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The solution of (d v)/(dt)+k/m v=-g is (...

The solution of `(d v)/(dt)+k/m v=-g` is (a) `( b ) (c) v=c (d) e^(( e ) (f) (g) k/( h )g( i ) (j) t (k))( l )-( m )(( n ) mg)/( o ) k (p) (q) (r)` (s) (b) `( t ) (u) v=c-( v )(( w ) mg)/( x ) k (y) (z) (aa) e^(( b b ) (cc) (dd) k/( e e ) m (ff) (gg) t (hh))( i i ) (jj)` (kk) (c) `( d ) (e) v (f) e^(( g ) (h) (i) k/( j ) m (k) (l) t (m))( n )=c-( o )(( p ) mg)/( q ) k (r) (s) (t)` (u) (d) `( v ) (w) v (x) e^(( y ) (z) (aa) k/( b b ) m (cc) (dd) t (ee))( f f )=c-( g g )(( h h ) mg)/( i i ) k (jj) (kk) (ll)` (mm)

A

`v=ce^(-k/mt)-(mg)/k`

B

`v=c-(mg)/ke^(-k/mt)`

C

`ve^(-k/mt)= c-(mg)/k`

D

`ve^(k/mt)=c-(mg)/k`

Text Solution

Verified by Experts

The correct Answer is:
A
`(dv)/(dt) +k/mv=-g`
or `(dv)/(dt) = -k/m(v+(mg)/k)`
or `(dv)/(v+mg//k)=-k/mdt`
or `log(v+(mg)/k)=-k/mt+logc`
or `v+(mg)/k = ce^(-k//mt)`
or `v=ce^(-k/mt)-(mg)/k`
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