quiz 10 - r a Plfljiqlflflfi-IIINWT-HSBSSZ Williams...

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Unformatted text preview: r a Plfljiqlflflfi-IIINWT-HSBSSZ - Williams Inhemet Explorer 1'1 ball an ll httpfldatum.arnporlum.vtzdulfilngssessmentistudent.actlon 0 Ca. ©b. g. The curvature of the curve x 2: +5t, y : 3t 75t71 with t> 0 1) . . . at the pomt wrth xecomponent 6 IS 0.0385 0.113 0.158 0.625 0.0559 0.0517 0.081 Find tfrom the x—component datum, compute v(t) and a(t), and then use the basic components of acceleration Note: formula to find the normal component of acceleration, or the cross product formula to find the cun/ature directly. Click here for further info. .lnternathrotededMongn {3‘ v £10096 v r j E P10_7_Q10.01—10042521]28-2953564 - Windows Inbemet Explorer ll imp/Mammmporium.vtgdulslsmssasmenustudamction You are on the engineering team for a NASCAR driver. The driver has to take a curve of radius 27 meters. Extensive tests have shown 7) that the tires will not slip if the normal acceleration does not exceed 18 meters per second squared. What is the highest constant speed the driver can take the cuwe without danger of slipping? O a. 225. meters per second 16.09 meters per second 18. meters per second 24.15 meters per second 31.75 meters per second 15. meters per second 486. meters per second 22.05 meters per second 'l'l— _____ .I .._..I LI..— _.._._L...._ .4.1...-:.. .. . Internet 1 Protected Mode: On r i —_ E P1o_1_Q10.02-1004252023-2953564 - Windows Imentet EKplDrer 1 1 LEE“ ll imp/Mammmporium.vtgdulslsmssasmenustudamction 6:28 PM ' M" IULSIZUIU (ii) IIaII = 3T + all} (iii) If the trajectory is along y : mx + b, then the curvature K : 0 (iv) v o N : 0 (v) If K is the curvature, then an) : ”V”2 K (vi) If the acceleration and velocity of the particle are always orthogonal, then the speed is constant. Only (iii) , (v) , and (vi). Only (v) , and (vi). Only (i) , and (iv). Only (i) , (ii), (iv) , and (vi). Only (ii) , (v), and (vi). Only (iii) , (iv) , (v) , and (vi). Write down the vector acceleration in terms of T and N, and write down .lnternethrotededMongn {3 v 9.100% ' .::‘ ' M" IULSIZUIU e meme-e “3...... “we...“ 5...... ,1 ‘ ‘ I- emu/mm," mam..." “a.“5.“Mammmummrm“mm .___._._, m..— are always orthogonal, then the Speed is constant. a a. only (iii) , (v) , and (vi). a h. Only (V) . and (vi). 0 u. only (i) , and (iv) - o d. Only (i) , (ii) , (iv) , and (vi) - o 9. Only (ii) , (v) , and (vi) . © r. Only (iii), (iv) , (v) , and (vi). Write dowh the Vector acceleration in terms of T ahd N, and Write down the formulae for the tangential and normal components of acceleration, Then think about What each statement is telling you. Click here for further info. Home: Ft E] E] El El 1?] El sue-mammary Lose-z. a idendipma Moo: 0.. (a v El 3 Est-Etna!“ M‘- r — e. g Plniziqmoa-mmsznzs-zgjasm - Windam IntenEt E-plmer ‘ ‘ [E ‘— .15; i. http:/l'datum.emporium.vlgdlzlfilngssEsmenb'stlldentaction A point moves along the curve y = x such that the xecomponent of '- AaBEiCC. Sublflle velocity is always 3. Find the tangential and normal components of 2) acceleration at the point with xecomponent %. (Hint: Start parameterizing the curve by setting X : ct and fixing c so that vX : 3.) o a. 34.2 and 45.6 respectively 0 b. 34.2 and 21.6 respectively 0 c. 16.2 and 35.6 respectively 0 d. 9.88 and 35.6 respectively o e. 9.88 and 21.6 respectively 0 f. 9.88 and 45.6 respectively 0 g. 34.2 and 35.6 respectively © h. 16.2 and 21.6 respectively Dune .llllcmcliPmttdchodaOll (3‘ v 3.100% v “5‘ fit a a a ”I El 1 ‘ - 9 "1° 2 cm” Mm ”555‘" Wimmms‘v‘mr; ‘ \ i- W”. -M...‘meat“........i,t.......:.r;:m .-, o c. 16.2 and 35.6 respectively a I 9.88 and 35.6 respectively 0 e. 8.88 and 21.6 respectively (3 r. 9.88 and 45.6 respectively a g. 34.2 and 35.6 respectively {Q h. 16.2 and 21.6 respectively (3 i. 16.2 and 45.6 respectively If you have x and y as a function of t, then you have Nun: the trajectory r(t) and can find v(t) and a(t). Click here for further info. E e E1 E1 @ El Dine-'5." Jug... o mime-i Mme-fl, Made 0.. (a v at 8 use EQI-fllitieiii oi... , ._ E P1u_6_Qloo4—10042521123-2953564 - Windows Imemet Explarerfi I hfimlldatum.2mportum.vtgdulfileAssEsmant/studentaction 6) Find the tangential and normal components of acceleration, aT and am, at time t:2 if the trajectory is r(t) : < e4, x/E t, at >. at : 6.54 and am : 0.26 Ga. at : 6.54 and am : 0.532 at : 6.54 and : 2.13 at : 7.25 and : 1.41 at : 7.25 and : 0.532 at : 7.25 and : 2.13 at : 7.25 and : 7.84 at : 6.54 and : 7.84 at : 6.54 and 21.41 at : 7.25 and : 0.26 Submit- end .41 and an] : 0.26 Flnd v(t) and an), end then use tne oeeic components of acceleration formulas. Click here for further info. i—im [.4 i—i..Lm r j E P10_5_Q10.06-10042521]28-2953564 - Windows Inbemet Explorer ll httplr'datum.amporium.vtgdulfilSLAssasment/studentaction 5) Find the curvature of the trajectory r(t) : <4 cos(3 t), 4 cos(3 r), 4 sin(3 t)> when t : 0. 0a. 1 2x5 ; 2x5 1 W 1 18 «E i 2 W2 Find v(t) and a(t), and then use the basic components of acceleration formula to find Note: the normal component of acceleration, or the cross product formula to find curvature directly. Click here for further info. . Internet l Protected Mode: On r i _ E P10_4_Q10.07-10042521]28-2953564 - Windows Inbemet Explorer ll httplr'datum.amporium.vtgdulfilSLAssasment/studentaction Incorrect The net force on a mass of 5 kilos causes it to follow the 4) trajectory r(t) : < e4, x/E t, e! >. Find the magnitude of the force on the mass perpendicular to its direction of motion at time t:2. 36.3 newtons. 31.3 newtons. 12.1 newtons. 0.895 newtons. 7.07 newtons. 3.22 newtons. Find v(t) and a(t), and then use the appropriate component of Note: acceleration formula. Use Newton's Law to find force from the acceleration. Click here for further info. Submit- . Internet l Protected Mode: On Submit- ' M" 49.5mm ...
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