A particle is released from rest at point O at time t= 0 and it falls vertically that its acceleration at is given g-kv,where k is positive constant, g is gravity,v is instantaneous velocity at time t.Express v in terms of g,k and t?

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A particle is released from rest at point O at time t= 0 and it falls vertically that its acceleration at is given g-kv,where k is positive constant, g is gravity,v is instantaneous velocity at time t.Express v in terms of g,k and t?

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Answer 1 · 2018-03-31T21:05:52

$V = \frac{g}{k} [1 - e^{- k t}]$ $V=g/k[1-e^[-kt]]$

Explanation:

Acceleration pf particle, $\frac{d v}{d t} = g - k v,$ $[dv]/[dt]=g-kv,$ .ie,

$\frac{d v}{d t} + k v = g$ $[dv]/[dt]+kv=g$ . This is a linear equation.............. $[1]$ $[1]$

Need to multiply by both sides of ..... $[1]$ $[1]$ by the integrating factor, given by $e^{\int [f t] d t}$ $e^[int[ft]dt$ = $e^{k t}$ $e^[kt]$ , since we are integrating with respect to time.

Therefore, $v e^{k t} = g \int e^{k t} d t + C$ $ve^[kt]=ginte^[kt]dt+C$ = $[g/ke^[-kt]+C]............$ [2] $.$ ...... Therefore $v=g/k+Ce^[-kt]$ .

Since the particle falls from rest, $v=0$ when $t=0,$

so $0=g/k+C$ ,....... therefore $C=-g/k$ .

And so $v=g/k[1-e^[-kt]]$ , ...[.as $e^[-kt]$ approaches $0$ as $t$ approaches infinity, the velocity approaches a fixed or terminal value $g/k$ , as $t$ increases]

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A particle is released from rest at point O at time t= 0 and it falls vertically that its acceleration at is given g-kv,where k is positive constant, g is gravity,v is instantaneous velocity at time t.Express v in terms of g,k and t?

1 Answer

Explanation:

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