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 = (2.30)i hat + (4.00 m/s)j, measured on a horizontal xy coordinate system. At time t2 = 9.00 s, a half-revolution later, its velocity is v with arrow2 = (-2.30 m/s)i hat + (-4.00 m/s)j.

Answer 1

Part a)

`a_c = 2.07 m/s^2`

Part b)

`a_(avg) = 1.32 m/s^2`

Step-by-step explanation:

As we know that it makes half revolution in given time interval

so we have

`(T)/(2) = t_2 - t_1`

`(T)/(2) = 9 - 2`

`T = 14 s`

now the angular speed is given as

`\omega = (2\pi)/(T)`

`\omega = (2\pi)/(14)`

`\omega = 0.448 rad/s`

now linear speed is given as

`v = √(2.30^2 + 4.00^2)`

`v = 4.61 m/s`

now we have

`v = R \omega`

`4.61 = R(0.448)`

`R = 10.3 m`

Now centripetal acceleration is given as

`a_c = \omega^2 R`

`a_c = 0.448^2 * 10.3`

`a_c = 2.07 m/s^2`

Part b)

Average acceleration of the cat is given as

`a_(avg) = (v_2 - v_1)/(\Delta t)`

`a_(avg) = (2v)/(\Delta t)`

`a_(avg) = (2(4.61))/(9 - 2)`

`a_(avg) = 1.32 m/s^2`