Question about derivation of kinematics equations

Question

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?

Answer 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.

Answer 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.

Answer 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.)

Answer 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}

Answer 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

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

Answer 6

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.