A point moves along a circle with a speed `v=kt` , where `k=0.5m//s^(2)`
Find the total acceleration of the point the moment when it has covered the `n^(th)` fraction of the circle after the begining of motion, where `n=(1)/(10)` .

Answer

`w=asqrt(1+(4pin)^2)=0.8m//s^2`

Explanation

The velocity of the particle `v=at`
So, `(dv)/(dt)=w_t=a` (1)
And `w_n=v^2/R=(a^2t^2)/(R)` (using `v=at`) (2)
From `s=int vdt`
`2piReta=a underset0 overset t int dt=1/2at^2`
So, `(4pieta)/(a)=t^2/R` (3)
From Eqs. (2) and (3) `w_n=4piaeta`
Hence `w=sqrt(w_t^2+w_n^2)`
`=sqrt(a^2+(4piaeta)^2)=asqrt(1+16pi^2eta^2)=0.8m//s^2`