At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (5.00 m/s2)i hat + (6.00 m/s2)j. It moves at constant speed. At time t2 = 5.00 s, its acceleration is (6.00 m/s2)i hat + (−5.00 m/s2)j. What is the radius of the path taken by the particle if

Question

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (5.00 m/s2)i hat + (6.00 m/s2)j. It moves at constant speed. At time t2 = 5.00 s, its acceleration is (6.00 m/s2)i hat + (−5.00 m/s2)j. What is the radius of the path taken by the particle if 
t2 − t1
 is less than one period

bobpursley · Answer

looking at the two accelerations, it appears the need to find the angle.
vi=5+6j
vf=6-5j

costheta=vi dot vj / (magvi * magvj)
maj vi, mag vf=sqrt(25+36)=sqrt61

vi dot vj=30-30=0
cosTheta=0
theta=90

but v=2PI r/12
v^2=4 PI^2 r^2/144

r=v^2/a=pi^2 r^2/sqrt61

r=sqrt61/pi^2
check that carefully. See this http://www.jiskha.com/display.cgi?id=1411303123