Calculus-Master

Particle Motion (Kinematics): Position, Velocity, and Acceleration Explained

A particle moving along a line is described by its position s(t)s(t), velocity v(t)v(t) and acceleration a(t)a(t). AP Calculus AB asks you to move between these three, decide when the particle moves left or right, whether it is speeding up or slowing down, and find displacement versus total distance traveled.

Core Theorem
v(t)=s′(t),a(t)=v′(t)=s′′(t)v(t) = s'(t), \qquad a(t) = v'(t) = s''(t)
Position from velocity: s(t)=s(t0)+∫t0tv(u) dus(t) = s(t_0) + \int_{t_0}^{t} v(u)\,du.
Displacement on [a,b][a, b]: ∫abv(t) dt\int_a^b v(t)\,dt. Total distance on [a,b][a, b]: ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt.
Speed is ∣v(t)∣|v(t)|. The particle is speeding up when vv and aa have the same sign, and slowing down when they have opposite signs.
Step-by-Step SOP
  1. 1

    1. Get v(t) and a(t)

    From position: v=s′v = s', a=v′a = v'. From velocity: a=v′a = v', and position needs an integral plus the starting position.
  2. 2

    2. Find where v(t) = 0

    These are the times the particle is at rest and the only places it can change direction.
  3. 3

    3. Sign chart for v (and a)

    v>0v > 0: moving right (or up). v<0v < 0: moving left (or down). Compare signs of vv and aa for speeding up or slowing down.
  4. 4

    4. Distance vs displacement

    Displacement: ∫abv dt\int_a^b v\,dt or s(b)−s(a)s(b) - s(a). Total distance: split at the turning points and add the absolute values, or use ∫ab∣v∣ dt\int_a^b |v|\,dt on a calculator.

Practice Exercises

Example 01Easy
A particle moves with velocity v(t)=t2−4v(t) = t^2 - 4 for 0≤t≤30 \le t \le 3. Find the total distance traveled.
NEED A HINT?
Find where v(t)=0v(t) = 0 to check for direction changes. t2−4=0t^2 - 4 = 0 at t=2t = 2.
SHOW DETAILED EXPLANATION

Step 1: Find the turning point

Set v(t)=0v(t) = 0: t2−4=0  ⟹  t=2t^2 - 4 = 0 \implies t = 2. Since t=2t = 2 is inside [0,3][0, 3], split the integral there.

Step 2: Direction on each piece

On [0,2][0, 2], v(1)=−3v(1) = -3 (moving left). On [2,3][2, 3], v(2.5)=2.25v(2.5) = 2.25 (moving right).

Step 3: Integrate each piece

An antiderivative is 13t3−4t\frac{1}{3}t^3 - 4t. From 00 to 22: 83−8=−163\frac{8}{3} - 8 = -\frac{16}{3}. From 22 to 33: (−3)−(−163)=73(-3) - \left(-\frac{16}{3}\right) = \frac{7}{3}.

Step 4: Add the absolute values

Total distance =163+73=233≈7.67= \frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.67. (The displacement is −163+73=−3-\frac{16}{3} + \frac{7}{3} = -3.)
Example 02AB/BC Standard
The position of a particle is s(t)=t3−6t2+9ts(t) = t^3 - 6t^2 + 9t for t≥0t \ge 0.
(a) Find v(t)v(t) and a(t)a(t).
(b) When is the particle at rest? When is it moving left?
(c) Is the particle speeding up or slowing down at t=2.5t = 2.5?
(d) Find the displacement and the total distance traveled on 0≤t≤40 \le t \le 4.
NEED A HINT?
Factor v(t)v(t) to find the turning points. For total distance, find the position at t=0t = 0, each turning point, and t=4t = 4.
SHOW DETAILED EXPLANATION

(a) Velocity and acceleration

v(t)=s′(t)=3t2−12t+9=3(t−1)(t−3)v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3) and a(t)=v′(t)=6t−12a(t) = v'(t) = 6t - 12.

(b) At rest and moving left

At rest when v(t)=0v(t) = 0: t=1t = 1 and t=3t = 3. Moving left when v(t)<0v(t) < 0: 1<t<31 < t < 3. Moving right on 0≤t<10 \le t < 1 and t>3t > 3.

(c) Speeding up or slowing down at t = 2.5

v(2.5)=3(1.5)(−0.5)=−2.25<0v(2.5) = 3(1.5)(-0.5) = -2.25 < 0 and a(2.5)=3>0a(2.5) = 3 > 0. The signs are opposite, so the particle is slowing down.

(d) Displacement

s(4)−s(0)=(64−96+36)−0=4s(4) - s(0) = (64 - 96 + 36) - 0 = 4.

(d) Total distance

Positions: s(0)=0s(0) = 0, s(1)=4s(1) = 4, s(3)=0s(3) = 0, s(4)=4s(4) = 4. Distance =∣4−0∣+∣0−4∣+∣4−0∣=12= |4 - 0| + |0 - 4| + |4 - 0| = 12.
Example 03Easy
Let s(t)=5t+ts(t) = 5t + \sqrt{t} be the position of a particle. Find the displacement on 0≤t≤90 \le t \le 9 and the velocity at t=4t = 4.
NEED A HINT?
Displacement is final position minus starting position. Velocity is s′(t)s'(t).
SHOW DETAILED EXPLANATION

Step 1: Displacement

s(9)−s(0)=(45+3)−0=48s(9) - s(0) = (45 + 3) - 0 = 48.

Step 2: Velocity function

v(t)=s′(t)=5+12tv(t) = s'(t) = 5 + \frac{1}{2\sqrt{t}}.

Step 3: Velocity at t = 4

v(4)=5+14=5.25v(4) = 5 + \frac{1}{4} = 5.25. Since v(t)>0v(t) > 0 for all t>0t > 0, the particle always moves right, so the total distance on [0,9][0, 9] is also 4848.
Example 04Medium
A particle moves along the x-axis with v(t)=2t−6v(t) = 2t - 6 and x(0)=5x(0) = 5. Find the position x(t)x(t) and the position at t=4t = 4.
NEED A HINT?
Position = starting position + accumulated change: x(t)=x(0)+∫0tv(u) dux(t) = x(0) + \int_0^t v(u)\,du.
SHOW DETAILED EXPLANATION

Step 1: Accumulate the velocity

∫0t(2u−6) du=t2−6t\int_0^t (2u - 6)\,du = t^2 - 6t.

Step 2: Add the starting position

x(t)=5+t2−6tx(t) = 5 + t^2 - 6t.

Step 3: Evaluate

x(4)=5+16−24=−3x(4) = 5 + 16 - 24 = -3.
Common Pitfalls
  • ⚠
    Distance vs. Displacement∫abv(t) dt\int_a^b v(t)\,dt is displacement (net change in position). Total distance needs ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt, or splitting at every point where vv changes sign.
  • ⚠
    Ignoring the turning pointsAlways solve v(t)=0v(t) = 0 first. If the particle changes direction and you don't split the interval, motion to the left cancels motion to the right.
  • ⚠
    Speeding up means a > 0 (it doesn't)Speeding up means vv and aa have the same sign. A particle with v<0v < 0 and a<0a < 0 is speeding up even though aa is negative.
  • ⚠
    Forgetting the starting position∫0tv du\int_0^t v\,du is only the change in position. Add the initial position x(0)x(0) to get where the particle actually is.
  • ⚠
    Calculation errors with negativesWhen evaluating F(b)−F(a)F(b) - F(a), wrap F(a)F(a) in parentheses. Sign slips are the most common lost point on particle-motion FRQs.
Frequently Asked Questions

How do you find velocity from a position function?

Differentiate: v(t)=s′(t)v(t) = s'(t). Then acceleration is a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t).

How do you know when a particle is moving left, right, up or down?

Look at the sign of v(t)v(t). On a horizontal line, v>0v > 0 means moving right and v<0v < 0 means moving left. For vertical motion, v>0v > 0 means moving up and v<0v < 0 means moving down. The particle can only change direction where v(t)=0v(t) = 0 (and vv changes sign).

How do you tell if a particle is speeding up or slowing down?

Compare the signs of velocity and acceleration at that time. Same sign: speeding up. Opposite signs: slowing down. On the AP exam, state both signs in your answer as the justification.

What is the difference between displacement and total distance?

Displacement is the net change in position, ∫abv(t) dt\int_a^b v(t)\,dt, and can be negative. Total distance counts every bit of motion, ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt, and is never negative.

How do you find the position of a particle from its velocity?

Use the starting position plus the accumulated velocity: x(t)=x(t0)+∫t0tv(u) dux(t) = x(t_0) + \int_{t_0}^{t} v(u)\,du.

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