A particle move in a semicircular of radius `R` from `O` to `A` . Then it moves parallel to z-axis covering distance `R` upto `B`. Finally it moves along `BC` parallel to y-axis throuth a distance `2R`. Find the ratio of `D//s`.
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Explanation

The distance `D`, that is, length of the actual path covered by time the particle `OA'ABC` as shown in
.
`D =` length of the semicircle `OA'A +` length
`AB +` length `BC`. This `D =pi R+ R+2R =(pi +3) R`.
Since `OA=2R, AB=R`, and `BC=2R`, the
coorkenates of `C` can be given as `C-=(2R.R 2R)`. Then the position of `C` is expresed as:
`vec r=2R hat j + 2R jat k`
As `vec v(=vec OC)=vec r_(c)-vec r_(0)`, Substituting `vec r_(c)` and `vec r_(0)=0 hat i+0 hat j+0 hat k`,
we obtains `vec s =(2 hat i+hat j+2 hat k) R`.
Its magnitude `|vec s|=(sqrt(2^(2)+1^(2)+2^(2))) R=3 R`
Hence, `(D)/(s)=((pi +3)R)/(3R)=(pi+3)/(3)`.