as needed. ' I'. X: · ;;;..() ~ D I t Jt" £!__, ,. zz, 795- , ?-;~ ' J j, t;I -~ J~09{e 7 ~ -

EF 157 - Fall 2018 - Exam 1 - Page 6 of 7 Section Name g = acceleration due to gravity= 9.81 m/s 2 = 32.2 ft/s\ 1 mile =-5-=--=2-:--80-ft--=--,-1-y-ar_d_=_3_ft_, _l_i_n_=_2-.5-4-cm-- Extra space for problem 5. b) ;t i - t I - z. L.L r < ,•. ✓ r~/~. 111 fr t - J/-1;:,2 ~ ~\ i -z. ·;tJ. .::. ~ -t l)f' · ., - .,., 1 cT ·t -=) "I.,.:) -,-t .:: ~rt 7U, L- ;J 'J t (. V t ":1 ' 2.. - -- 1 ') co,,,. 2 ,- .f - - 7 0 c-:, /'I,(· ' .(.. -> .- 1.&· .) , / L ·- ~ • T - '") (J f L · . r II= 39.r:-7 c) r ' 'ff.I). (; ·- ·~ ------- 'J .,. JJ l ( ,,- .,., 't1f J\ ~-·v~\ - - = [z2,79J ·tr · - '-12s-2/.1 .4- 1 ] - [ · 1 1r 1 , . 1 J . -G ' 1,,.: - t .,, ~ A. TU < :) ... -..l .... , ' ..J :..- _,J • . 5 - i:; 0 :: 3s-l 195- -~ __ .l"- -- t/Z;52C:• l ~ ;· t..£/, :..Ji~ j ·:; ,~ 79.} ~) \ (1 z. rz u.r;;),. - S/,.;, 5.5· 't 9 ./is --------.1 r---,.. I I ·1-/~½ 2: ~~> -ri5

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EF 157 - Fall 2018 – Exam 1 - Page 1 of 7 Section ___ Name _____________________________________ g = acceleration due to gravity = 9.81 m/s 2 = 32.2 ft/s 2 , 1 mile = 5280 ft, 1 yard = 3 ft, 1 in = 2.54 cm 6) BONUS (5 points max) Define “ weekly average velocity ” as your average velocity between one Thursday when you are in your EF 157 recitation to the next Thursday when you are again in your EF 157 recitation. Estimate your average “weekl y average velocity ” over the four week period between Aug 23 (your first EF 157 recitation) and Sep 20 (today). Explain . You must have a numeric answer AND you must explain your answer in order to get credit. You may use words, diagrams, and equations to explain your answer. Your “weekly average velocity” for 1 week will equal your “ weekly average displacement ” Δ𝑟 ?𝑒𝑒𝑘 ( i.e ., the vector that connects your position during one Thursday recitation to your position in the next Thursday recitation) divided by 1 week: 𝑣 𝑎?𝑔 𝑤𝑒𝑒𝑘 = Δ𝑟 𝑤𝑒𝑒𝑘 1 week . Your average “weekly average velocity” over the four-week span will equal the sum of the 4 Δ𝑟 ?𝑒𝑒𝑘 vectors divided by 4 weeks. Due to the nature of vector addition, this sum will equal the vector that connects your position in week on Aug 23 to your position on Sep 20. Assuming you attended recitation on Aug 23 and Sep 20, the sum of your 4 Δ𝑟 ?𝑒𝑒𝑘 vectors would be a vector that lies inside the classroom. If you were in the same exact position these 2 days, this vector would have a magnitude of zero and your estimated weekly average velocity 𝚫? 𝒘𝒆𝒆𝒌 would also be zero. If you were not exactly in the same position on Aug 23 and Sep 20, the sum of the 4 Δ𝑟 ?𝑒𝑒𝑘 vectors and the average 𝚫? 𝒘𝒆𝒆𝒌 would be non-zero . I t’s reasonable to estimate that the difference in your location between Sep 20 and Aug 23 would be on the order of 16 ft in a northerly direction. Therefore, your weekly average velocity would be 𝚫? 𝒘𝒆𝒆𝒌 = 𝟏𝟔 ?? ??? 𝐍𝐨??𝐡 𝟒 ???𝐤? = 𝟒 ?? ???𝐤 ⁄ ??? 𝐍𝐨??𝐡 .