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

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 ) = ( 5 xx 5 )/(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)`