Deriving equations of motion using integration

Question

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?

Answer 1

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: $$ x(t)-x(t_0) = \int_{t_0}^t d\alpha\,v(\alpha) = \int_{t_0}^td\alpha\int_{t_0}^{\alpha} d\beta \,a(\beta) $$

Answer 2

I use the equations below all the time so solve problems where acceleration is a function of velocity or distance only.

$$ t = \int \frac{1}{a(u)}\,{\rm d}u $$ $$ x = \int \frac{u}{a(u)}\,{\rm d}u $$ $$ \frac{1}{2} u^2 = \int a(x)\,{\rm d}x $$ $$ t = \int \frac{1}{u}\,{\rm d}x $$

Example

If acceleration is $a(u) = 1-u/10$ with initial $x_0=0$ and $v_0=0$ find the time and distance to reach $u=5$.

$$ t = \int \frac{1}{1-u/10}\,{\rm d}u + K_0 $$ which is solved by $$ t = \left. \mbox{-} \frac{10 \log\left(1-u/10\right)}{1} \right\} t = 10\log(2) $$

$$ x = \int \frac{u}{1-u/10}\,{\rm d}u+K_1 $$ which is solved by $$ x = -100\log(1-u/10)-10u = 100\log(2)-50 $$

You can also get the general equation, by substituting $u(t)$ into $x(u)$.

$$ x = 100 \exp^{\mbox{-}t/100}+10t-100 $$

Answer 3

VERY CLASSICAL PROBLEM OF THEORETICAL MECHANICS:

The equation

$a=\frac{dv}{dt}$

can be written even for cases in which the force, hence acceleration, is a function of $x$, by simply using the chain rule as

$a(x)=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$.

So you have the differential equation

$a(x)=v\frac{dv}{dx}$,

separate variables and write it

$a(x)dx=vdv$,

integrate (no need to put limits, it is an indefinite integration, you add a constnant of integration instead)

$\int{a(x)dx}=\int{vdv}+C$,

and you get

$\int {a(x)dx}=\frac{1}{2}v^2+C$ ……..(1)

where $C$ is a constant of the integration to be determined by the initial conditions. In the case of constant acceleration this gives the famous equation of uniformly accelerated motion:

$ax=\frac{1}{2}v^2+C$.

Now, assume that at $x=0$ $v(0)=u$, where u is the initial speed so that $C=-\frac{1}{2}u^2$. Therefore we get the well known equation

$2ax=v^2-u^2$.

So equation (1) is a very general equation and applies to numerous mechanics problems in which force is a function of distance, $x$.

Answer 4

Let an object be moving with uniform acceleration a moving with initial velocity $u$ to cover a displacement $s$ in time $t$ with final velocity $v$.

$$v=u+at$$ $$a=\mbox dv/\mbox dt$$
$$\Rightarrow a\mbox{ d}t=\mbox{d} v $$
Integrating both sides,
$$\int a\mbox{ d} t=∫\mbox{d}v $$

$$a[t]= [v] $$ Putting limits $0$ to $t$ and $u$ to $v$, $$at=v-u$$ $$v=u+at$$

Q.E.D.