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Physics / Chapter 2: Motion Along a Straight Line / Section 2-5

Acceleration — Question 2

Section 2-5 · Physics pages 214–215

Question 2 Read acceleration from velocity slope, not graph height The graph gives the complete east-positive velocity history. Its vertices are (0, 2), (3, 8), (5, 8), and (9, −4), with time in seconds and velocity in m/s. v (m/s) t (s) 3 5 9 −4 2 8 Exact piecewise-linear velocity history. Corners at 3 s and 5 s are not smoothed in this idealized model. Instantaneous acceleration is the slope of the velocity graph wherever that slope exists. When velocity is nonzero, speed increases if velocity and acceleration have the same sign, and decreases if their signs differ. (a) Find acceleration on each open straight segment and the average acceleration over 0–9 s. (b) Find when the cart reverses direction. Classify the intervals of increasing speed, decreasing speed, and constant speed. Explain how a single negative-acceleration segment can include both slowing and speeding up. (c) A student calls the 3–5 s segment “rest” because its slope is zero. What is actually zero, and what is the cart doing? (d) Does instantaneous acceleration exist at 3 s or 5 s in this exact model? Compare one-sided slopes rather than assigning a convenient value at each corner.
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Question 2 - Solution Read acceleration from velocity slope, not graph height v (m/s) t (s) 3 5 9 −4 2 8 (a) The graph’s vertical quantity is velocity. Its three slopes are a = +2 m/s2 (0 < t < 3) , a = 0 (3 < t < 5) , a = −3 m/s2 (5 < t < 9) . Over the full interval, ¯a = (−4 −2)/9 = −2/3 m/s2 . (b) Locate zero velocity and compare signs on either side. ds dt = sgn(v)a (v̸ = 0); v = 8 −3(t −5) = 0 ⇒t = 23 3 s. On the last segment, v = 8 −3(t −5) in m/s with t numerical seconds. It crosses zero at t = 23/3 s and changes from eastward to westward motion. Speed increases on 0 < t < 3, is constant on 3 < t < 5, decreases on 5 < t < 23/3, then increases on 23/3 < t < 9. The same negative acceleration first reduces a positive velocity toward zero, then makes an already negative velocity more negative. The direction of velocity determines whether this change reduces or increases its magnitude. (c) A flat velocity graph means constant velocity. During 3–5 s the cart moves east at 8 m/s. Its acceleration is zero, but its velocity and speed are not. Rest would require zero velocity, not merely zero velocity slope. (d) A corner has unequal one-sided derivatives. a(3−) = 2̸ = 0 = a(3+), a(5−) = 0̸ = −3 = a(5+) (m/s2). At 3 s the slopes are +2 and 0 m/s2; at 5 s they are 0 and −3 m/s2. Hence acceleration is undefined at each corner under the two-sided derivative definition. Velocity itself is continuous there. Changing a single assigned acceleration value cannot change these unequal limiting slopes.

Question 2

Read acceleration from velocity slope, not graph height

The graph gives the complete east-positive velocity history. Its vertices are (0,2)(0,2), (3,8)(3,8), (5,8)(5,8), and (9,−4)(9,-4), with time in seconds and velocity in m/s.

Diagram shown in the original SVG worksheet.

Instantaneous acceleration is the slope of the velocity graph wherever that slope exists. When velocity is nonzero, speed increases if velocity and acceleration have the same sign, and decreases if their signs differ.

  1. Find acceleration on each open straight segment and the average acceleration over 0–9 s.

  2. Find when the cart reverses direction. Classify the intervals of increasing speed, decreasing speed, and constant speed. Explain how a single negative-acceleration segment can include both slowing and speeding up.

  3. A student calls the 3–5 s segment “rest” because its slope is zero. What is actually zero, and what is the cart doing?

  4. Does instantaneous acceleration exist at 3 s or 5 s in this exact model? Compare one-sided slopes rather than assigning a convenient value at each corner.

Question 2 - Solution

Read acceleration from velocity slope, not graph height

Diagram shown in the original SVG worksheet.

(a) The graph’s vertical quantity is velocity.

Its three slopes are a=+2m/s2(0<t<3),a=0(3<t<5),a=−3m/s2(5<t<9).\boxed{a=+2\ \mathrm{m/s^2}\ (0<t<3)},\quad \boxed{a=0\ (3<t<5)},\quad \boxed{a=-3\ \mathrm{m/s^2}\ (5<t<9)}. Over the full interval, a‾=(−4−2)/9=−2/3m/s2\boxed{\bar a=(-4-2)/9=-2/3\ \mathrm{m/s^2}}.

(b) Locate zero velocity and compare signs on either side.

dsdt=sgn⁡(v)a(v≠0);v=8−3(t−5)=0⇒t=233s.\frac{ds}{dt}=\operatorname{sgn}(v)a\quad(v\ne0);\qquad v=8-3(t-5)=0\Rightarrow t=\frac{23}{3}\ \mathrm{s}.

On the last segment, v=8−3(t−5)v=8-3(t-5) in m/s with tt numerical seconds. It crosses zero at t=23/3s\boxed{t=23/3\ \mathrm{s}} and changes from eastward to westward motion. Speed increases on 0<t<30<t<3, is constant on 3<t<53<t<5, decreases on 5<t<23/35<t<23/3, then increases on 23/3<t<923/3<t<9.

The same negative acceleration first reduces a positive velocity toward zero, then makes an already negative velocity more negative. The direction of velocity determines whether this change reduces or increases its magnitude.

(c) A flat velocity graph means constant velocity.

During 3–5 s the cart moves east at 8 m/s. Its acceleration is zero, but its velocity and speed are not. Rest would require zero velocity, not merely zero velocity slope.

(d) A corner has unequal one-sided derivatives.

a(3−)=2≠0=a(3+),a(5−)=0≠−3=a(5+)(m/s2).a(3^-)=2\ne0=a(3^+),\qquad a(5^-)=0\ne-3=a(5^+)\quad(\mathrm{m/s^2}).

At 3 s the slopes are +2+2 and 0m/s20\ \mathrm{m/s^2}; at 5 s they are 00 and −3m/s2-3\ \mathrm{m/s^2}. Hence acceleration is undefined at each corner under the two-sided derivative definition. Velocity itself is continuous there. Changing a single assigned acceleration value cannot change these unequal limiting slopes.

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