Another Look at Constant Acceleration — Question 9

Question 9

Midpoint velocity and midpoint position obey different identities

For a constant-acceleration trajectory on $0\leq t\leq T$, derive results from $$v(t)=v_0+at,\qquad x(t)=x_0+v_0t+\tfrac12at^2.$$ A particular trajectory has $T=4\ \mathrm{s}$, $x(0)=2\ \mathrm{m}$, $x(2\ \mathrm{s})=8\ \mathrm{m}$ and $x(4\ \mathrm{s})=18\ \mathrm{m}$.

1. Integrate the linear velocity to prove $\bar v=v(T/2)$ exactly. Does this imply $x(T/2)=[x(0)+x(T)]/2$?

2. Derive the difference between the endpoint-position mean and the actual middle-time position. Use the measured difference to infer $a$ and then $v_0$.

3. Determine the middle-time velocity and compare it with the full-interval average. Explain why averaging endpoint positions and averaging endpoint velocities behave differently here.

4. Would either midpoint identity be guaranteed for an arbitrary smooth, variable-acceleration trajectory? Identify the specific polynomial structure used in the derivations.

Question 9 - Solution

Midpoint velocity and midpoint position obey different identities

(a) The integral mean of an affine velocity equals its middle value.

$$\bar v=\frac1T\int_0^T(v_0+at)\,dt=v_0+\tfrac12aT=\boxed{v(T/2)}.$$ This is a velocity identity; it does not assert that the middle-time position is the mean of the endpoint positions.

(b) Subtract the two position expressions.

The endpoint mean is $x_0+v_0T/2+aT^2/4$, while the middle-time position is $x_0+v_0T/2+aT^2/8$. Thus $$\boxed{\frac{x(0)+x(T)}2-x(T/2)=\frac{aT^2}{8}}.$$ The measured difference is $(2+18)/2-8=2\ \mathrm{m}$. Hence $\boxed{a=1\ \mathrm{m/s^2}}$. Using $18-2=4v_0+\tfrac12(1)(4)^2$ then gives $\boxed{v_0=2\ \mathrm{m/s}}$.

(c) Check the velocity mean without substituting a position mean.

$$x(t)=2+2t+\tfrac12t^2,\quad v(t)=2+t;\qquad x(2)=8\ne10=\frac{x(0)+x(4)}2\quad\text{(SI)}.$$

At 2 s, $v=2+1(2)=\boxed{4\ \mathrm{m/s}}$, equal to $(18-2)/4=4\ \mathrm{m/s}$. The velocity function is affine in time, while the position function contains a quadratic term. That term causes the actual middle position, 8 m, to differ from the endpoint-position mean, 10 m.

(d) State the scope of the identities.

The results follow from linear velocity and quadratic position under one constant acceleration. Arbitrary smooth variable-acceleration histories need not have either property, so neither identity is guaranteed in that broader class. Agreement for a special nonconstant history could occur accidentally and would not prove constant acceleration.