Another Look at Constant Acceleration — Question 6

Question 6

A dimensionless quadratic still needs two scale factors

A simulation stores dimensionless time $\tau=t/T_*$ and dimensionless position $\xi=x/L_*$, where $$T_*=0.50\ \mathrm{s},\qquad L_*=2.0\ \mathrm{m}.$$ Its exact output is $\xi(\tau)=3-2\tau+4\tau^2$. Thus the number 4 is a coefficient in a dimensionless position polynomial, not an acceleration supplied in SI units.

You may use the chain rule and $d\tau/dt=1/T_*$.

1. Derive expressions for $v$ and $a$ in terms of derivatives of $\xi$ with respect to $\tau$. Track the units of both scale factors.

2. Determine physical initial position, initial velocity and constant acceleration. Write $x(t)$ with dimensional coefficients.

3. At $t=0.25\ \mathrm{s}$, determine position, velocity and speed. Has the point already reversed from its initial direction? Locate the reversal to justify your answer.

4. Diagnose a report that labels the quadratic coefficient 4 as $4\ \mathrm{m/s^2}$. Identify both the derivative factor and the unit conversion it omits.

Question 6 - Solution

A dimensionless quadratic still needs two scale factors

(a) Each time derivative introduces another inverse time scale.

Since $x=L_*\xi(t/T_*)$, $$\boxed{v=\frac{L_*}{T_*}\frac{d\xi}{d\tau}},\qquad \boxed{a=\frac{L_*}{T_*^2}\frac{d^2\xi}{d\tau^2}}.$$ The scale factors have units m/s and $\mathrm{m/s^2}$, respectively; the derivatives of the dimensionless quantities are dimensionless.

(b) Differentiate the polynomial before converting units.

Here $\xi'=-2+8\tau$ and $\xi''=8$. Therefore $$\boxed{x_0=6\ \mathrm{m},\quad v_0=-8\ \mathrm{m/s},\quad a=64\ \mathrm{m/s^2}}.$$ Substituting $\tau=t/(0.50\ \mathrm{s})$ yields $$\boxed{x(t)=6\ \mathrm{m}-(8\ \mathrm{m/s})t+(32\ \mathrm{m/s^2})t^2}.$$ Twice the quadratic coefficient gives the physical acceleration, as direct differentiation confirms.

(c) Evaluate at the physical event, then check the turnaround.

$$\tau_{\mathrm{turn}}=\frac28=\frac14,\quad t_{\mathrm{turn}}=T_*\tau_{\mathrm{turn}}=0.125\ \mathrm{s},\quad x_{\mathrm{turn}}=2(3-\tfrac12+\tfrac14)=5.5\ \mathrm{m}.$$

At $t=0.25\ \mathrm{s}$, $\tau=0.5$, so $\xi=3-1+1=3$ and $\xi'=2$. Thus $$\boxed{x=6\ \mathrm{m},\quad v=+8\ \mathrm{m/s},\quad s=8\ \mathrm{m/s}}.$$ The derivative vanishes at $\tau=1/4$, or $\boxed{t=0.125\ \mathrm{s}}$. Velocity is negative before this event and positive afterward, so reversal has already occurred. Returning to the initial coordinate at 0.25 s does not imply unchanged velocity.

(d) A coefficient is not automatically a physical derivative.

Twice differentiating $4\tau^2$ gives 8, not 4. That dimensionless second derivative must then be multiplied by $L_*/T_*^2=8\ \mathrm{m/s^2}$, giving 64 $\mathrm{m/s^2}$. The erroneous report omits both the factor of two from differentiation and the physical scale conversion.