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

A particle is moving on a circular path of 10 m radius. At any instant

1.	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
Answer» `5 MS^(-2)` `2 ms^(-2)` `3.2ms^(-2)` `4.3 ms^(-2)` Solution :Here `R=10 m.v =5 ms^(-1) , a_t= 2 ms^(-2)` ` a_r= (v^2)/(r) = ( 5 xx 5 )/(10 ) = 2.5ms^(-2) ` Thenetaccelerationis ` a = sqrt( a_(r )^(2) + a_(t)^(2)) =sqrt((2.5)^2+2^2 ) = sqrt(10 .25) = 3.2ms^(-2)`