Power of the only force acting on a particle of mass m = 1 kg moving in a straight line depends on its velocity as `P=v^(2)` where v is in `ms^(-1)` and P is in watt. If the initial velocity of the particle is `1ms^(-1)`, then the displacement of the particle in ln(2) seconds will be

Options

A. `(ln2-1)m`
B. `(ln2)^(2)m`
C. 1m
D. 2m

Explanation

P=Fv
`v^(2)=Fv`
Ma=v
`1xxv(dv)/(dx)=v`
`int_(1)^(v)dv=int_(0)^(x)dx`
v-1=x
v=x+1
`(dx)/(dt)=x+1`
`int_(0)^(x)(dx)/(x+1)=int_(0)^(t) dt`
`ln(x+1)-ln(0+1)=t`
`x+1=e^(t)`
`x=e^(t)-1`
at `t=ln2`
`x=2-1=1 m`