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

Options

A. `5ms^(-2)`
B. `2ms^(-2)`
C. `3.2ms^(-2)`
D. `4.3ms^(-2)`

Answer

C

Explanation

`"Here", r=10m, v=5ms^(-1), a_(t)=2m^(-2)`
`a_(r)=(v^(2))/(r)=(5xx5)/(10)=2.5ms^(-2)`
The net acceleration is `a=sqrt(a_(r)^(2)+a_(t)^(2))+sqrt((2.5)^(2)+2^(2))=sqrt(10.25)=3.2ms^(-2)`