AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/8/8-2)
Unit 8 · Topic 8.2
8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals
Integrals undo the derivatives from straight-line motion. The integral of velocity gives displacement, the integral of speed gives total distance traveled, and adding displacement to a starting position gives a later position.
Key terms
- displacement
- total distance traveled
- speed
- initial position
- velocity
- acceleration
The chain: position, velocity, acceleration
For a particle moving along a line, x(t) is position, v(t) = x′(t) is velocity and a(t) = v′(t) is acceleration. In Unit 4 you went down the chain by differentiating. Now you go up by integrating:
x(b) = x(a) + ∫ₐᵇ v(t) dt, and v(b) = v(a) + ∫ₐᵇ a(t) dt.
Both are the net change idea from 6.7: final value = starting value + accumulated change.
Displacement vs. total distance
Displacement is the change in position: ∫ₐᵇ v(t) dt. If the particle moves right 5 units and then left 3, its displacement is +2.
Total distance traveled counts every bit of motion as positive: ∫ₐᵇ |v(t)| dt. In the same example, it's 5 + 3 = 8. The quantity |v(t)| is speed.
Without a calculator, find where v(t) = 0 and changes sign, split the interval there, integrate v on each piece and add the absolute values. With a calculator, enter ∫ₐᵇ |v(t)| dt directly.
Displacement and distance are equal only when the particle never changes direction on the interval. If they differ, the particle turned around at least once.
| You want | Integrate | Notes |
|---|---|---|
| displacement (change in position) | v(t) | can be negative |
| position at time b | x(a) + ∫ₐᵇ v(t) dt | needs a starting position |
| total distance traveled | speed (the absolute value of v) | never negative |
| change in velocity | a(t) | then add v(a) for v(b) |
Reading motion from v
- v(t) > 0: moving right (or up, or in the positive direction).
- v(t) < 0: moving left.
- The particle changes direction where v changes sign.
- Speeding up when v and a have the same sign; slowing down when they have opposite signs.
- The particle is farthest right on an interval at an endpoint or where v changes from positive to negative. Compare positions with the Candidates Test.
Working from a graph of v
If you're given a velocity graph made of lines and semicircles, the displacement is the signed area and the total distance is the total (unsigned) area. The particle's position at any time is its starting position plus the signed area so far. It's the accumulation function idea from 6.5 with v in place of f.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Position, displacement and distance by hand
A particle moves along the x-axis with velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 4, and x(0) = 2. Find (a) the displacement, (b) x(4) and (c) the total distance traveled.
Show the solutionHide the solution
- Step 1: Antiderivative: t³/3 − 2t² + 3t.
- Step 2: (a) ∫₀⁴ v(t) dt = 64/3 − 32 + 12 − 0 = 4/3.
- Step 3: (b) x(4) = x(0) + 4/3 = 2 + 4/3 = 10/3.
- Step 4: (c) v(t) = (t − 1)(t − 3), which is positive on (0, 1), negative on (1, 3), positive on (3, 4).
- Step 5: ∫₀¹ v dt = 1/3 − 2 + 3 = 4/3. ∫₁³ v dt = (9 − 18 + 9) − 4/3 = −4/3. ∫₃⁴ v dt = 4/3 − 0 = 4/3.
- Step 6: Total distance = 4/3 + |−4/3| + 4/3 = 4.
Answer: (a) 4/3 (b) 10/3 (c) 4
- Example 2Calculator allowed
Calculator: distance and speeding up
A particle moves along a line with velocity v(t) = e^(sin t) − 1.2 for 0 ≤ t ≤ 5. (a) Find the displacement and the total distance traveled. (b) Is the particle speeding up or slowing down at t = 4?
Show the solutionHide the solution
- Step 1: (a) Displacement = ∫₀⁵ v(t) dt ≈ 1.189.
- Step 2: Total distance = ∫₀⁵ |v(t)| dt ≈ 3.764. (These differ because v is negative on part of the interval.)
- Step 3: (b) v(4) = e^(sin 4) − 1.2 ≈ −0.731.
- Step 4: a(t) = v′(t) = cos t · e^(sin t), so a(4) ≈ −0.307.
- Step 5: v(4) and a(4) are both negative, so the speed is increasing.
Answer: (a) Displacement ≈ 1.189; distance ≈ 3.764 (b) Speeding up, since v(4) and a(4) are both negative.
- Example 3
Trap: from acceleration, two constants
A particle has acceleration a(t) = 6t − 4, with v(0) = −2 and x(0) = 5. Find x(2).
Show the solutionHide the solution
- Step 1: v(t) = 3t² − 4t + C. v(0) = −2 gives C = −2, so v(t) = 3t² − 4t − 2.
- Step 2: x(t) = t³ − 2t² − 2t + D. x(0) = 5 gives D = 5.
- Step 3: x(2) = 8 − 8 − 4 + 5 = 1.
- Step 4: Common error: forgetting the −2 in v(t), which changes every later answer.
Answer: x(2) = 1
Common mistakes
- Reporting ∫ v(t) dt as total distance when v changes sign.
- Giving the displacement when asked for the position. Position needs the starting value added.
- Saying the particle speeds up because a > 0. Speeding up depends on whether v and a share a sign.
- Forgetting the constant when going from a to v or from v to x.
On the exam
- Motion questions are a staple of free response, often with a calculator. Write each integral before giving its value, like x(5) = 2 + ∫₀⁵ v(t) dt.
- For “farthest left/right” or “change direction” questions, use the sign of v and compare positions at the candidates.
Connected topics
Videos
Check yourself
4 questions on 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals. Pick an answer to see if you got it, and why.
A particle moves along the x-axis. Its velocity at time t ≥ 0 is given by v(t) = t² − 6t + 8, and its position at time t = 0 is x(0) = 3.
Described motion
What is the position of the particle at time t = 5?
What is the total distance traveled by the particle from t = 0 to t = 5?
A particle moves along a line so that its velocity at time t is v(t) = (t − 2) cos(t/2) for 0 ≤ t ≤ 5. What is the total distance traveled by the particle over this time interval?
A particle moves along the x-axis with acceleration a(t) = 6t − 4 for t ≥ 0. At time t = 0, the velocity of the particle is −2 and its position is 1. What is the position of the particle at time t = 2?
0 of 4 answered