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Displacement — Definition, Formula & Examples

Displacement is how far an object ends up from where it started, along with the direction. It counts motion forward as positive and motion backward as negative, so trips in opposite directions can cancel out.

For an object moving along a line with position function s(t)s(t) and continuous velocity v(t)=s′(t)v(t) = s'(t), the displacement on the interval [a,b][a, b] is the signed quantity s(b)−s(a)s(b) - s(a). By the Fundamental Theorem of Calculus, this equals the definite integral ∫abv(t) dt\int_a^b v(t)\,dt. Total distance traveled on the same interval is ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt, which is always nonnegative and is at least the absolute value of the displacement.

Key Formula

Displacement=∫abv(t) dt=s(b)−s(a)\text{Displacement} = \int_a^b v(t)\,dt = s(b) - s(a)
Where:
  • v(t)v(t) = velocity at time t (positive means moving forward, negative means moving backward)
  • s(t)s(t) = position at time t, an antiderivative of v(t)
  • a,ba, b = start and end times of the interval

How It Works

To find displacement, integrate the velocity function over the time interval. Do not split anything up or take absolute values. Regions where the velocity graph lies above the tt-axis add to the result, and regions below the axis subtract from it. For total distance, find where v(t)=0v(t) = 0, split the interval at those times, and add the absolute value of each piece. To get a final position, add the displacement to the starting position: s(b)=s(a)+∫abv(t) dts(b) = s(a) + \int_a^b v(t)\,dt.

Worked Example

Problem: A particle moves along a line with velocity v(t)=2t−4v(t) = 2t - 4 meters per second for 0≤t≤50 \le t \le 5 seconds. Find the particle's displacement and its total distance traveled.
Step 1: Displacement is the integral of velocity. An antiderivative of 2t−42t - 4 is t2−4tt^2 - 4t.
∫05(2t−4) dt=[t2−4t]05\int_0^5 (2t - 4)\,dt = \left[t^2 - 4t\right]_0^5
Step 2: Evaluate at the bounds and subtract.
(25−20)−(0−0)=5(25 - 20) - (0 - 0) = 5
Step 3: For distance, find where the particle changes direction by setting the velocity equal to zero. The velocity is negative on [0,2)[0, 2) and positive on (2,5](2, 5].
2t−4=0  ⟹  t=22t - 4 = 0 \implies t = 2
Step 4: Integrate the absolute value by splitting at t=2t = 2. On [0,2][0, 2] the particle moves backward 4 m. On [2,5][2, 5] it moves forward 9 m.
∫02(4−2t) dt+∫25(2t−4) dt=4+9=13\int_0^2 (4 - 2t)\,dt + \int_2^5 (2t - 4)\,dt = 4 + 9 = 13
Answer: The displacement is 5 meters (the particle ends 5 m forward of its start). The total distance traveled is 13 meters.

Visualization

Why It Matters

Displacement versus total distance appears on nearly every AP Calculus AB and BC exam, usually in a particle-motion free-response question. The same idea of a net signed change shows up in physics (net work, net change in momentum) and in economics, where integrating a rate of change gives a net gain or loss.

Common Mistakes

Mistake: Integrating velocity directly when the question asks for total distance, or integrating ∣v(t)∣|v(t)| when it asks for displacement.
Correction: Displacement uses ∫abv(t) dt\int_a^b v(t)\,dt with no absolute value. Distance uses ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt, so you must split the interval wherever v(t)v(t) changes sign. In the example, ∫05v dt=5\int_0^5 v\,dt = 5, but the distance is 13.
Mistake: Reporting the displacement as the final position.
Correction: Displacement is a change in position, not a location. If the particle above starts at s(0)=1s(0) = 1, its final position is s(5)=1+5=6s(5) = 1 + 5 = 6, not 5.

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