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?

  • $\begingroup$ The equation you have written is used very often in mechanics problems, where the speed of a particle is taken to be a function of the distance travelled. Once you write the diffrential equation of motion down then you need to separate the variables, x and t, in your differential equation and then integrate. This method applies for any type of motion in which the force depends on x, it can be used in 3-D as well. $\endgroup$ – JKL Mar 25 '13 at 16:23

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

  • $\begingroup$ is that apostrophe for a first and second derivative? $\endgroup$ – Saurabh Raje Mar 25 '13 at 13:22
  • $\begingroup$ thank you for the very prompt reply. dont you think that if the book were implying that acceleration be taken as a function of time and then solved, then they should have given it, and not written that after solving the integral as if it were constant? Moreover, can you please write those useful relations without using the apostrophe? I am sorry, but I cant understand it.. $\endgroup$ – Saurabh Raje Mar 25 '13 at 13:27
  • $\begingroup$ The apostrophes were just being used to denote dummy variables of integration; I changed it to variables $\alpha$ and $\beta$ for the dummy integration variables so as to (hopefully) be less confusing. As for what the book is talking about; all I we can do is guess because it's totally unclear (to me anyway) what precisely the book is referring to from the context. $\endgroup$ – joshphysics Mar 25 '13 at 22:30
  • $\begingroup$ So alpha and beta are variables that are substituted in place of the second and first derivatives of position as a function of time? $\endgroup$ – Saurabh Raje Mar 26 '13 at 10:03

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

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


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 $x(u)$.

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



The equation


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


So you have the differential equation


separate variables and write it


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


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:


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


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

  • $\begingroup$ 1)How is C= -1/2u^2? $\endgroup$ – Saurabh Raje Mar 27 '13 at 15:59
  • $\begingroup$ 2)how can a(x) be integrated ax? $\endgroup$ – Saurabh Raje Mar 27 '13 at 16:02
  • $\begingroup$ @SaurabhRaje 1) In the equation $ax=\frac{1}{2}v^2+C$, you use the initial condition of the problem. I.e the object has some speed $u$ when $x=0$ so that you get $0=\frac{1}{2}u^2+C$ and this gives $C=-\frac{1}{2}u^2.$ 2) In the case where the acceleration is constnat,$a(x)=a$, the integration reduces to $ax$. The constant of integration has been added on the RHS of the equation. I.e. no need to write $a\times(x_2-x_1)$ on the left, but you can if you want to, in which case you can put $x_2-x_1=x$ for simplicity. $\endgroup$ – JKL Mar 27 '13 at 17:31
  • $\begingroup$ The first line of your answer says,"when acceleration is not constant. Then how can you say that it is constant and a(x)=a? $\endgroup$ – Saurabh Raje Mar 28 '13 at 9:34
  • 1
    $\begingroup$ @SaurabhRaje No it doesn't get stuck. You need to know the force. If you know the force is for example $F=-kx$, then $a(x)=-(k/m)x$ and you integrate that. If you don't know what kind of force acting on the object, you will not know how it is going to move. $\endgroup$ – JKL Mar 29 '13 at 23:38

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 $v$, $$at=v-u$$ $$v=u+at$$


  • $\begingroup$ please read the question once more. It is more specific than you think $\endgroup$ – Saurabh Raje Aug 11 '13 at 14:45

protected by Qmechanic Aug 11 '13 at 19:10

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