Another Look at Constant Acceleration — Question 7

Question 7

Endpoint agreement does not prove the differential equation

For $0\leq t\leq2$, consider two smooth position histories, with $t$ numerical seconds and $x$ numerical metres: $$x_A(t)=1+2t+\tfrac12t^2,\qquad x_B(t)=x_A(t)+\tfrac14t^2(t-2)^2.$$ A report claims that both describe constant acceleration $+1\ \mathrm{m/s^2}$ because they agree in position and velocity at both endpoints. You may differentiate polynomials twice.

1. Check the endpoint positions and velocities of both histories. Explain why the added term and its first derivative vanish at both endpoints.

2. Evaluate both positions and accelerations at $t=1\ \mathrm{s}$. Does the endpoint report establish the claimed constant acceleration for B?

3. Let two twice-differentiable histories $x_1,x_2$ actually satisfy $x_1''=x_2''=a$ throughout an interval and share initial position and velocity. Apply two integrations to $y=x_1-x_2$ to prove that their positions must agree everywhere in that interval.

4. Which hypothesis of that uniqueness argument fails for A and B? Explain why an endpoint fit is weaker than verifying the differential equation throughout the motion.

Question 7 - Solution

Endpoint agreement does not prove the differential equation

(a) Double endpoint zeros preserve both position and first derivative.

The added term has factors $t^2$ and $(t-2)^2$, so it and its first derivative vanish at 0 and 2. Directly, both histories give $$\boxed{x(0)=1\ \mathrm{m},\ v(0)=2\ \mathrm{m/s}},\qquad \boxed{x(2)=7\ \mathrm{m},\ v(2)=4\ \mathrm{m/s}}.$$ Agreement of these four numbers says nothing by itself about the second derivative at intermediate times.

(b) Test the acceleration rather than inferring it from the endpoint fit.

The perturbation expands to $\tfrac14(t^4-4t^3+4t^2)$, whose second derivative is $3t^2-6t+2$ in numerical SI units. Thus $$a_A=1\ \mathrm{m/s^2},\qquad a_B=(3t^2-6t+3)\ \mathrm{m/s^2}.$$ At 1 s, $$\boxed{x_A=3.5\ \mathrm{m},\ x_B=3.75\ \mathrm{m}},\qquad \boxed{a_A=1\ \mathrm{m/s^2},\ a_B=0\ \mathrm{m/s^2}}.$$ B has variable acceleration and does not satisfy the claimed constant-acceleration law.

(c) Actual dynamics plus initial data give uniqueness.

$$y''=0\Rightarrow y'=C_1\Rightarrow y=C_1t+C_2;\qquad y(0)=y'(0)=0\Rightarrow y\equiv0.$$

If $y=x_1-x_2$, then $y''=0$. Integrating once gives $y'=C_1$. Equal initial velocities imply $y'=0$ initially, hence $C_1=0$. Integrating again gives $y=C_2$, and equal initial positions imply $C_2=0$. Therefore $\boxed{x_1(t)=x_2(t)}$ throughout the interval.

(d) Endpoint agreement cannot replace a condition at all times.

A and B have the required common initial data, but they do not share the same acceleration throughout the interval. The differential-equation hypothesis fails. Extra endpoint agreement does not repair that failure: a function can be altered between observations while leaving all supplied endpoint values and slopes intact.