Another Look at Constant Acceleration — Question 10

Question 10

An average-position measurement can determine a trajectory

A sensor observes a point for $T=4\ \mathrm{s}$ on an east-positive line. It records $x(0)=0$ and final velocity $v(T)=4\ \mathrm{m/s}$. Its other output is the time-averaged position, $$\langle x\rangle=\frac1T\int_0^T x(t)\,dt=4\ \mathrm{m}.$$ This output has units of length; it is not average velocity. Assume one unknown constant acceleration $a$ and one unknown initial velocity $u$ throughout the interval. Integrating the motion equations gives $v=u+at$ and $x=ut+\tfrac12at^2$.

1. Integrate the position function to express $\langle x\rangle$ in terms of $u,a,T$. Combine that result with the final-velocity observation to determine $u$ and $a$.

2. Write the resulting motion functions and verify all three observations, including the sensor’s integral output.

3. Find final position, average velocity, average speed, and middle-time position. Compare the measured time-averaged position with the endpoint-position mean and the middle-time position.

4. If the average-position measurement were removed, would initial position and final velocity alone determine $u$ and $a$? Explain how the integral observation supplies independent information in this model.

Question 10 - Solution

An average-position measurement can determine a trajectory

(a) Integrate position, then divide by the observation duration.

$$\langle x\rangle=\frac1T\left[\tfrac12ut^2+\tfrac16at^3\right]_0^T =\boxed{\tfrac12uT+\tfrac16aT^2}.$$ With numerical SI values, the two measurements require $$4=u+4a,\qquad4=2u+\tfrac83a.$$ Substitution gives $4=8-\tfrac{16}{3}a$, so $\boxed{a=0.75\ \mathrm{m/s^2}}$ and $\boxed{u=1\ \mathrm{m/s}}$.

(b) Verify the differential and measurement constraints.

$$\boxed{v(t)=1\ \mathrm{m/s}+(0.75\ \mathrm{m/s^2})t},\qquad \boxed{x(t)=(1\ \mathrm{m/s})t+(0.375\ \mathrm{m/s^2})t^2}.$$ These give $x(0)=0$, $v(4\ \mathrm{s})=4\ \mathrm{m/s}$ and constant acceleration 0.75 $\mathrm{m/s^2}$. With $t$ numerical seconds, $$\frac14\int_0^4(t+0.375t^2)\,dt =\frac14(8+8)=\boxed{4\ \mathrm{m}},$$ matching the sensor.

(c) Different averages use different functions and operations.

Final position is $\boxed{x(4)=10\ \mathrm{m}}$. Velocity stays positive, so traveled distance is 10 m. Thus $\boxed{\bar v=\bar s=2.5\ \mathrm{m/s}}$. The endpoint-position mean is $(0+10)/2=\boxed{5\ \mathrm{m}}$, while $\boxed{x(2)=3.5\ \mathrm{m}}$.

Thus the time-averaged position (4 m), endpoint-position mean (5 m), and middle-time position (3.5 m) are distinct. None is the average velocity, whose units and definition are different.

(d) The integral constraint resolves a remaining degree of freedom.

$$\langle x\rangle=\frac{T}{2}[v(T)-aT]+\frac{aT^2}{6}=\frac{Tv(T)}2-\frac{aT^2}{3}.$$

Without it, $u+aT=v(T)$ gives only $u=4\ \mathrm{m/s}-(4\ \mathrm{s})a$, a family of possible pairs. Initial position fixes a position integration constant but not these two parameters. Substituting this family into the position average gives $\langle x\rangle=Tv(T)/2-aT^2/3$, which has a nonzero coefficient of $a$ for $T>0$. The measured average therefore selects one acceleration and then one initial velocity under the stated constant-acceleration model.