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.

Answer

(a) `v=v_0//(1+v_0t//R)=v_0e^(-8//R)`; (b) `w=sqrt2v_0^2//Re^(2s//R)=sqrt2v^2//R`.

Explanation

According to the problem
`|w_t|=|w_n|`
For `v(t)`, `(-dv)/(dt)=(v^2)/(R)`
Integrating this equation from `v_0levlev` and `0letlet`
`-underset(v_0)oversetvint(dv)/(v^2)=1/Runderset0oversettint dt` or, `v=(v_0)/((1+(v_0t)/(R)))`
Now for `v(s)`, `-(vdv)/(ds)=v^2/R`, Integrating this equation from `v_0levlev` and `0lesles`
So, `underset(v_0)overset(v)int(dv)/(v)=-1/Runderset(0)overset(s)intds` or, `1n(v)/(v_0)=-s/R`
Hence `v=v_0e^(-s//R)`
(b) The normal acceleration of the point
`w_n=(v^2)/(R)=(v^2e^(-2s//R))/(R)` (using 2)
And as accordance with the problem
`|w_t|=|w_n|` and `w_thatu_t_|_w_nhatu_n`
so, `w=sqrt2w_n=sqrt2(v_0^2)/(R)e^(-2s//R)=sqrt2(v^2)/(R)`