Motion in Two and three dimensions?

Question

A cat rides a merr-go-round tuning with uniform circular motion. At time t1 = 2.00s, the cat's velocity is v1 = (3.00 m/s)i + (4.00 m/s)j, measured on a horizontal xy coordinate system. At t2 = 5.00s, its velocity is v2 = (-3.00m/s)i + (-4.00m/s)j. What are (a) the magnitude of the cat's centeripetal acceleration and (b) the cat's average acceleration during the time interval t2 -t1?

Answer 1

(a)
at t=2.00s v1 = (3.00 m/s)i + (4.00 m/s)j then
|v1|=5m/s with an angle of arctan(4/3)=52.1 degree

at t2 = 5.00s, its velocity is v2 = (-3.00m/s)i + (-4.00m/s)j.
|v2|=5m/s with an angle of arctan(4/3)=233.1 degree

So we have a constant velocity of 5 m/s and the
the centripetal acceleration ac=v^2/r where
v-velocity
r-radius of the trajectory

The radius is not given but can be estimated since
the we can assume that the cat traveled distance S and that distance was ½ of the circumference of the trajectory.
r=S/pi and s=(t2-t1)v
then r=(t2-t1)v/pi
and ac=v^2/r = v^2/(t2-t1)v/pi=
ac=pi v/(t2-t1)= 5.24m/s^2

(b) And the cat's average acceleration during the time interval t2 -t1 is 0 since the velocity has not changed.

Answer 2

first, u need 2 find the magnitude of the initial velocity and final velocity of the merry go round respectively ,then u convert it into the angular velocity for both velocity,
then u use the equation of the angular velocity
(omega)2=(omega)1+(alpha)(t2-t1)

Answer 3

dR/dx = 2v^2/g * cos (2x) permit dR/dx = 0 2v^2/g * cos(2x) = 0 cos(2x) = 0 2x = ? /2 x = ? /4 At x = ? /4, dR/dx = 0 yet this may well be correspond to minimum to boot as for the optimal fee. Therfore discover d/ d x of dR/dx d/ d x of dR/dx = ? 4 v^2/g * sin (2x) and because it is damaging x = ? /4 corresponds to the optimal fee.