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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.

Qmechanic
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user3104311
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2 Answers2

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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.

NLambert
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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.

Simon S
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