Question

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 ​​ ?

Answer 1

a

`v =   7.69 \  m/s`

b

`a =  19.86 \  m/s^2`

Step-by-step explanation:

From the question we are told that

The radial velocity is
`v_r =  4.30 \  m/s`

The angular velocity is
`w = 2.14 \ rad/s`

The distance considered is
`r =  2.98 \  m`

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

`v_t  =  r *  w`

=>
`v_t  =  2.98 *  2.14`

=>
`v_t  =  6.38 \ m/s`

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

`v =  √(v_r^2 + v_t^2)`

=>
`v =  √(  4.30^2 + 6.38^2)`

=>
`v =   7.69 \  m/s`

Generally the acceleration is mathematically represented as

`a =  (v^2)/(r)`

=>
`a =  (7.69^2)/(2.98)`

=>
`a =  19.86 \  m/s^2`