Please refer to my school textbook pg48 (of the book, and not the pdf counter) here: http://ncertbooks.prashanthellina.com/class_11.Physics.PhysicsPartI/ch-3.pdf
My doubt is in this context: (right side column)
$a = dv/dt = v dv/dx$
then integrating $v$ with respect to $dv$, and $a$ with respect to $dx$.
Now, when we integrate $a$, either we say it is a constant, and give $a\times(x-x_o)$ as a result of the integration, or we say it is not constant, and is a function of time, in which case, it can not be integrated like this. But the book integrates it like this, and then it is written:
The advantage of this method is that it can be used for motion with non-uniform acceleration also.
Can you please explain how it can be used with non uniform acceleration while it has been assumed while integrating that $a$ is constant? And moreover, does this line seem to apply to only this derivation, or does it apply to other two before it as well?
Best Answer
I think that the book is simply referring to the fact that, even in the case of non-constant acceleration, calculus can be used to find the position as a function of time if the acceleration as a function of time is known. In particular, whether or not the acceleration is constant, the definitions of acceleration in terms of velocity and of velocity in terms of position give $x$ in terms of $a$ as follows: $$ x(t)-x(t_0) = \int_{t_0}^t d\alpha\,v(\alpha) = \int_{t_0}^td\alpha\int_{t_0}^{\alpha} d\beta \,a(\beta) $$