Another Look at Constant Acceleration — Question 3

Question 3

Boundary data can determine the missing integration constants

A point has one unknown constant acceleration throughout $1\leq t\leq4\ \mathrm{s}$. Measurements give $$x(1\ \mathrm{s})=3\ \mathrm{m},\quad x(4\ \mathrm{s})=15\ \mathrm{m},\quad v(4\ \mathrm{s})=7\ \mathrm{m/s}.$$ Let $\tau=t-1\ \mathrm{s}$ and $u=v(1\ \mathrm{s})$. Starting from $dv/d\tau=a$ and $dx/d\tau=v$, integration gives $v=u+a\tau$ and $x=3\ \mathrm{m}+u\tau+\tfrac12a\tau^2$.

1. Use both final observations to determine $u$ and $a$. Write the full position and velocity functions and verify all three observations.

2. Find the velocity at the middle time, 2.5 s, and compare it with the average velocity from 1 to 4 s. Explain the equality from the derived functions.

3. Remove the final-velocity observation but retain both positions and the constant-acceleration hypothesis. Express the allowed initial velocity as a function of $a$. Give two different constant-acceleration histories matching the two positions.

4. Explain why two position observations do not supply the same information as an initial position and velocity when the acceleration itself is also unknown.

Question 3 - Solution

Boundary data can determine the missing integration constants

(a) Apply both endpoint constraints to the integrated functions.

At $\tau=3\ \mathrm{s}$, $$7=u+3a,\qquad12=3u+\tfrac92a,$$ with numerical SI values. Substituting $u=7-3a$ gives $12=21-\tfrac92a$, so $$\boxed{a=2\ \mathrm{m/s^2}},\qquad\boxed{u=1\ \mathrm{m/s}}.$$ The dimensional functions are $$\boxed{v=1\ \mathrm{m/s}+(2\ \mathrm{m/s^2})\tau},\qquad \boxed{x=3\ \mathrm{m}+(1\ \mathrm{m/s})\tau+(1\ \mathrm{m/s^2})\tau^2}.$$ At $\tau=0$, $x=3\ \mathrm{m}$; at $\tau=3\ \mathrm{s}$, $x=15\ \mathrm{m}$ and $v=7\ \mathrm{m/s}$, as required.

(b) Linear velocity makes the middle value equal the interval mean.

$$\bar v=\frac1T\int_0^T(u+a\tau)\,d\tau=u+\frac{aT}{2}=v(T/2).$$

At $\tau=1.5\ \mathrm{s}$, $v=1+2(1.5)=\boxed{4\ \mathrm{m/s}}$. The average is $(15-3)/3=\boxed{4\ \mathrm{m/s}}$. Algebraically, integrating $u+a\tau$ over $0\leq\tau\leq T$ and dividing by $T$ gives $u+aT/2$, the velocity at the middle time.

(c) Without the third datum, one parameter remains free.

The position constraint alone gives $12=3u+\tfrac92a$, or $$\boxed{u=4\ \mathrm{m/s}-(1.5\ \mathrm{s})a}.$$ For $a=0$, $u=4\ \mathrm{m/s}$ and the position is $3+4\tau$ in numerical SI units. For $a=2\ \mathrm{m/s^2}$, $u=1\ \mathrm{m/s}$ and the position is the quadratic above. Both fit the two positions but have different final velocities, 4 and 7 m/s.

(d) Count the unknown physical information, not only the observations.

The initial position is known, but both the initial velocity and constant acceleration are unknown. The second position imposes only one relation between them. A further independent datum, here final velocity, is needed to fix both. Calling acceleration constant specifies its time dependence, not its numerical value.