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

Acceleration — Question 5

Section 2-5 · Physics pages 220–221

Question 5 A common position tangent does not imply common acceleration Two points have the following position histories on east-positive axes for 1 ≤t ≤3: xA = 4 + 2(t −2) + (t −2)2, xB = 4 + 2(t −2) −(t −2)2. Here t is numerical seconds and positions are numerical metres. Both curves pass through P at t = 2 s and share the tangent x = 2t there. t (s) x (m) 1 2 3 1 4 7 P A B Solid A and dashed B are position histories. The dotted line is their common tangent at P, not a third history. (a) Find both positions and velocities at P. What information about the two motions is captured by the common tangent? (b) Differentiate again to find both accelerations at P. Which point is speeding up and which is slowing down at that event? (c) Describe how the slope of each position curve changes near P. Explain why acceleration corresponds to that change of slope, rather than graph height or tangent slope alone. (d) A report includes only x = 4 m and v = 2 m/s at 2 s and claims to determine acceleration. Use these histories to assess that claim. Would shifting both position origins change either acceleration?
Show solutionHide solution
Question 5 - Solution A common position tangent does not imply common acceleration t (s) x (m) 1 2 3 1 4 7 P A B (a) Match the position and its first derivative. At 2 s both positions equal 4 m . Differentiating gives vA = 2 + 2(t −2) and vB = 2 −2(t −2) in m/s. Hence both velocities at P are +2 m/s . The common tangent records the same event coordinate and the same local position slope. (b) The second derivatives have opposite signs. dsA dt t=2 = +2 m/s2, dsB dt t=2 = −2 m/s2 (vA = vB > 0). aA = +2 m/s2 , aB = −2 m/s2 . Both points move east at P. A’s positive acceleration increases its speed; B’s negative acceleration decreases its speed. Equal instantaneous velocities do not require equal accelerations. (c) Inspect how nearby tangent slopes vary. xA −xB = 2(t −2)2, vA −vB = 4(t −2), aA −aB = 4 m/s2. A’s slope becomes more positive as time increases, so its position curve bends upward. B’s slope becomes less positive, so its position curve bends downward. Acceleration is the derivative of velocity, hence the rate of change of the position slope. Graph height gives position, and a single tangent slope gives velocity; neither alone gives acceleration. (d) A single position-and-velocity pair is insufficient. The report is compatible with both supplied curves even though their accelerations are opposite. It therefore cannot determine acceleration without additional nearby-time information or a restrictive motion model. Adding the same fixed constant to either position history changes its height but neither its first nor its second derivative. A fixed origin shift cannot change these accelerations.

Question 5

A common position tangent does not imply common acceleration

Two points have the following position histories on east-positive axes for 1≤t≤31\leq t\leq3: xA=4+2(t−2)+(t−2)2,xB=4+2(t−2)−(t−2)2.x_A=4+2(t-2)+(t-2)^2,\qquad x_B=4+2(t-2)-(t-2)^2. Here tt is numerical seconds and positions are numerical metres. Both curves pass through P at t=2st=2\ \mathrm{s} and share the tangent x=2tx=2t there.

Diagram shown in the original SVG worksheet.

  1. Find both positions and velocities at P. What information about the two motions is captured by the common tangent?

  2. Differentiate again to find both accelerations at P. Which point is speeding up and which is slowing down at that event?

  3. Describe how the slope of each position curve changes near P. Explain why acceleration corresponds to that change of slope, rather than graph height or tangent slope alone.

  4. A report includes only x=4mx=4\ \mathrm{m} and v=2m/sv=2\ \mathrm{m/s} at 2 s and claims to determine acceleration. Use these histories to assess that claim. Would shifting both position origins change either acceleration?

Question 5 - Solution

A common position tangent does not imply common acceleration

Diagram shown in the original SVG worksheet.

(a) Match the position and its first derivative.

At 2 s both positions equal 4m\boxed{4\ \mathrm{m}}. Differentiating gives vA=2+2(t−2)v_A=2+2(t-2) and vB=2−2(t−2)v_B=2-2(t-2) in m/s. Hence both velocities at P are +2m/s\boxed{+2\ \mathrm{m/s}}. The common tangent records the same event coordinate and the same local position slope.

(b) The second derivatives have opposite signs.

dsAdt|t=2=+2m/s2,dsBdt|t=2=−2m/s2(vA=vB>0).\left.\frac{ds_A}{dt}\right|_{t=2}=+2\ \mathrm{m/s^2},\qquad\left.\frac{ds_B}{dt}\right|_{t=2}=-2\ \mathrm{m/s^2}\quad(v_A=v_B>0).

aA=+2m/s2,aB=−2m/s2.\boxed{a_A=+2\ \mathrm{m/s^2}},\qquad \boxed{a_B=-2\ \mathrm{m/s^2}}. Both points move east at P. A’s positive acceleration increases its speed; B’s negative acceleration decreases its speed. Equal instantaneous velocities do not require equal accelerations.

(c) Inspect how nearby tangent slopes vary.

xA−xB=2(t−2)2,vA−vB=4(t−2),aA−aB=4m/s2.x_A-x_B=2(t-2)^2,\quad v_A-v_B=4(t-2),\quad a_A-a_B=4\ \mathrm{m/s^2}.

A’s slope becomes more positive as time increases, so its position curve bends upward. B’s slope becomes less positive, so its position curve bends downward. Acceleration is the derivative of velocity, hence the rate of change of the position slope. Graph height gives position, and a single tangent slope gives velocity; neither alone gives acceleration.

(d) A single position-and-velocity pair is insufficient.

The report is compatible with both supplied curves even though their accelerations are opposite. It therefore cannot determine acceleration without additional nearby-time information or a restrictive motion model. Adding the same fixed constant to either position history changes its height but neither its first nor its second derivative. A fixed origin shift cannot change these accelerations.

← Section questionsChapter contents