Acat rides a merry-go-round while turning with uniform circular motion. at time t1 = 2.00 s, the cat's velocity is v with arrow1 = (1.70)i hat + (4.00 m/s)j, measured on a horizontal xy coordinate system. at time t2 = 4.00 s, a half-revolution later, its velocity is v with arrow2 = (-1.70 m/s)i hat + (-4.00 m/s)j. (a) what is the magnitude of the cat's centripetal acceleration?
m/s2

(b) what is the magnitude of the cat's average acceleration during the time interval t2 - t1?
m/s2

(a)

(b)

Explanation:

Let's calculate the magnitude of cat's velocity at time t1 and time t2:

For now:

So, we can assume that cat's tangential velocity is constant. Now, considering that the time T required for one complete revolution is called the period. For  constant speed is given by:

 (1)

Where:

The problem tell us that at time t2=4 the cat has completed a half-revolution, so we can conclude that the cat complete a revolution in 8s, T=8s. Replacing the data in (1) and isolating r:

Now the centripetal acceleration is given by:

Finally, The average acceleration is the final velocity minus the initial velocity per time taken.

prolly a beacuse the momentuem the car has increases while going down hill

double the distance = divide strength by 4