Mod 9 Discussion My CIN A=4 B=2 1. C=6 05/20/2016 D=9 E=5 The acceleration of a particle is given by A=4 x t. If V=5m/ S at t=1 and S=12 m at 1 sec. Find S at t=12 sec a 2t dv 2t dt dv 2tdt v 2tdt t2 C Step 1 - V= 5, use this fact to find constant as 5 12 C 51 C C 4 C=4, velocity becomes V=t2+4 V=t2+4 ds/dt=t2 +4 S=(t2+4)dt =t3/3 +4t+C Step 2When t=1, S =12 use this fact to find constant C as 12=13/3 +4(1)+C 13/3+4(1)+c=12 1/3+4+C=12 C=23/3 Step 3 Using C=23/3, distance function becomes s=t3/3 +8t+23/3 Plug in t=2 into equation s=t2/3 +4t+23/3 to find distance as s=t3/3 +4t+23/3 s=22/3 +4(2) +23/3 s=67/3 s=17 2. Find the area between the curves for the equations below and round answer to one decimal place. A=4 B=2 C=6 D=9 E=5 Step 1-- find the intersection by setting one to equal the other y=x2+4x-4 and y=x+2 x2+4x-4 = x+2 x2+3x-6=0 x=-3 33/2=1.3722, -4.3722 A=-4.371.37[(x+2)-(x2+4x-4)]dx A=31.6