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Physics / Chapter 2: Motion Along a Straight Line / Section 2-4

Instantaneous Velocity and Speed — Question 1

Section 2-4 · Physics pages 192–193

Question 1 Shrink the interval without dividing by zero A point moves on an east-positive line according to x(t) = (2 m) + (3 m/s)t + (1 m/s2)t2. Instantaneous velocity at an interior time t0 is defined, when the limit exists, by v(t0) = lim h→0 x(t0 + h) −x(t0) h . Here h is a nonzero time increment, which may approach zero from either sign. Instantaneous speed is s(t0) = |v(t0)|. t x t0 t0 + h P Q Solid: the supplied position curve. Dashed: one secant through P and a later Q; Q approaches P as positive h shrinks. No numerical axis scale is shown. (a) At t0 = 2 s, expand and simplify the quotient before taking the limit. Give its values for h = +1, +0.1, −0.1, −1 s. (b) Determine the instantaneous velocity and speed at 2 s, including direction for velocity. Explain why setting h = 0 in the original quotient is not the calculation of a limit. (c) Find the average velocity from 0 to 2 s. Why need it not equal the instantaneous velocity at the final event? (d) Explain what negative h means here. Does it mean the point is moving west, or that time physically runs backward?
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Question 1 - Solution Shrink the interval without dividing by zero t x t0 t0 + h P Q (a) Cancel the time increment only while it is nonzero. At 2 s the position is 12 m. Expanding the position at 2 s + h gives x(2 s + h) −x(2 s) = (7 m/s)h + (1 m/s2)h2. For h̸ = 0, the quotient is 7 m/s + (1 m/s2)h . For the requested increments its values are, in order, 8, 7.1, 6.9, 6 m/s . Positive and negative increments approach the same value. (b) The limit describes the behavior near zero. v(2 s) = +7 m/s (east) , s(2 s) = 7 m/s . Direct substitution into the original quotient would produce 0/0, which is undefined. The simplified nonzero-increment expression has a well-defined limiting value as the increment shrinks. A limit does not require assigning a value to the original quotient at zero. (c) An interval average uses separated endpoints. Since x(0) = 2 m and x(2 s) = 12 m, ¯v0→2 = 12 −2 2 = +5 m/s . This summarizes the complete 2 s interval. Instantaneous velocity instead uses arbitrarily nearby events around the selected instant. The two quantities can differ when the slope of the position history varies. (d) A negative increment selects an earlier event. x(t + h) −x(t) h = x(t) −x(t + h) −h (h < 0). For h < 0, 2 s + h is before the event at 2 s. Both numerator and denominator are then negative in this example, giving a positive quotient. Reordering both endpoint differences leaves the usual earlier-to-later average unchanged. The sign of the increment is a mathematical choice of neighboring event, not a direction of physical motion or a reversal of time.

Question 1

Shrink the interval without dividing by zero

A point moves on an east-positive line according to x(t)=(2m)+(3m/s)t+(1m/s2)t2.x(t)=(2\ \mathrm{m})+(3\ \mathrm{m/s})t+(1\ \mathrm{m/s^2})t^2. Instantaneous velocity at an interior time t0t_0 is defined, when the limit exists, by v(t0)=limh→0x(t0+h)−x(t0)h.v(t_0)=\lim_{h\to0}\frac{x(t_0+h)-x(t_0)}{h}. Here hh is a nonzero time increment, which may approach zero from either sign. Instantaneous speed is s(t0)=|v(t0)|s(t_0)=|v(t_0)|.

Diagram shown in the original SVG worksheet.

  1. At t0=2st_0=2\ \mathrm{s}, expand and simplify the quotient before taking the limit. Give its values for h=+1,+0.1,−0.1,−1sh=+1,+0.1,-0.1,-1\ \mathrm{s}.

  2. Determine the instantaneous velocity and speed at 2 s, including direction for velocity. Explain why setting h=0h=0 in the original quotient is not the calculation of a limit.

  3. Find the average velocity from 0 to 2 s. Why need it not equal the instantaneous velocity at the final event?

  4. Explain what negative hh means here. Does it mean the point is moving west, or that time physically runs backward?

Question 1 - Solution

Shrink the interval without dividing by zero

Diagram shown in the original SVG worksheet.

(a) Cancel the time increment only while it is nonzero.

At 2 s the position is 12 m. Expanding the position at 2s+h2\ \mathrm{s}+h gives x(2s+h)−x(2s)=(7m/s)h+(1m/s2)h2.x(2\ \mathrm{s}+h)-x(2\ \mathrm{s})=(7\ \mathrm{m/s})h+(1\ \mathrm{m/s^2})h^2. For h≠0h\ne0, the quotient is 7m/s+(1m/s2)h.\boxed{7\ \mathrm{m/s}+(1\ \mathrm{m/s^2})h}. For the requested increments its values are, in order, 8,7.1,6.9,6m/s\boxed{8,\ 7.1,\ 6.9,\ 6\ \mathrm{m/s}}. Positive and negative increments approach the same value.

(b) The limit describes the behavior near zero.

v(2s)=+7m/s (east),s(2s)=7m/s.\boxed{v(2\ \mathrm{s})=+7\ \mathrm{m/s}\text{ (east)}},\qquad \boxed{s(2\ \mathrm{s})=7\ \mathrm{m/s}}. Direct substitution into the original quotient would produce 0/00/0, which is undefined. The simplified nonzero-increment expression has a well-defined limiting value as the increment shrinks. A limit does not require assigning a value to the original quotient at zero.

(c) An interval average uses separated endpoints.

Since x(0)=2mx(0)=2\ \mathrm{m} and x(2s)=12mx(2\ \mathrm{s})=12\ \mathrm{m}, v‾0→2=12−22=+5m/s.\boxed{\bar v_{0\to2}=\frac{12-2}{2}=+5\ \mathrm{m/s}}. This summarizes the complete 2 s interval. Instantaneous velocity instead uses arbitrarily nearby events around the selected instant. The two quantities can differ when the slope of the position history varies.

(d) A negative increment selects an earlier event.

x(t+h)−x(t)h=x(t)−x(t+h)−h(h<0).\frac{x(t+h)-x(t)}h=\frac{x(t)-x(t+h)}{-h}\quad(h<0).

For h<0h<0, 2s+h2\ \mathrm{s}+h is before the event at 2 s. Both numerator and denominator are then negative in this example, giving a positive quotient. Reordering both endpoint differences leaves the usual earlier-to-later average unchanged. The sign of the increment is a mathematical choice of neighboring event, not a direction of physical motion or a reversal of time.

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