Another Look at Constant Acceleration — Question 4

Question 4

Derive the squared-velocity relation without dividing by velocity

For a smooth one-dimensional motion with constant $a$, define $I(t)=v(t)^2-2ax(t)$. Use $dx/dt=v$, $dv/dt=a$ and the product rule. A differentiable function whose derivative is zero throughout an interval is constant there.

As an example, take $x(0)=-2\ \mathrm{m}$, $v(0)=+6\ \mathrm{m/s}$ and $a=-3\ \mathrm{m/s^2}$ for 4 s, allowing reversal.

1. Differentiate $I$ with respect to time to derive $v^2=v_0^2+2a(x-x_0)$. Why does this derivation remain valid when $v=0$?

2. Find the invariant’s value and the turning position and time in the example. Check the squared-velocity equation at the turn.

3. Some derivations use $dv/dx=a/v$. Explain why that expression cannot be used directly at the turn, and why velocity is not a single-valued function of position over the entire trajectory.

4. Do the two signs of $v$ on the curve represent two different constant-acceleration laws? Explain what selects the branch at a particular event.

Question 4 - Solution

Derive the squared-velocity relation without dividing by velocity

(a) Differentiate with respect to time, without dividing.

$$\frac{dI}{dt}=2v\frac{dv}{dt}-2a\frac{dx}{dt}=2va-2av=0.$$ Thus $I(t)=I(0)$ and $\boxed{v^2=v_0^2+2a(x-x_0)}$. At $v=0$ both terms in the derivative vanish; there is no division by zero. Smooth $x(t)$ and $v(t)$ suffice even at a turnaround.

(b) Evaluate the invariant and the turning event.

$$I=36-2(-3)(-2)=\boxed{24\ \mathrm{m^2/s^2}}.$$ Consequently $v^2=24-6x$ in numerical SI units. Setting $v=0$ gives $\boxed{x=4\ \mathrm{m}}$. The velocity $6-3t$ vanishes at $\boxed{t=2\ \mathrm{s}}$. At the turn the invariant is $0-2(-3)(4)=24\ \mathrm{m^2/s^2}$, unchanged.

(c) A spatial reparametrization can fail while the time motion is smooth.

Away from a turn, the chain rule allows $dv/dx=(dv/dt)/(dx/dt)=a/v$. At the turn its denominator vanishes, so that formula cannot be evaluated there. The point visits positions between $-2$ and 4 m once eastward and once westward. At a repeated position the two velocities have opposite signs. Therefore a single-valued $v(x)$ cannot describe both branches at once.

(d) The branches belong to one time history.

$$v(t)=6-3t;\quad v=+\sqrt{24-6x}\ (t<2),\qquad v=-\sqrt{24-6x}\ (t>2)\quad\text{(SI)}.$$

Both signs lie on the same invariant curve for the same constant acceleration. Time evolution and the initial condition select the branch: $v>0$ before 2 s, $v=0$ at 2 s, and $v<0$ afterward. The invariant constrains allowed states but does not label their times or replace the signed equation $v=v_0+at$.