Question

1) (25 pts) Given the following table of velocity data t (s) v(m/s) a) (6 pts) Estimate the position at 2 seconds (use 1/3 Simpson's rule) b) (6 pts) Estimate the acceleration at 2 seconds (use first order centered finite difference) c) (6 pts) Estimate the position at 3 seconds (use 3/8 Simpson's rule ) d) (7 pts) Estimate the acceleration at 3 seconds (use first order centered finite difference) 7 5 0 1 2 3 4 3 3.5 4 5 3 4.5 3 3.5 Show your work &describe your steps in the space provided on this page. Please work neatly to ensure

Answer 1

We have ie t=0 Veto) = 7,31 Vali) = 3 +z = 2 V(2) = 3.5 t; = 3 vs) = 3 s vlly) - 4.5 ts=s ults) : s te6 víts) = 4 4737 v(tt) = 3.5 te = % Vite) = 3 a) 10 find position at t=2 second we know that Va d x 1 dt 7 dx : vot 7 t - svat

since body starts from rest=0 » distance at t= 2 see | : x - y do By simpson's 1 red rule we know that 1 S + x) dx = b-a [ + (x) + uk +1+;) + 2 2 2 J. 2 In i = euen 1 i: odd b-2 a= 0 here flu) - Vit) let us take n=2 fa vedt - 3[vita) + 4x[v4.)] + 2x vt)] [ 0 + 4x3 + 2x3.5) (12+ 2)

at a second . for acceleration we know that = v't) at + 2 sec centred finite difference i forward) By first orden - we know that vitz) = V1 +241) - Y42) Ot let us take ot=1 ville) = V14) - VERD 3 - 3.5 = -0.5 m/s²

by (C) Now position at x = 3 sec simpson % sule x = { V (t) at ut n=3 then h= 3-0 = 1 How by simpsons % rule we have x3 = 5 vt) at +262 Blixt] S no vale veta? o valle as vet, - Ven) to) + V(t) + + V 2(1) 3 (v PP???? = 3! [0+ 3 + 0 + 3 x(3 + 3.5)] = ( 3+ 19.53 = x 22.5 :54.5 m

acceleration at ] second ag = v'itz) Hapon By first order centred finite difference (for word) az - vilty) - vist1) - V(+3) Dt hen otal = v(tu) - V (to) - 4.5 - 3 = 1.5 m/s2