A bead moves along straight horizonal wire of length `L`, starting from the left end with a velocity `v_(0)`. Its retardation is proportional to the distance from the right end of the wire. Find the initial retardation (in `m//s^(2)`) (at let end of the wire) if the bead reaches the right end of the wire with a velocity `(v_(0))/2`. (given `v_(0)=5m//s` and `L=1m`)

Answer

`00018.75`

Explanation

`v(dv)/(dx)=-k(L-x)` where `k` is a constant.
`int_(v_(0))^(v_(0)/2) vdv=-kint_(0)^(L)(L-x)dx`
`implies3/4 v_(0)^(2)=KL^(2)`
So initial retardation `=KL=3/4 (v_(0)^(2))/L=3/4xx25=18m//s^(2)`