Constant Acceleration - A Special Case — Question 1

Question 1

Endpoint averaging needs a motion model

For one-dimensional motion with constant acceleration $a$ over elapsed time $T$, use $$v_f=v_i+aT,\qquad \Delta x=v_iT+\tfrac12aT^2,\qquad \bar v=\frac{\Delta x}{T}.$$ A cart starts at $x_i=10\ \mathrm{m}$ with $v_i=+4\ \mathrm{m/s}$ and has $a=+3\ \mathrm{m/s^2}$ for 2 s on an east-positive line.

1. Find final velocity, displacement, final position and average velocity. Explain why $v_fT$ does not give the displacement here.

2. Eliminate $aT$ between the two supplied equations to prove $\bar v=(v_i+v_f)/2$ for this constant-acceleration interval.

3. Consider a different complete history $x(t)=10+4t+\tfrac12t^3$, with $t$ numerical seconds and $x$ numerical metres, over 0–2 s. Differentiate to check its endpoint velocities and its acceleration. Find its actual displacement and average velocity.

4. Both motions have the same duration and endpoint velocities. Explain why only one is entitled to use the endpoint-velocity arithmetic mean on the basis of its stated model.

Question 1 - Solution

Endpoint averaging needs a motion model

(a) Distinguish a final rate from an interval average.

$$v_fT-\Delta x=(v_i+aT)T-\left(v_iT+\tfrac12aT^2\right)=\tfrac12aT^2=6\ \mathrm{m}.$$

$$\boxed{v_f=4+3(2)=10\ \mathrm{m/s}},\qquad \boxed{\Delta x=4(2)+\tfrac12(3)(2)^2=14\ \mathrm{m}}.$$ Thus $\boxed{x_f=24\ \mathrm{m}}$ and $\boxed{\bar v=7\ \mathrm{m/s}}$. Multiplying $v_fT$ would give 20 m, as though the cart had traveled at its final velocity for the entire interval. It actually starts more slowly.

(b) The constant-acceleration equations imply the endpoint mean.

Divide displacement by $T>0$: $$\bar v=v_i+\tfrac12aT =v_i+\tfrac12(v_f-v_i)=\boxed{\frac{v_i+v_f}{2}}.$$ This follows from the constant-acceleration model, not merely from knowing two velocities.

(c) Check the different history directly.

For the cubic position history, differentiation gives $$v(t)=4+\tfrac32t^2\ \mathrm{m/s},\qquad a(t)=3t\ \mathrm{m/s^2},$$ with the stated numerical-time convention. Its endpoint velocities are also 4 and 10 m/s, but acceleration is not constant. Its positions are $x(0)=10$ and $x(2)=22\ \mathrm{m}$, so $$\boxed{\Delta x=12\ \mathrm{m}},\qquad\boxed{\bar v=6\ \mathrm{m/s}}.$$

(d) Matching endpoint data do not establish the intervening model.

The arithmetic mean of the endpoint velocities is 7 m/s in both cases, but it equals the actual average only for the first supplied history. The cubic example disproves any general inference of displacement from duration and endpoint velocities using this shortcut. Some nonconstant histories may happen to give the same mean; the point is that endpoint data alone do not guarantee it.