Instantaneous Velocity and Speed — Question 8

Question 8

Changing coordinates does not change the physical motion

A point has east-positive position $$x(t)=(4\ \mathrm{m})+(2\ \mathrm{m/s})t-(1\ \mathrm{m/s^2})t^2.$$ A second observer uses the fixed west-positive coordinate $y=10\ \mathrm{m}-x$. A recorder uses a display coordinate $q=100\ \mathrm{tick}+(2\ \mathrm{tick/s})t$. The tick is a display unit; 2 ticks represent 1 second.

You may differentiate the polynomial in $t$. For any two nearby events, use $\Delta y=-\Delta x$ and $\Delta q=(2\ \mathrm{tick/s})\Delta t$ to relate derivative ratios.

1. At the physical event $t=3\ \mathrm{s}$, find $x$, $y$, $q$, the east-positive velocity $dx/dt$, and the physical speed.

2. Find $dy/dt$ at that event and state the physical direction it represents. Has reversing the axis changed the speed in m/s?

3. Find $dx/dq$ and $dy/dq$, with units. Convert each to m/s and explain why a slope reported in metres per tick must not be labeled metres per second.

4. Explain separately the effects of a fixed position-origin shift, axis reversal, clock offset, and clock scale change on a reported derivative.

Question 8 - Solution

Changing coordinates does not change the physical motion

(a) Locate the event and differentiate in physical seconds.

At 3 s, $$\boxed{x=1\ \mathrm{m},\quad y=9\ \mathrm{m},\quad q=106\ \mathrm{tick}}.$$ The position derivative is $dx/dt=2\ \mathrm{m/s}-(2\ \mathrm{m/s^2})t$, yielding $$\boxed{dx/dt=-4\ \mathrm{m/s}},\qquad\boxed{s=4\ \mathrm{m/s}}.$$ The point moves west at this event.

(b) The component sign follows the chosen positive direction.

Because $\Delta y=-\Delta x$, dividing by the same time increment and taking the limit gives $dy/dt=-dx/dt$. Hence $$\boxed{dy/dt=+4\ \mathrm{m/s}}.$$ Positive $y$ points west, so this is the same physical westward motion. Taking magnitudes gives the same speed of 4 m/s.

(c) A scale change changes the numerical derivative and its unit.

For nonzero time increments, $$\frac{\Delta x}{\Delta q}=\frac{\Delta x/\Delta t}{2\ \mathrm{tick/s}}.$$ Taking the limit yields $\boxed{dx/dq=-2\ \mathrm{m/tick}}$ and $\boxed{dy/dq=+2\ \mathrm{m/tick}}$. Multiplying by $2\ \mathrm{tick/s}$ recovers $-4$ and $+4\ \mathrm{m/s}$, respectively. Labeling a value of 2 m/tick as 2 m/s would omit the display calibration.

(d) Separate offsets, signs and scales.

Write a general fixed spatial transformation with $\sigma=\pm1$ and a clock calibration with nonzero scale $\alpha$: $$\widetilde x=\sigma x+c,\quad\widetilde t=\alpha t+\beta\quad\Rightarrow\quad\frac{d\widetilde x}{d\widetilde t}=\frac\sigma\alpha\frac{dx}{dt}\quad(\alpha\ne0).$$

A fixed spatial offset cancels in position differences. Reversing the spatial axis changes the derivative’s sign. A constant clock offset cancels in time differences. A clock scale change rescales the numerical denominator and hence the derivative’s numerical value, with its unit changed accordingly. None of these coordinate changes alters the actual motion or the correctly converted physical speed.