A point moves with decleration along the circle of radius R so that at any moment of time its tangential and normal accelerations
are equal in moduli. At the initial moment `t=0` the velocity of the point equals `v_0`. Find:
(a) the velocity of the point as a function of time and as a function of the distance covered `s_1`,
(b) the total acceleration of the point as a function of velocity and the distance covered.

Explanation

`a_(t)=a_(c)implies(dv)/(dt)=-(v^(2))/(R )`
`int_(v_(0))^(v)v^(-2)dv=(1)/(R )int_(0)^(t)dtimplies|(v^(-1))/(-1)|_(v_(0))^(v)=(1)/(R )t`
`|(1)/(v)|_(v_(0))^(v)=(t)/(R )implies(1)/(v)-(1)/(v_(0))=(-t)/(R )`
`(1)/(v)=(1)/(v_(0))-(t)/(R )=(R-v_(0)t)/(v_(0)R)impliesv=(v_(0)R)/(R-v_(0)t)`
`a_(t)=a_(c)impliesv(dv)/(ds)=-(v^(2))/(R )`
`int_(v_(0))^(v) (dv)/(v)=-(1)/(R )int_(0)^(s)dsimplies|log_(e)v|_(v_(0))^(v)=-(1)/(R )s`
`log_(e)v-log_(e)v_(0)=-(s)/(R )`
`implieslog_(e)((v)/(v_(0)))=-(s)/(R )`
`v=v_(0)e^(-(s)/(R )`