Question

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The velocity v(t) of a sprinter on a straight track is shown in the graph below. 11 9 7 5 v(t) 3 1 -1 -3 0 2 8 10 t Determine the evaluaton of each definite integral. 564 v(t) dt = 547 v(t) dt = 5610 v(t) dt = 0 5,10 v(t) dt =- Submit Answer Incorrect. Tries 2/8 Previous Tries If s'(t)=v(t), then s(t) is the position of the runner at time t. Let s(0)=-4, determine the following values. s(4) = s(5) = s(7) - s(9) =

Answer 1

و و و موم 11 dom (4/10) | - 7 (2,6) 5 T 7 VLE) 3 10,02! cgoy (6,712 (10-1) -2 17,-2) 1 Ica,-2) -3 6 7 8 9 ia t .) Yverdt = }vetsdt + Juets de In the interal oto 2 the v(t) eanation is. 42-41 X₂X1 y-4,5 m (X-XI) m=sope (Coordinats) two points are coo) & (216) X2 42. y-0 = 6-0 (2-0). 2-0 (gavets y=v y = 32 V(+) =34 10-6 (x-27 Interval Oto u Points a Coordinates (210) 8 (1110 y-6z y 6 = 2CX-2) =) Y = 2x +2

Act)=2++2] gultat = Jult) dt. + 2011 "Sult) at gst at & Jetta *[39 9191 +22497472 at 2-[+]," 119[4] #2[1-4]+2(u=2] j Jet+2 det 6 + 12th Jut) at U 22 O 2) Jv (t) det gultat + Is Ult) dit u 10 Interval u to 5 coordinates =) (6-ai yo (4,1o) as (5,0) 12 G2 92. y-y = m (x-x1. GX k) on y-10 = 0-10 (AC-u) 5-u y-10 = -10(X-u) = -lox +50 V(t) = -106 +501 Interal 5 to 7 coordinates =) (5,0) 6 (7,-2) x so X2 y-0 = -2-0 (x-5) 7-5 y=-1(X-5) y = -x +5 =) (+) -E+5. Ste

5 My Wet) dt = s gultat + Auctionet -156-104 € sijae] 1,3l+tes) de +501413]+((- 6) + 5173 +50(5-4377 ['cs=[245 +50)+f2 +0] - fioletu el=F10(954) (-49 +2?]+10 23 Ult)dt = 3 8) Juct) dit quit) att guch.de ㅋ 7 9 Interval 7 to 9 Coordinates (7,-2) (9,-2) 외 y-yi= m (22-XV) y-(-2) -2+2 9-7 (x-7) 4+2 y=-2 -) vlt) = -2 () Interval 9 to 10 coordinar =) (9,-2) & (10;o) 12 X2 Y-, = m (7-XV) y-(-2) = 0+2 (x-a) 10~9 Y+2 = 2(x-a). a) Y=2x - 20 VLE=24 -20 11 mm

10 10 - 2 detit dt. Con 7 10 O gult) at = 9G-2 Jet-20) a =:-(633*** (€1,9-2014 11:13 +269-77 +2 (40-44)-20110-4] I! = -2(?) +2 [?] -2011]. (or = -u +19-20 10 JULE)dt -5 7 C12 1. u) of vie de sucede + I vits det & Juctsat 22 + 3 + (-5) Jutat e 2o - tot se o] 0 Given にー s' (t) = VE) S!lt) = a(s) dut $(t) = Ss'lt) at = Sult) at conder the given graph. SCH) -) area seu) = SVE dt 2-4 18(5) - SV64)odt = 22-4 =18 -u = Ivet.de-+ svardit E 22+ 5-U = 23

$(7) = VH) at u give sjuride + Fultat usulident Juridt + 7 Jultidet - 22 + 5+ (-2)-4 (517) 21 S(9) 9 0 Su (t)dt - 4 sertlat + gult) de I Sult) at tasultide at t 22+5+ (-2) + (-u) - av - 21– 2 – – 4 27 - 10 s(a) = 17 SO) slu) sie an S(5)=23(q) =213 s(9)–17;
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