Question

The velocity v, of a parachute falling vertically satisfies the equation `"v" (dv)/(dx) = "g"(1 - v^2/k^2)` where g and k are constants. If v and are both initially zero, find v in terms of x

Answer 1

Given equation is `"v" (dv)/(dx) = "g"(1 - v^2/k^2)`

⇒  `v (dv)/(dx)= "g"((k^2 - v^2)/k^2)`

The given equation can be written as

`(v  dv)/(k^2 - v^2) = "g"/k^2  "d"x`

Multiply by – 2 on both sides, we get

`(- 2v)/(k^2 - v^2)  "d"v = (- 2"g")/(k^2)  "d"x`

Taking integrating on both sides, we get

`int (-2v)/(k^2 - v^2)  "d"v = int (- 2"g")/k^2  "d"x`

`log ("k"^2 - "v"^2) = (- 2"g"x)/k^2 + log "C"`

`log ("k"^2 - "v"^2) - log "C" = - (2"g"x)/k^2`

`log((k^2 - v^2)/"C") = - (2"g"x)/k^2`

`(k^2 - v^2)/"C" = "e"^(- (2gx)/k^2)` 

k² – v² = `"Ce"^((-2gx)/k^2)`  .......(1)

Initial condition:

Given v = 0

when x = 0

we get k²(0)² = `"Ce" (-2g(0))/k^2`

k² = Ce°

k² = C

(1) ⇒ k² – v² = `"k"^2"e"^((-2gx)/"k"^2)`

`"k"^2 - "k"^2"e" (-2gx)/"k"^2` = v²

`"k" [1 - "e" (-2gx)/"k"^2]` = v²