A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5m s^(-1)` and the speed is increasing at a rate of `2ms^(-2)` The magnitude of net acceleration at this instant is

Options

A. `3.2ms^(-2)`
B. `2ms^(-2)`
C. `1.2ms^(-2)`
D. `4.3ms^(-2)`

Answer

A

Explanation

`a=sqrt(a_(t)^(2)+a_(n)^(2))`
`a_(t)` = rate of change of speed =`2 ms^(-2)`
`a_(n)=(v^(2))/(R)=((5)^(2))/(10)=2.5ms^(-2)`
`therefore" "a=sqrt(a_(t)^(2)+a_(n)^(2))=sqrt((2)^(2)+(2.5)^(2))=3.2ms^(-2)`