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Power of the only force acting on a particle of mass m=1 kg moving in straight line depends on its velocity as `P=v^(2)` where v is in m/s and P is in watt. If initial velocity of the particle is 1m/s, then the displacement of the particle in `ln2` second will be :A. `(ln2-1)m`B. `(ln2)^(2)m`C. 1mD. 2m |
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Answer» 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` |
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