Question

The motion of a body is given by the equation `(dv)/(dt)=4-2v`, where v is the speed in m/s and t in second. If the body was at rest at t = 0, then find
(i) The magnitude of initial acceleration
(ii) Speed of body as a function of time

Answer 1

Given that, `(dv)/(dt)=4-2v`
(i) At t = 0, v = 0, so initial acceleration is `4-0=4m//s^(2)`
(ii) `dv=(4-2v)dt`
`or, (dv)/(4-2v)=dt`
or, `dv=int_(0)^(v)(dv)/(4-2v)=int_(0)^(t)dt`
`(log_(e)(4-2v))/(-2):|_(0)^(v)=(t)_(0)^(t)=t`
`log_(e){4-2v}-log_(e)(4)=-2t`
`log_(e){(4-2v)/(4)}=-2t`
`rArr (4-2v)/(4)=e^(-2t) rArr (4)/(4)-(2v)/(4)=e^(-2t)`
`rArr 1-(2v)/(4)=e^(-2t)`
`rArr (2v)/(4)=1-e^(-2t)`
or `v=2(1-e^(-2t))`