Instantaneous Velocity and Speed — Question 5

Question 5

A symmetric quotient can hide an undefined velocity

A mathematical motion model is $$x(t)=(3\ \mathrm{m/s})\,|t-2\ \mathrm{s}|,\qquad0\leq t\leq4\ \mathrm{s}.$$ Use the two-sided derivative definition for velocity at an interior time. In this problem, instantaneous speed is defined as $s(t)=|v(t)|$ only where that velocity exists.

1. Find velocity and speed for $t<2\ \mathrm{s}$ and $t>2\ \mathrm{s}$ within the stated interval.

2. At 2 s, evaluate the quotient $[x(2\ \mathrm{s}+h)-x(2\ \mathrm{s})]/h$ for positive and negative $h$. Does a single instantaneous velocity exist at the corner?

3. For $h>0$, compute the symmetric quotient $[x(2\ \mathrm{s}+h)-x(2\ \mathrm{s}-h)]/(2h)$. Explain why its value does not establish zero velocity at 2 s.

4. The speed just before and just after the corner tends to 3 m/s. Under the stated definition, is speed at the corner therefore automatically defined as 3 m/s? Distinguish a limiting or assigned value from a value furnished by $|v(2\ \mathrm{s})|$.

Question 5 - Solution

A symmetric quotient can hide an undefined velocity

(a) Differentiate each straight branch away from the corner.

$$v(t)=\begin{cases}-3\ \mathrm{m/s},&t<2\ \mathrm{s},\\+3\ \mathrm{m/s},&t>2\ \mathrm{s},\end{cases}\qquad s(t)=3\ \mathrm{m/s}\quad(t\ne2\ \mathrm{s}).$$

For $t<2\ \mathrm{s}$, position decreases by 3 m each second, so $\boxed{v=-3\ \mathrm{m/s}}$. For $t>2\ \mathrm{s}$, it increases at the same rate, so $\boxed{v=+3\ \mathrm{m/s}}$. On both open branches $\boxed{s=3\ \mathrm{m/s}}$.

(b) The two one-sided quotients disagree.

$$\lim_{h\to0^-}\frac{3|h|}{h}=-3\ne+3=\lim_{h\to0^+}\frac{3|h|}{h}\quad(\mathrm{m/s}).$$

At 2 s the position is zero. The quotient is $(3\ \mathrm{m/s})|h|/h$, equal to $+3\ \mathrm{m/s}$ for positive $h$ and $-3\ \mathrm{m/s}$ for negative $h$. The two-sided limit does not exist. Thus $\boxed{v(2\ \mathrm{s})\text{ is undefined in this model}}$.

(c) Symmetric cancellation is not the derivative test.

$$\frac{x(2+h)-x(2-h)}{2h}=\frac{3h-3h}{2h}=0\quad(h>0).$$

The two symmetric positions are equal, so the centered quotient is zero for every allowed $h>0$. This describes zero average velocity between those paired endpoints. It does not make the left and right slopes at the central event agree. A centered quotient alone is therefore insufficient to establish differentiability or zero instantaneous velocity.

(d) Specify the convention at an idealized corner.

The neighboring speeds have common limiting value 3 m/s, which could be assigned as an extension. However, under the question’s definition $s=|v|$ where $v$ exists, neither velocity nor its magnitude is defined at 2 s. Calling an assigned extension an existing $|v(2\ \mathrm{s})|$ would conceal the failed derivative. A smooth turnaround would require a different local position model; the exact V-shaped graph does not supply one.