Question

A cat 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

Answer 1

(a)

`a_c=3.41(m)/(s^2)`

(b)

`a_a_v_g=0(m)/(s^2)`

Step-by-step explanation:

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

`At\hspace{3}t_1`

`|v_1|=\sqrt{(v_x)^(2)+(v_y)^(2)  } =\sqrt{(1.70)^(2)+(4)^(2)  } =4.346262762m/s`

`At\hspace{3}t_2`

`|v_2|=\sqrt{(v_x)^(2)+(v_y)^(2)  } =\sqrt{(-1.70)^(2)+(-4)^(2)  } =4.346262762m/s`

For now:

`|v_1|=|v_2|`

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:

`T=(2*\pi *r)/(v)` (1)

Where:

`r=radius\hspace{3}of\hspace{3}curvatrue`

`v=tangential\hspace{3}velocity`

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:

`r=(T*v)/(2*\pi) =(8*4.346262762)/(2*\pi)=5.533833621m`

Now the centripetal acceleration is given by:

`a_c=(v^(2) )/(r) =((4.346262762)^(2) )/(5.53383621) =3.413546791m/s^2`

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

`a_a_v_g=(\Delta v)/(\Delta t) =(v_2-v_1)/(t_2-t_1) =(4.346262762-4.346262762)/(4-2)=(0)/(2) =0`