Another Look at Constant Acceleration — Question 8

Question 8

Bound the error caused by treating acceleration as constant

A nominal model uses $x_0=0$, $v_0=3\ \mathrm{m/s}$ and constant $a_0=-1\ \mathrm{m/s^2}$. The actual acceleration is continuous and satisfies $$a(t)=a_0+\delta a(t),\qquad |\delta a(t)|\leq\varepsilon=0.20\ \mathrm{m/s^2}$$ throughout $0\leq t\leq T=4\ \mathrm{s}$. Actual and nominal motions have exactly the same initial position and velocity.

You may use the fundamental theorem of calculus and the integral bound $|\int_0^t f(s)\,ds|\leq\int_0^t|f(s)|\,ds$. These are hard bounds, not statistical uncertainties.

1. Subtract the two acceleration equations and integrate to prove $|v(t)-v_{\mathrm{nom}}(t)|\leq\varepsilon t$.

2. Integrate the velocity difference to prove $|x(t)-x_{\mathrm{nom}}(t)|\leq\tfrac12\varepsilon t^2$. Explain why using the final velocity-error bound across the whole interval would give a looser estimate.

3. Find the nominal final state and tight intervals for actual final velocity and position. Which final motion direction is guaranteed? Is that consistent with the possible final position signs?

4. Exhibit admissible acceleration functions attaining both ends of each interval. Can the extreme position and velocity errors be freely paired as though independent?

Question 8 - Solution

Bound the error caused by treating acceleration as constant

(a) The shared initial velocity removes the integration constant.

Writing $\Delta v=v-v_{\mathrm{nom}}$ gives $$\Delta v(t)=\int_0^t\delta a(s)\,ds,\qquad |\Delta v(t)|\leq\int_0^t\varepsilon\,ds=\boxed{\varepsilon t}.$$

(b) Use the time-dependent bound inside the second integral.

Since the initial positions also agree, $$\Delta x(t)=\int_0^t\Delta v(s)\,ds,\qquad |\Delta x(t)|\leq\int_0^t\varepsilon s\,ds=\boxed{\tfrac12\varepsilon t^2}.$$ Using the final bound $\varepsilon t$ at every intermediate time would instead give $\varepsilon t^2$, twice as large. The velocity error begins at zero and has a tighter bound $\varepsilon s$ at each earlier time $s$.

(c) Apply the error bounds to the nominal state.

At 4 s, $v_{\mathrm{nom}}=3-4=-1\ \mathrm{m/s}$ and $x_{\mathrm{nom}}=3(4)-\tfrac12(4)^2=4\ \mathrm{m}$. The maximum errors are 0.8 m/s and 1.6 m, so $$\boxed{v(4)\in[-1.8,-0.2]\ \mathrm{m/s}},\qquad \boxed{x(4)\in[2.4,5.6]\ \mathrm{m}}.$$ Final motion is certainly westward, even though the final position is certainly east of the initial origin. Location and direction of change are different quantities.

(d) Constant extreme deviations attain the bounds.

$$\delta a=\pm\varepsilon\Rightarrow(\Delta v(T),\Delta x(T))=(\pm\varepsilon T,\pm\tfrac12\varepsilon T^2),\quad\text{with linked signs}.$$

Choosing $\delta a(t)=+0.20\ \mathrm{m/s^2}$ throughout gives both upper endpoints; choosing $\delta a(t)=-0.20\ \mathrm{m/s^2}$ gives both lower endpoints. These continuous functions are allowed, so each stated interval is tight.

The two errors come from the same acceleration deviation and are not independent. For example, attaining the upper velocity error requires $\delta a=+\varepsilon$ throughout under the continuity and bound assumptions; it then also yields the upper position error, not the lower one. Separate marginal bounds do not make every pair of their endpoints jointly feasible.