A particle is moving in a plane with constant radial velocity \dot{r} = 4.30~\mathrm{m/s} r ˙ =4.30 m/s, having started at the origin. It also has a constant angular velocity \dot{\theta} = 2.14~\mathrm{rad/s} θ ˙ =2.14 rad/s. When the particle is 2.98~\mathrm{m}2.98 m from the origin, what is the magnitude of its acceleration in \mathrm{m/s^2}m/s 2 ?

a

b

Explanation:

From the question we are told that

  The radial  velocity is  

   The angular velocity is  

  The distance considered is  

Converting the angular velocity to its equivalent linear velocity (tangential velocity )  we have that

=>    

=>  

Generally the radial and the tangential velocity are perpendicular to each other and their resultant velocity is mathematically represented as

=>  

=>  

Generally the acceleration is mathematically represented as

=>    

=>      

reserve is the answer