# Question about derivation of kinematics equations

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?

• 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 $v(t) = at$"? – Alfred Centauri Sep 3 at 1:46
• @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 Sep 3 at 1:51
• 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? – ChemSniper Sep 3 at 2:03
• 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? – Alfred Centauri Sep 3 at 2:17
• 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. – knzhou Sep 3 at 20:33

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.

• I think you may have forgotten the put a d in front of the final t in your integral :) ( in order to have $\int v(t)dt$ ) – Q.Reindeerson Sep 3 at 11:46
• @Q.Reindeerson indeed, fixed, thanks! – Allure Sep 3 at 11:47

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.

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.)

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:

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

• Isn't it $x(t) = \int_0^t\mathrm{d}\tau\,v(\tau) + x(0)$? – Alfred Centauri Sep 3 at 1:52
• 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? – ChemSniper Sep 3 at 1:54
• @AlfredCentauri yes, it is. I will correct. – Eduardo Sep 3 at 2:01
• @ChemSniper try to think about it graphically to see why multiplying by not constant v(t) is wrong. – Eduardo Sep 3 at 2:02
• 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. – Alfred Centauri Sep 3 at 2:35

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

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

Each derivation rests on the assumptions used. The standard kinematics equations you mention first, depend on the assumption of constant acceleration.

My problem is that your friend hasn't stated what assumptions he used to get to $$x(t) = x_0 + v(t)\,t$$.

So let us differentiate both sides to see what kind of acceleration is needed (using the product rule)

$$\frac{{\rm d}}{{\rm d}t} x(t) = t \frac{{\rm d}}{{\rm d}t} v(t) + v(t) \frac{{\rm d}}{{\rm d}t} t$$ $$v(t) = t a(t) + v(t)$$ $$a(t) = 0 \;\text{or}\; t=0$$

So this equation is only valid for zero acceleration (or constant velocity). This makes his third line inconsistent to the first two lines.