Question

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?

Answer 1

`V=g/k[1-e^[-kt]]`

Explanation:

Acceleration pf particle, `[dv]/[dt]=g-kv,`.ie,

`[dv]/[dt]+kv=g`. This is a linear equation..............`[1]`

Need to multiply by both sides of .....`[1]` by the integrating factor, given by `e^[int[ft]dt` =`e^[kt]`, since we are integrating with respect to time.

Therefore, `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]

Hope this was helpful..