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

Answer

C

Explanation

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