Answer:
Part a)
[tex]a_c = 2.07 m/s^2[/tex]
Part b)
[tex]a_{avg} = 1.32 m/s^2[/tex]
Explanation:
As we know that it makes half revolution in given time interval
so we have
[tex]\frac{T}{2} = t_2 - t_1[/tex]
[tex]\frac{T}{2} = 9 - 2[/tex]
[tex]T = 14 s[/tex]
now the angular speed is given as
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{14}[/tex]
[tex]\omega = 0.448 rad/s[/tex]
now linear speed is given as
[tex]v = \sqrt{2.30^2 + 4.00^2}[/tex]
[tex]v = 4.61 m/s[/tex]
now we have
[tex]v = R \omega[/tex]
[tex]4.61 = R(0.448)[/tex]
[tex]R = 10.3 m[/tex]
Now centripetal acceleration is given as
[tex]a_c = \omega^2 R[/tex]
[tex]a_c = 0.448^2 \times 10.3[/tex]
[tex]a_c = 2.07 m/s^2[/tex]
Part b)
Average acceleration of the cat is given as
[tex]a_{avg} = \frac{v_2 - v_1}{\Delta t}[/tex]
[tex]a_{avg} = \frac{2v}{\Delta t}[/tex]
[tex]a_{avg} = \frac{2(4.61)}{9 - 2}[/tex]
[tex]a_{avg} = 1.32 m/s^2[/tex]
