A point moves in the plane so that its tangential acceleration `w_tau=a`, and its normal acceleration `w_n=bt^4`, where a and b are positive constants, and t is time. At the moment `t=0` the point was at rest. Find how the curvature radius R of the point's trajectory and the total acceleration w depend on the distance covered s.

Explanation

As `w_t=a` and at `t=0`, the point is at rest
So, `v(t)` and `s(t)` are, `v=at`, and `s=1/2at^2` (1)
Let R be the curvature radius, then
`w_n=(v^2)/(R)=(a^2t^2)/(R)=(2as)/(R)` (using 1)
But according to the problem
`w_n=bt^4`
So, `bt^4=(a^2t^2)/(R)` or, `R=(a^2)/(bt^2)=(a^2)/(2bs)` (using 1) (2)
Therefore `w=sqrt(w_t^2+w_n^2)=sqrt(a^2+(2as//R^2))=sqrt(a^2+(4bs^2//a^2)^2)` (using 2)
Hence `w=asqrt(1+(4bs^2//a^3)^2)`