Motion from Velocity

One of the most natural applications of integration is reconstructing the motion of an object from information about its rate of change. Velocity tells us how quickly position is changing, and acceleration tells us how quickly velocity is changing. Because integration is the reverse of differentiation, if we know an object's velocity function v(t), we can integrate it to recover the position function s(t) — up to a constant determined by an initial condition. This idea underpins everything from tracking a spacecraft's trajectory to computing how far a car travels during a braking maneuver. Engineers and physicists routinely measure acceleration (using accelerometers) and then integrate twice to obtain position, since position sensors are often more expensive or impractical. Understanding this chain — acceleration → velocity → position — is a cornerstone of kinematics and of calculus-based physics. In this article you will learn how to find velocity from acceleration using indefinite integration, how to find position from velocity, how initial conditions pin down the constant of integration, and how definite integrals give the net displacement (which can differ from total distance traveled). These ideas tie together the Fundamental Theorem of Calculus with real physical motion.

Key takeaways

The Fundamental Chain
Using Initial Conditions
Displacement vs. Distance Traveled
Connecting to the Fundamental Theorem

The Fundamental Chain

The position of an object moving in one dimension is a function s(t). Its velocity is the derivative v(t) = ds/dt, and acceleration is a(t) = dv/dt = d²s/dt². Integration reverses differentiation, so: v(t) = ∫ a(t) dt and s(t) = ∫ v(t) dt. Each integration introduces an arbitrary constant C, which is resolved using initial conditions such as v(0) = v₀ or s(0) = s₀.

Using Initial Conditions

Suppose a(t) = 6t − 2. Integrating: v(t) = ∫(6t − 2) dt = 3t² − 2t + C₁. If v(0) = 5, then C₁ = 5, giving v(t) = 3t² − 2t + 5. Integrating again: s(t) = ∫(3t² − 2t + 5) dt = t³ − t² + 5t + C₂. If s(0) = 0, then C₂ = 0, so s(t) = t³ − t² + 5t. Every initial condition eliminates one constant of integration.

Displacement vs. Distance Traveled

The definite integral ∫[a to b] v(t) dt gives net displacement: the signed change in position. It counts motion in the negative direction as subtracting from the total. To find total distance traveled, you must integrate the absolute value of velocity: ∫[a to b] |v(t)| dt. To evaluate this, find all zeros of v(t) in [a, b] and split the integral at those points, flipping the sign of negative sub-intervals.

Connecting to the Fundamental Theorem

The Fundamental Theorem of Calculus guarantees that if v is continuous, then s(b) − s(a) = ∫[a to b] v(t) dt. This elegant statement means the area under the velocity–time curve equals the net displacement. Geometrically, area above the t-axis represents forward motion and area below represents backward motion. The theorem gives us a rigorous way to compute exact displacement without needing a formula for s(t) in advance.

The Key Equation

s(t) = s_0 + \int_{t_0}^{t} v(\tau)\, d\tau

s(t) — Position (meters (m)): Location of the object on the number line at time t.
s_0 — Initial Position (meters (m)): Known position of the object at the starting time t₀.
v(\tau) — Velocity function (meters per second (m/s)): The velocity of the object as a function of the dummy integration variable τ.
t_0 — Initial time (seconds (s)): The time at which the initial condition is specified.
t — Current time (seconds (s)): The time at which we want to know the position.

Worked Example

Problem. A particle moves along the x-axis with velocity v(t) = t² − 4t + 3 m/s for 0 ≤ t ≤ 4 s. Its initial position is s(0) = 2 m. (a) Find the position function s(t). (b) Find the net displacement over [0, 4]. (c) Find the total distance traveled over [0, 4].

s(t) = \frac{t^3}{3} - 2t^2 + 3t + 2;\quad \Delta s = \frac{4}{3}\text{ m};\quad d_{\text{total}} = 4\text{ m}

Answer. s(t) = t³/3 − 2t² + 3t + 2; Net displacement = 4/3 m; Total distance = 4 m

Figure

Diagram

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · English

Newton developed calculus (which he called 'the method of fluxions') largely motivated by problems of motion. He introduced the derivative to describe instantaneous velocity and recognized that integration recovers position from velocity. His laws of motion and universal gravitation, combined with his calculus, allowed the precise prediction of planetary orbits — a direct application of integrating acceleration to find position. Newton reportedly developed the core ideas of calculus during 1665–1666 while the University of Cambridge was closed due to an outbreak of plague. He worked in isolation at his family home in Woolsthorpe, calling it his 'year of wonders' (annus mirabilis).

Definite Integrals and Area — The net displacement formula is a direct application of the definite integral as signed area under a curve.
Fundamental Theorem of Calculus — The connection s(b) − s(a) = ∫v(t)dt is a direct consequence of the Fundamental Theorem, which links differentiation and integration.
Derivatives and Rates of Change — Velocity and acceleration are defined as first and second derivatives of position, so understanding derivatives is essential before reversing the process with integration.
Antiderivatives and Indefinite Integrals — Finding position from velocity requires computing an antiderivative, and understanding the +C constant is crucial for applying initial conditions.
Initial Value Problems — Using s(0) = s₀ and v(0) = v₀ to determine constants of integration is the classic setup of an initial value problem in differential equations.