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I'm an engineering student and none of my professors ever explained why do we use derivations and/or integrations in physics. So I have this task, it goes like:

The object is moving in a positive direction $x$, where the velocity of the body is changed according to the law: $v(x)=b\times\sqrt{x}$, where $b$ is a positive constant. Define the following dependencies:

a) the speed and acceleration of the object on the time $t$,

b) the average velocity of the object on the path-$x$.

Now, I'm not asking you to solve it, I just need an explanation why do I need to derive/integrate that equation in order to get speed and acc.

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closed as off-topic by ja72, BMS, JamalS, Prahar, Jim Nov 10 '14 at 14:19

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    $\begingroup$ Do you recall the definition of the derivative in calculus? Now compare to the definition of velocity in the limit of a short time interval. Yes, the instructor or the text should mention these things but you are also expected to recall what has gone before and think about how they related to one another. $\endgroup$ – dmckee Nov 2 '14 at 18:57
  • $\begingroup$ possible duplicate of How to get distance when acceleration is not constant? $\endgroup$ – ja72 Nov 3 '14 at 1:18
  • $\begingroup$ Do you understand the concept of independent vs. dependent quantities? Differential calculus describes the changes in a system undergoing infinitesimal changes of the independent variables. $\endgroup$ – ja72 Nov 3 '14 at 1:20
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The job of calculus is to handle quantities that vary over the domain of the problem at hand. Often, and particularly in introductory physics, we care about quantities that vary in time. We cannot put them into our equations as constants.

We also often care about the interrelationship of these quantities. So for example now velocity is given by $v=\frac{dx}{dt}$ and acceleration by $a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$. This is telling is that velocity is the rate of change of position with respect to time. Acceleration is the rate of change of velocity with respect to time. We knew this anyway, but writing it like this allows us to establish an algebraic relationship between these quantities. If we know $x(t)$, we also know $v(t)$ and $a(t)$.

In the case of the problem you are facing, instead of writing $x=v t_f$, (where $t_f$ is the time elapsed) we can now write

$$x=\int_0^{t_f}v(t)dt=\int_0^{t_f}\frac{dx}{dt}dt$$

which allows us to handle $v(t)$ which varies over time.

In short, in a world where quantities are rarely constant - they vary with time, or space, or energy, or any of a thousand other parameters - calculus is the engine which drives all of physics.

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How else can you recover displacement from a non-linear velocity without integrating? How else would you calculate acceleration of a non-linear velocity without differentiating?


ADDED: If your question is how to solve $\displaystyle v(t) = \frac{dx}{dt} = b\sqrt{x}$, let us know.

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  • $\begingroup$ Well I don't know, we didn't do integrals and derivations, so I don't know their purpose... $\endgroup$ – user3104311 Nov 2 '14 at 18:08
  • $\begingroup$ In general, if a particle moving in 1D has position $x(t)$, then its velocity is $v(t) = dx/dt$ and acceleration $a(t) = dv/dt = d^2x/dt^2$. In your problem you have that $dx/dt = b\sqrt{x}$. You'll have to solve that ordinary differential equation to find a general expression for $x(t)$. $\endgroup$ – Simon S Nov 2 '14 at 18:10
  • $\begingroup$ No, not really, I was just wondering what is the purpose of differentials, and when do I need to integrate/derive an equation.(response to ADDED :) ) $\endgroup$ – user3104311 Nov 2 '14 at 18:22

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