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
On
It's easy just keep on substituting the given relation of 'v' and 's' in 3 general equation and you will get displacement=8m.
On
$v=u+at$
Putting the value of v from the given equation($v^2=4+4s$) in this and substituting $t=0$ and $s=0$,we get $v^2=4$
Thus $v=2=u$ (negative is not possible since initial velocity is positive thus displacement could never be negative).
Putting in $v^2=u^2+2as$,we get
$4+4s=4+2as$
$4s=2as$
Thus,$a=2$
Putting these obtained values in $s=ut+(1/2)at^2$ and $t=2$,
$s=2(2)+(1/2)(2)(4)$
$ =4+4$
$ =8$
Thus we get the required displacement along with other variables like acceleration(it'll help to find these variables if it is like a paragraph based question or something like that) Hope you like the 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.