Instantaneous Velocity and Speed — Question 4

Section 2-4 · Physics pages 198–199

Question 4 Zero velocity does not always mark a reversal Two points move on separate east-positive straight tracks. In the formulas below, t is numerical time in seconds and positions are numerical metres. Near t = 2 their exact histories are xA(t) = 5 + (t −2)3, xB(t) = 5 + (t −2)2. You may use d[(t −c)n]/dt = n(t −c)n−1. Instantaneous velocity is the position derivative and instantaneous speed its magnitude. t t = 2 x = 5 A t t = 2 x = 5 B Schematic local position shapes. Both have a horizontal tangent at the marked event, but their neighboring positions differ. (a) Find the position, velocity and speed of both points at 2 s. (b) Examine velocity signs immediately before and after 2 s. Which point reverses direction, and which continues in the same direction through a zero-velocity instant? (c) A student claims that every horizontal tangent is a maximum or minimum of position. Use these two histories to evaluate the claim. (d) Does either point remain at rest over a nonzero-duration interval around 2 s? Give a third position history that agrees with both points’ position and velocity at 2 s but remains at rest throughout 1 ≤t ≤3 s. (e) Explain why position and velocity at one instant cannot, by themselves, distinguish all three behaviors.
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Question 4 - Solution Zero velocity does not always mark a reversal t t = 2 x = 5 A t t = 2 x = 5 B (a) Evaluate the derivative, not only the coordinate. vA(t) = 3(t −2)2 m/s, vB(t) = 2(t −2) m/s. At 2 s both points have x = 5 m, v = 0 m/s, s = 0 m/s . (b) Reversal requires a change in direction across the instant. For A, the square is positive on both sides of 2 s, so motion is eastward before and after, with an isolated zero velocity at 2 s. There is no reversal. For B, the velocity is negative before 2 s and positive afterward. B approaches 5 m from greater positions, reaches its minimum, and reverses from westward to eastward motion. (c) A zero slope is not sufficient for an extremum. xA(2 + h) −5 = h3, xA(2 −h) −5 = −h3; xB(2 ± h) −5 = h2 > 0 (h > 0). A has a horizontal tangent at 2 s, but positions just before are below 5 m and positions just after are above 5 m. Thus it has neither a local maximum nor a local minimum there. B does have a minimum. The existence of a horizontal tangent alone does not decide whether a turning point occurs. (d) Rest over an interval is a statement about every time in it. Neither A nor B is constant on any interval around 2 s. Their speeds vanish only at the isolated instant. A suitable third history is xC(t) = 5 m for all 1 ≤t ≤3 s . Its velocity is zero throughout the interior of that interval. (e) A single local value does not specify the surrounding history. All three agree in position and velocity at 2 s. Yet A passes through that position in the same direction, B reverses there, and C remains there for an interval. Distinguishing these cases requires nearby-time information, such as the velocity signs or whether position stays constant, not merely the shared zero velocity.

Question 4

Zero velocity does not always mark a reversal

Two points move on separate east-positive straight tracks. In the formulas below, tt is numerical time in seconds and positions are numerical metres. Near t=2t=2 their exact histories are xA(t)=5+(t−2)3,xB(t)=5+(t−2)2.x_A(t)=5+(t-2)^3,\qquad x_B(t)=5+(t-2)^2. You may use d[(t−c)n]/dt=n(t−c)n−1d[(t-c)^n]/dt=n(t-c)^{n-1}. Instantaneous velocity is the position derivative and instantaneous speed its magnitude.

Diagram shown in the original SVG worksheet.

  1. Find the position, velocity and speed of both points at 2 s.

  2. Examine velocity signs immediately before and after 2 s. Which point reverses direction, and which continues in the same direction through a zero-velocity instant?

  3. A student claims that every horizontal tangent is a maximum or minimum of position. Use these two histories to evaluate the claim.

  4. Does either point remain at rest over a nonzero-duration interval around 2 s? Give a third position history that agrees with both points’ position and velocity at 2 s but remains at rest throughout 1≤t≤3s1\leq t\leq3\ \mathrm{s}.

  5. Explain why position and velocity at one instant cannot, by themselves, distinguish all three behaviors.

Question 4 - Solution

Zero velocity does not always mark a reversal

Diagram shown in the original SVG worksheet.

(a) Evaluate the derivative, not only the coordinate.

vA(t)=3(t−2)2m/s,vB(t)=2(t−2)m/s.v_A(t)=3(t-2)^2\ \mathrm{m/s},\qquad v_B(t)=2(t-2)\ \mathrm{m/s}. At 2 s both points have x=5m,v=0m/s,s=0m/s\boxed{x=5\ \mathrm{m},\ v=0\ \mathrm{m/s},\ s=0\ \mathrm{m/s}}.

(b) Reversal requires a change in direction across the instant.

For A, the square is positive on both sides of 2 s, so motion is eastward before and after, with an isolated zero velocity at 2 s. There is no reversal.

For B, the velocity is negative before 2 s and positive afterward. B approaches 5 m from greater positions, reaches its minimum, and reverses from westward to eastward motion.

(c) A zero slope is not sufficient for an extremum.

xA(2+h)−5=h3,xA(2−h)−5=−h3;xB(2±h)−5=h2>0(h>0).x_A(2+h)-5=h^3,\quad x_A(2-h)-5=-h^3;\qquad x_B(2\pm h)-5=h^2>0\quad(h>0).

A has a horizontal tangent at 2 s, but positions just before are below 5 m and positions just after are above 5 m. Thus it has neither a local maximum nor a local minimum there. B does have a minimum. The existence of a horizontal tangent alone does not decide whether a turning point occurs.

(d) Rest over an interval is a statement about every time in it.

Neither A nor B is constant on any interval around 2 s. Their speeds vanish only at the isolated instant. A suitable third history is xC(t)=5mfor all 1≤t≤3s.\boxed{x_C(t)=5\ \mathrm{m}\quad\text{for all }1\leq t\leq3\ \mathrm{s}}. Its velocity is zero throughout the interior of that interval.

(e) A single local value does not specify the surrounding history.

All three agree in position and velocity at 2 s. Yet A passes through that position in the same direction, B reverses there, and C remains there for an interval. Distinguishing these cases requires nearby-time information, such as the velocity signs or whether position stays constant, not merely the shared zero velocity.