Question about derivation of kinematics equations

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Solution 1

Given velocity $v(t)$, the distance moved after a certain time $t$ is not $v(t)t$ - this formula works at constant velocity, but when the velocity is changing, the correct expression is $\int^{t_f}_{t_0} v(t) dt$. Therefore your friend's third line is incorrect.

Solution 2

The error is just that $v(t)t$ is not the anti-derivative of $at$. This is easily checked by just taking the derivative. $$\frac{\text d}{\text dt}\left(v(t)\cdot t\right)=v(t)\cdot\frac{\text d}{\text dt}(t)+t\cdot\frac{\text d}{\text dt}(v(t))=v(t)+at\neq at$$

It's a simple calculus mistake.

Solution 3

To give a purely qualitative answer consider the meaning of your friends third line $$ x(t) = x_0 + v(t) \cdot t \;, \tag{1} $$ (where I've made the multiplication explicit).

This claims that you find the position at moment $t$ by taking the initial position ($x_0$) and adding to that the elapsed time times the velocity the particle has at moment $t$.

So, your friends procedure ignore the fact that the particle had different speeds in between the start of the clock and time $t$. The factor of $1/2$ on the quadratic term in the correct formula accounts for the fact that velocity has been changing all along. (Note that it is a factor of $1/2$ only because the acceleration was constant. Allow acceleration to change and you get still more complicated expressions.)

Solution 4

The displacement is only the velocity multiplied by the elapsed time if velocity is constant as you suggested. To derive the equation for varying velocity you must consider the infinitesimal case where the elapsed time is so small that you can consider velocity constant. In this case, a small displacement $dx$ is given by the product of velocity v(t) by small increment of time $dt$. So, the correct derivation would be:

\begin{equation} dx = v \, dt \end{equation}

\begin{equation} \int_{x_0}^{x(t)} dx = \Delta x = x(t) - x_0 = \int_{t_0}^{t_F} v(t) \, dt = \int_{t_0}^{t_F} v_0 + a\, t \, dt \end{equation}

\begin{equation} x(t) = x_0 + v_ 0 \, \Delta t + a \frac{\Delta t^2}{2} \end{equation}

with $\Delta t$ = $t_F$ - $t_0$.

You can visualize this graphically by thinking on the velocity by time graph. As velocity increases linearly with time, the area below the graph will be a triangle with height $a\, \Delta t$ and width $\Delta t$ plus a rectangle with height $v_0$ and width $\Delta t$. So the area is:

\begin{equation} Area = \Delta x = x(t) - x_0 = v_0 \, \Delta t + \frac{(a \, \Delta t) \, \Delta t}{2} \end{equation}

Solution 5

I think I understand it now, mathematically speaking, but is there a more conceptual answer?

OP evidently seeks a conceptual answer to why $x(t) \ne x_0 + v(t)\cdot t$ when $v(t) = v_0 + at$ and $a$ is a constant.

Consider the simple case that the initial position and initial velocity are zero. Stipulate that $v(t) = at$ where $a$ is a constant and it follows that the average velocity over the time from $t=0$ to t is

$$\bar{v} =\frac{1}{2}at < at$$

But, intuitively,

$$\Delta x = \bar{v}\cdot\Delta t$$

and so

$$x = \frac{1}{2}at^2$$

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Wanna-be philosopher, especially ethics. Currently reading about Human Rights. Other hobbies include coding (Java, but learning C++), language (Amo Linguam Latinam), and chemistry. Current HS LD resolution: Immigration ought to be recognized as a human right. Ehh...better than carbon tax, worse than secession. Words to Live By: "Eschew Surplusage" - Mark Twain "Everything should be made as simple as possible, but not simpler." - Einstein "Immigrants, we get the job done!" - Marquis de Lafayette &amp; Alexander Hamilton (Just you wait...)

Updated on August 01, 2022

Comments

  • ChemSniper
    ChemSniper over 1 year

    Apologies if this has been asked before, but I browsed the sub and couldn't find something specific.

    I understand the derivation for one of the equations as follows: \begin{gather} \frac{dv}{dt} = a \\ v(t) = v_0 + at \\ \frac{dx}{dt} = v_0 + at \\ x(t) = x_0 + v_0t + \frac{1}{2} at^2 \end{gather}

    But my friend today used a different derivation for a problem: \begin{gather} \frac{dv}{dt} = a \\ v(t) = v_0 + at \\ x(t) = x_0 + v(t)t \\ x(t) = x_0 + v_0t + at^2 \end{gather}

    Why exactly is my friend's third line incorrect?

    • Alfred Centauri
      Alfred Centauri about 4 years
      Welcome New contributor ChemSniper! I don't understand why you're asking "What exactly is wrong in the derivation" when you've already identified what is wrong in the derivation. You've stated that you don't know how to justify it properly. Is your question "why isn't $\Delta x = v(t)\cdot \Delta t$ when $v(t) = at$"?
    • ChemSniper
      ChemSniper about 4 years
      @AlfredCentauri I think so - I didn't know that I identified correctly because I do not see an error in their work. There must be, since they arrive at the wrong answer, but I don't understand what is wrong
    • ChemSniper
      ChemSniper about 4 years
      Yes, I see, thank you...But don't you get \dot{x}(t) = at + v(t) or 0 = at? I think I understand it now, mathematically speaking, but is there a more conceptual answer?
    • Alfred Centauri
      Alfred Centauri about 4 years
      I'm still not sure what conceptual answer you're looking for. Assume, for simplicity, the initial position and velocity are zero and the acceleration is the constant $a\,\mathrm{m/s^2}$. The velocity started at zero and, at the end of one second, the velocity is $a\,\mathrm{m/s}$ correct? Think about it ... what is the average velocity over this time? Isn't it $\bar{v} = \frac{1}{2}a\,\mathrm{m/s}$? What does that imply about the distance traveled in one second?
    • knzhou
      knzhou about 4 years
      The amount you have moved is not equal to $v(t) t$. For example, that says that if you stop your car at a traffic light, so $v(t) = 0$ for a moment, then you must be back where you started your trip.
    • alephzero
      alephzero about 4 years
      The third line would be right if $v(t)$ was constant. But asking "why this is wrong" is impossible to answer, because we have no idea what your friend was thinking when he/she wrote it.. $x(t) = x_0 +a \sin(\frac{v(t)}{2\pi})$ is also wrong, and nobody can explain why that is wrong either!
  • Alfred Centauri
    Alfred Centauri about 4 years
    Isn't it $x(t) = \int_0^t\mathrm{d}\tau\,v(\tau) + x(0)$?
  • ChemSniper
    ChemSniper about 4 years
    Thank you for your quick answer - I understand that that is the correct derivation, as I originally showed. I am asking for an explanation for why substituting for v(t)*t as my friend did was incorrect. Why does v(t) not being constant make a difference?
  • Eduardo
    Eduardo about 4 years
    @AlfredCentauri yes, it is. I will correct.
  • Eduardo
    Eduardo about 4 years
    @ChemSniper try to think about it graphically to see why multiplying by not constant v(t) is wrong.
  • Alfred Centauri
    Alfred Centauri about 4 years
    Eduardo, you have a function of time $t$ equal to an integral of a function of $t$ over all time which is nonsense. You need a dummy variable (like $\tau$) to integrate over, and you need the variable $t$ to be a limit of integration. See my first comment.
  • Q.Reindeerson
    Q.Reindeerson about 4 years
    I think you may have forgotten the put a d in front of the final t in your integral :) ( in order to have $\int v(t)dt$ )
  • Allure
    Allure about 4 years
    @Q.Reindeerson indeed, fixed, thanks!