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

Acceleration — Question 1

Section 2-5 · Physics pages 212–213

Question 1 An eastward change in velocity need not mean eastward motion A cart moves on an east-positive straight track. At clock reading ti = 3 s its velocity is vi = −12 m/s; at tf = 7 s its velocity is vf = −4 m/s. These are exact observations. Average acceleration over a positive-duration interval is ¯a = vf −vi tf −ti . Instantaneous acceleration, when it exists, is the derivative a = dv/dt. A velocity change has a sign; speed is the nonnegative magnitude of velocity. ti = 3 s, vi = −12 m/s tf = 7 s, vf = −4 m/s east positive Both measured velocities point west. The arrows represent velocities, not positions. (a) Find the velocity change, elapsed time, and average acceleration. Explain the unit m/s2 as a velocity change per unit time. (b) Find the initial and final speeds, and calculate their difference divided by elapsed time. Why is this not the same signed quantity as ¯a? (c) A student says that positive average acceleration proves the cart moved east throughout the interval. Diagnose the claim using the endpoint velocities. (d) For a separate, fully specified model, suppose velocity varies linearly between the two observations. State the direction of motion and whether speed increases or decreases throughout the interval. Does the endpoint record alone establish that entire history?
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Question 1 - Solution An eastward change in velocity need not mean eastward motion ti = 3 s, vi = −12 m/s tf = 7 s, vf = −4 m/s east positive (a) Subtract signed velocities and use elapsed time. ∆v = −4 −(−12) = +8 m/s , ∆t = 7 −3 = 4 s . Therefore ¯a = +2 m/s2 , an eastward average change in velocity. The unit means metres per second of velocity change, per second of elapsed time. It does not mean a distance of 2 m is covered each second. (b) Speed removes direction before taking the difference. The endpoint speeds are 12 and 4 m/s, so their change per elapsed time is 4 −12 4 = −2 m/s2 . The speed has decreased, while the signed velocity has increased from a more negative to a less negative value. Taking magnitudes before subtraction changes the question being answered. Average acceleration uses the change in signed velocity, not the change in speed. (c) Acceleration direction is not necessarily motion direction. Both observed velocities are westward. Their positive difference says that the final velocity is 8 m/s more eastward than the initial one; it does not say the final velocity is itself eastward. Positive average acceleration cannot establish eastward motion throughout the interval and is compatible with westward motion at both observed endpoints. (d) Apply the additional linear-history assumption explicitly. v(t) = −12 m/s + (2 m/s2)(t −3 s) < 0, d|v| dt = −2 m/s2. A straight velocity history between −12 and −4 m/s remains negative and increases throughout. In that model the cart moves west and continuously loses speed, with instantaneous acceleration +2 m/s2 on the interval’s interior. The two observations alone do not establish linear variation or monotonic speed. For example, a continuous velocity history could pass through +8 m/s at 5 s before returning to −4 m/s at 7 s. Its average acceleration is still +2 m/s2, but its intervening directions and speed changes differ.

Question 1

An eastward change in velocity need not mean eastward motion

A cart moves on an east-positive straight track. At clock reading ti=3st_i=3\ \mathrm{s} its velocity is vi=−12m/sv_i=-12\ \mathrm{m/s}; at tf=7st_f=7\ \mathrm{s} its velocity is vf=−4m/sv_f=-4\ \mathrm{m/s}. These are exact observations.

Average acceleration over a positive-duration interval is a‾=vf−vitf−ti.\bar a=\frac{v_f-v_i}{t_f-t_i}. Instantaneous acceleration, when it exists, is the derivative a=dv/dta=dv/dt. A velocity change has a sign; speed is the nonnegative magnitude of velocity.

Diagram shown in the original SVG worksheet.

  1. Find the velocity change, elapsed time, and average acceleration. Explain the unit m/s2\mathrm{m/s^2} as a velocity change per unit time.

  2. Find the initial and final speeds, and calculate their difference divided by elapsed time. Why is this not the same signed quantity as a‾\bar a?

  3. A student says that positive average acceleration proves the cart moved east throughout the interval. Diagnose the claim using the endpoint velocities.

  4. For a separate, fully specified model, suppose velocity varies linearly between the two observations. State the direction of motion and whether speed increases or decreases throughout the interval. Does the endpoint record alone establish that entire history?

Question 1 - Solution

An eastward change in velocity need not mean eastward motion

Diagram shown in the original SVG worksheet.

(a) Subtract signed velocities and use elapsed time.

Δv=−4−(−12)=+8m/s,Δt=7−3=4s.\Delta v=-4-(-12)=\boxed{+8\ \mathrm{m/s}},\qquad \Delta t=7-3=\boxed{4\ \mathrm{s}}. Therefore a‾=+2m/s2\boxed{\bar a=+2\ \mathrm{m/s^2}}, an eastward average change in velocity. The unit means metres per second of velocity change, per second of elapsed time. It does not mean a distance of 2 m is covered each second.

(b) Speed removes direction before taking the difference.

The endpoint speeds are 12 and 4 m/s, so their change per elapsed time is 4−124=−2m/s2.\frac{4-12}{4}=\boxed{-2\ \mathrm{m/s^2}}. The speed has decreased, while the signed velocity has increased from a more negative to a less negative value. Taking magnitudes before subtraction changes the question being answered. Average acceleration uses the change in signed velocity, not the change in speed.

(c) Acceleration direction is not necessarily motion direction.

Both observed velocities are westward. Their positive difference says that the final velocity is 8 m/s more eastward than the initial one; it does not say the final velocity is itself eastward. Positive average acceleration cannot establish eastward motion throughout the interval and is compatible with westward motion at both observed endpoints.

(d) Apply the additional linear-history assumption explicitly.

v(t)=−12m/s+(2m/s2)(t−3s)<0,d|v|dt=−2m/s2.v(t)=-12\ \mathrm{m/s}+(2\ \mathrm{m/s^2})(t-3\ \mathrm{s})<0,\qquad\frac{d|v|}{dt}=-2\ \mathrm{m/s^2}.

A straight velocity history between −12-12 and −4m/s-4\ \mathrm{m/s} remains negative and increases throughout. In that model the cart moves west and continuously loses speed, with instantaneous acceleration +2m/s2+2\ \mathrm{m/s^2} on the interval’s interior.

The two observations alone do not establish linear variation or monotonic speed. For example, a continuous velocity history could pass through +8m/s+8\ \mathrm{m/s} at 5 s before returning to −4m/s-4\ \mathrm{m/s} at 7 s. Its average acceleration is still +2m/s2+2\ \mathrm{m/s^2}, but its intervening directions and speed changes differ.

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