Another Look at Constant Acceleration — Question 5

Question 5

The zero-acceleration case exposes lost information

A derivation of the constant-acceleration equations begins with $$v=v_0+at,\qquad x-x_0=v_0t+\tfrac12at^2.$$ It then substitutes $t=(v-v_0)/a$ to obtain $v^2=v_0^2+2a(x-x_0)$. A student says the final relation is a complete replacement for both original equations, even when $a=0$.

Consider a smooth motion with $a=0$, $v_0=-4\ \mathrm{m/s}$, and $x_0=3\ \mathrm{m}$ over 0–5 s.

1. Identify the restriction on the substitution step. Independently integrate $dv/dt=0$ and $dx/dt=v$ to determine the actual motion when $a=0$.

2. Find the final position and velocity in the example. What does the squared-velocity relation alone say about them when $a=0$?

3. Why is choosing $v=+4\ \mathrm{m/s}$ from that squared relation not a valid later state for the given smooth zero-acceleration motion?

4. Explain how the squared relation can remain true at $a=0$ even though the displayed division-based derivation does not cover that case. Distinguish a necessary relation from a complete trajectory description.

Question 5 - Solution

The zero-acceleration case exposes lost information

(a) Division by acceleration requires nonzero acceleration.

The substitution $t=(v-v_0)/a$ is valid only for $a\ne0$. For $a=0$, direct integration gives $v=C_1$. The initial velocity fixes $C_1=v_0$, and a second integration gives $$\boxed{v(t)=v_0,\qquad x(t)=x_0+v_0t}.$$ These equations require no division by acceleration.

(b) Retain elapsed time to determine position.

At 5 s, $$\boxed{v=-4\ \mathrm{m/s}},\qquad\boxed{x=3-4(5)=-17\ \mathrm{m}}.$$ The squared equation reduces only to $v^2=v_0^2=16\ \mathrm{m^2/s^2}$. It allows the algebraic values $v=\pm4\ \mathrm{m/s}$ and contains no position or elapsed-time constraint at all when $a=0$.

(c) The differential equation and initial condition fix the sign.

$$v(t)-v(0)=\int_0^t0\,ds=0\quad\Rightarrow\quad v(t)=-4\ \mathrm{m/s}\ \text{throughout}.$$

Since $dv/dt=0$ throughout the interval, velocity is constant and remains its initial value $-4\ \mathrm{m/s}$. A continuous change to $+4\ \mathrm{m/s}$ would require intervening velocity changes; a jump would violate the stipulated smooth motion. Squaring has discarded information that the original evolution equation retains.

(d) An incomplete derivation need not make its final identity false.

One can independently differentiate $v^2-2ax$ in time for constant $a$ to obtain $2va-2av=0$, including $a=0$. This proves that the squared relation is necessary in that case too, without the invalid division.

However, the relation alone is not sufficient to reconstruct the full motion, particularly when $a=0$. A valid identity can omit timing, position or sign information. Keep the initial conditions and an appropriate time-evolution equation when selecting a physical trajectory.