The retardation `alpha` of a particle in rectilinear motion is proportional to the square root of its velocity v. Assume that the constant A of proportionality is positive. The initial velocity of the particle is `v_(0)` . How far would the particle move before coming to rest? What would be the time required to travel that distance?

Explanation

`a prop - sqrt(v)`
`:. " " a = -Asqrt(v) = -Av^(1//2) " ""or," (dv)/(dt) = - Av^(1//2)`
or,`" " v^(-1//2)` dv = -Adt
`:. " " int v^(-1//2)dv = - A int dt + c` [c = integration constant ]
or,`" " 2v^(1//2) = -At + c " "cdots(1)`
Given, at t = 0, `v = v_(0).` Putting in (1), c = `2 v_(0)^(1//2)`
`:." " 2v^(1//2) = - At + v_(0)^(1//2)`
or,`" " 2(sqrt(v_(0))- sqrt(v))` = At `" "cdots(2)`
When the particle comes to rest after a time T, we have v = 0 at t= T. From (2),
`2sqrt(v_(0))=At " ""or",T=(2)/(A)sqrtv_(0)`
Now, `a = (dv)/(dt) = (dv)/(dx)(dx)/(dt) = (dv)/(dx) v`
or,`" " dx = (1)/(alpha)vdv = - (1)/(Asqrt(v))vdv = (1)/(A)v^(1//2) dv`
`:. " " int dx = -(1)/(A) int v^(1//2) dv +k` [ k = integration constant ]
or, `" " x = (1)/(A).(2)/(3)v^(3//2) +k = -(2)/(3A) v^(3//2) +k " " cdots(3)`
At start , x= 0 and v = `v_(0)` , Putting in (3),
`0 = (2)/(3A)v_(0)^(3//2) +k " ""or", k = (2)/(3A)v_(0)^(3//2)`
So, equation (3) becomes,
`x = (2)/(3A)(v_(0)^(3/2)-v_(2)^(3/2)) " " cdots (4)`
When the particle comes to rest v =0. Then the total distance travelled is,
`x_(0) = (2)/(3A)v_(0)^(3//2)`