Average Velocity and Average Speed — Question 9

Question 9

Average velocities add across moving references; speeds need not

A platform moves east along a fixed straight ground axis. Its origin has ground position $X_O=2t$, with metres and seconds understood. A cart’s east-positive coordinate on the platform is $x'$. At each instant, $$X_{\mathrm{cart}}=X_O+x'.$$ During 0–5 s, $x'$ changes linearly from 0 to $-15\ \mathrm{m}$; during 5–10 s it changes linearly from $-15$ to $-10\ \mathrm{m}$. These are complete motion histories in an idealized model.

1. Find the cart’s ground positions at 0, 5, and 10 s. Determine its ground displacement and distance over 0–10 s.

2. Find the cart’s average velocity and average speed relative to the platform, and relative to the ground.

3. Derive the relation between the two average velocities over a shared time interval. Test it numerically here.

4. Would adding the platform’s average speed to the cart’s platform-relative average speed give the ground average speed? Diagnose the failure using this example.

Question 9 - Solution

Average velocities add across moving references; speeds need not

(a) Add positions at the same event.

At 0, 5, and 10 s, the platform origin is at 0, 10, and 20 m. Adding the local cart coordinates gives ground positions $\boxed{0,-5,10\ \mathrm{m}}$. The sum of two linear functions is linear on each stage, so there are no hidden ground reversals within those stages. Thus $\Delta X=10\ \mathrm{m}$ and $D_{\mathrm{ground}}=5+15=20\ \mathrm{m}$.

(b) Use the distance appropriate to each reference.

Locally, the cart has displacement $-10\ \mathrm{m}$ and distance $15+5=20\ \mathrm{m}$ in 10 s. Therefore $$\boxed{\bar v'=-1\ \mathrm{m/s},\quad\bar s'=2\ \mathrm{m/s}}.$$ For the ground history, $$\boxed{\bar v_X=+1\ \mathrm{m/s},\quad\bar s_X=2\ \mathrm{m/s}}.$$ The equal numerical speeds in this example do not imply equal displacements or equal paths.

(c) Endpoint differences add, then share one duration.

Subtracting endpoint coordinate relations gives $\Delta X=\Delta X_O+\Delta x'$. Dividing by the same positive $T$ gives $$\boxed{\bar v_X=\bar v_O+\bar v'}.$$ Here $+1=+2+(-1)$ in m/s, as required.

(d) Path lengths do not obey the same signed addition rule.

$$D_X=|10-15|+|10+5|=20\ \mathrm{m}\ne D_O+D'=20+20=40\ \mathrm{m}.$$

Adding the two proposed speeds would give $2+2=4\ \mathrm{m/s}$, whereas the actual ground average speed is 2 m/s. Magnitudes discard direction before combining motions. During the first stage local westward change partly offsets platform eastward change; during the second stage they reinforce. Compute the ground path before taking its traveled length. The velocity relation does not justify a corresponding addition of scalar average speeds.