Particle is moving on a straight line and where velocity varies with its displacement as $v=\sqrt{4+4s}$. Find displacement at t = 2 s if s=0 at t=0. I am not able to figure out how to approach this.
PS. I don't know methods like Integration by parts as I am in class XI
Best Answer
As you have been told in the various comments and answer the diffrential equation is $$\frac{ds}{dt} = \sqrt{4+4s}$$ This equation is separable if you rewrite it as $$\frac{dt}{ds} =\frac{1}{\sqrt{4+4s}}=\frac{1}{2}\frac{1}{\sqrt{1+s}}=\frac{1}{2}(1+s)^{-\frac{1}{2}}$$ I suppose that you recognize that the rhs is just the derivative of $(1+s)^{\frac{1}{2}}$. So, you have $$dt=\frac{1}{2}(1+s)^{-\frac{1}{2}}=(\sqrt{1+s})'$$ Integrating both sides then gives $$t+c=\sqrt{1+s} $$ and you know this initial condition ($s=0$ if $t=0$), which can only happen if $c=1$.
Squaring now both sides then gives you $$1+s=(t+1)^2$$ that is to say $$s=2t+t^2$$
I hope and wish this becomes clearer to you. I am sure that you can take from here.