A particle falls in a medium whose resistance is propotional to the square of the velocity of the particles. If the differential equation of the free fall is `(dv)/(dt) = g-kv^(2)` (k is constant) then

Options

A. `v=2sqrt(g/k)(e^(2tsqrt(g//t))+1)/(e^(2rsqrt(g//k))-1)`
B. `v=sqrt(g/k)(e^(2tsqrt(gk))-1)/(e^(2tsqrt(gk))+1`
C. `v to 0` as `t to infty`
D. `v to sqrt(g/k)` as `t to infty`

Answer

B, D

Explanation

`(dv)/(dt) = g-kv^(2)`
`rArr (dv)/(g-kv^(2))`
`rArr (1/k int(dv)/((g/k)-v^(2))) = int(dt+C)`
`rArr 1/(2sqrt(gk)) log|(sqrt(g/k)+v)/(sqrt(g//k)-v)|=t+C`
At t=0, v=0 `rArr C=0`
`rArr v=sqrt(g/k) (1-1/(e^(2tsqrt(gk))/(1-1/(e^(2tsqrt(gk))))))`
Clearly when `v to sqrt(g/k)` as `t to infty`