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Displacement and Average Velocity from Vector Integrals

Use component-by-component definite integration to find displacement, and use the elapsed time—not the final clock reading—to calculate average velocity.

Advanced 9 min read

What You'll Learn

  • Define displacement as the change in position over a time interval
  • Evaluate a vector-valued definite integral component by component
  • Use the Fundamental Theorem of Calculus to connect velocity and position
  • Calculate average velocity using the interval’s elapsed time
  • Distinguish net displacement from motion that reverses direction
  • Check vector answers, signs, and units

From Velocity to Net Change in Position

A velocity vector tells both how quickly position changes and in which direction it changes. When velocity varies over time, its value at a single instant does not describe the total change in position across an interval. To find that change, accumulate the velocity over the interval with a definite integral.

The result is displacement: the final position minus the initial position. Dividing displacement by the elapsed time gives average velocity. Both quantities are vectors, so their components and directions matter.

Key idea: If \(\mathbf{v}(t)\) is velocity on the interval from \(t=a\) to \(t=b\), then displacement is \(\int_a^b \mathbf{v}(t)\,dt\), and average velocity is that displacement divided by \(b-a\), assuming \(b>a\).

Definitions and Conditions

Let an object move in two or three dimensions, with position vector \(\mathbf{r}(t)\). Its velocity is the derivative of its position: \(\mathbf{v}(t)=\mathbf{r}'(t)\). The time interval considered will be \([a,b]\), where \(a<b\), so its elapsed time is \(b-a>0\).

Definition: The displacement of an object from time \(a\) to time \(b\) is the change in its position vector: $$ \Delta\mathbf{r}=\mathbf{r}(b)-\mathbf{r}(a). $$ If its velocity is integrable on \([a,b]\), its displacement can be calculated as \(\int_a^b\mathbf{v}(t)\,dt\).

For a two-dimensional velocity \(\mathbf{v}(t)=\langle v_x(t),v_y(t)\rangle\), the definite integral is evaluated component by component:

$$ \int_a^b\mathbf{v}(t)\,dt = \left\langle \int_a^b v_x(t)\,dt,\, \int_a^b v_y(t)\,dt \right\rangle. $$

The same rule applies to three components. Each component integral gives the net change in the corresponding coordinate. A negative component means the net change in that coordinate is in the negative direction; it does not mean the time interval or the whole displacement is negative.

Definition: For \(a<b\), the average velocity on \([a,b]\) is displacement divided by elapsed time: $$ \mathbf{v}_{\mathrm{avg}} = \frac{\Delta\mathbf{r}}{b-a} = \frac{1}{b-a}\int_a^b\mathbf{v}(t)\,dt. $$

For example, if position is measured in meters and time in seconds, displacement has units of meters and average velocity has units of meters per second. Dividing by the length of the interval is essential: average velocity describes net position change per unit of elapsed time.

Why the Integral Gives Displacement

The connection follows from the Fundamental Theorem of Calculus applied to each position component. Suppose \(\mathbf{r}(t)\) is differentiable and \(\mathbf{v}(t)=\mathbf{r}'(t)\) is continuous on \([a,b]\). Write \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\) and \(\mathbf{v}(t)=\langle v_x(t),v_y(t)\rangle\). Then \(x'(t)=v_x(t)\) and \(y'(t)=v_y(t)\).

1
Apply the Fundamental Theorem to the first component.
\(\int_a^b v_x(t)\,dt=\int_a^b x'(t)\,dt=x(b)-x(a)\).
2
Apply it to the second component.
\(\int_a^b v_y(t)\,dt=\int_a^b y'(t)\,dt=y(b)-y(a)\).
3
Combine the component changes.
\(\int_a^b\mathbf{v}(t)\,dt=\langle x(b)-x(a),y(b)-y(a)\rangle=\mathbf{r}(b)-\mathbf{r}(a)\).

The same reasoning works with a third component. Thus the integral adds the signed coordinate changes over time, and those net changes together form the displacement vector. Dividing the identity by \(b-a\) gives the formula for average velocity. This also shows why average velocity points in the direction of the net displacement, not necessarily in the direction of the velocity at every instant.

Worked Examples

Worked Example: Integrate a Two-Component Velocity

An object has velocity \(\mathbf{v}(t)=\langle 3t^2-2,4t\rangle\) meters per second for \(0\leq t\leq2\). Find its displacement and average velocity.

Integrate each component to obtain displacement:

$$ \begin{aligned} \Delta\mathbf{r} &=\int_0^2\langle 3t^2-2,4t\rangle\,dt\\ &=\left\langle\int_0^2(3t^2-2)\,dt,\int_0^2 4t\,dt\right\rangle\\ &=\left\langle[t^3-2t]_0^2,[2t^2]_0^2\right\rangle\\ &=\langle(8-4)-(0-0),(8-0)\rangle\\ &=\langle4,8\rangle\text{ m}. \end{aligned} $$

The elapsed time is \(2-0=2\) seconds. Therefore:

$$ \mathbf{v}_{\mathrm{avg}} =\frac{\langle4,8\rangle\text{ m}}{2\text{ s}} =\langle2,4\rangle\text{ m/s}. $$

Check: Antiderivatives of the two velocity components are \(t^3-2t\) and \(2t^2\). Differentiating gives \(3t^2-2\) and \(4t\), respectively, so the component integrals are correct. The displacement units are meters, and dividing by two seconds gives meters per second.

Worked Example: A Component Can Have Zero Net Displacement

Suppose \(\mathbf{v}(t)=\langle2t-3,1\rangle\) meters per second for \(0\leq t\leq3\). Find displacement and average velocity.

Integrate both components on the full interval:

$$ \begin{aligned} \Delta\mathbf{r} &=\left\langle\int_0^3(2t-3)\,dt,\int_0^3 1\,dt\right\rangle\\ &=\left\langle[t^2-3t]_0^3,[t]_0^3\right\rangle\\ &=\langle(9-9)-(0-0),3-0\rangle\\ &=\langle0,3\rangle\text{ m}. \end{aligned} $$

The elapsed time is \(3-0=3\) seconds, so:

$$ \mathbf{v}_{\mathrm{avg}} =\frac{\langle0,3\rangle\text{ m}}{3\text{ s}} =\langle0,1\rangle\text{ m/s}. $$

The horizontal velocity changes sign: \(2t-3=0\) at \(t=1.5\). The object’s horizontal motion first contributes negative change and later positive change. These contributions cancel in the integral, producing zero net horizontal displacement. This does not say that the object never moved horizontally.

Check: Differentiating \(t^2-3t\) gives \(2t-3\), and differentiating \(t\) gives \(1\). Evaluating those antiderivatives at \(0\) and \(3\) yields the stated displacement. Dividing by \(3\) seconds gives \(\langle0,1\rangle\) meters per second.

Worked Example: Use a Nonzero Starting Time

An object has velocity \(\mathbf{v}(t)=\langle t,2-t\rangle\) meters per second from \(t=1\) to \(t=4\) seconds. Find its displacement and average velocity.

Integrate component by component, keeping both endpoints in each evaluation:

$$ \begin{aligned} \Delta\mathbf{r} &=\left\langle\int_1^4 t\,dt,\int_1^4(2-t)\,dt\right\rangle\\ &=\left\langle\left[\frac{t^2}{2}\right]_1^4, \left[2t-\frac{t^2}{2}\right]_1^4\right\rangle\\ &=\left\langle\frac{16}{2}-\frac{1}{2}, \left(8-\frac{16}{2}\right)-\left(2-\frac{1}{2}\right)\right\rangle\\ &=\left\langle\frac{15}{2},0-\frac{3}{2}\right\rangle\\ &=\left\langle\frac{15}{2},-\frac{3}{2}\right\rangle\text{ m}. \end{aligned} $$

The elapsed time is \(4-1=3\) seconds, not \(4\) seconds. Thus:

$$ \mathbf{v}_{\mathrm{avg}} =\frac{1}{3}\left\langle\frac{15}{2},-\frac{3}{2}\right\rangle =\left\langle\frac{5}{2},-\frac{1}{2}\right\rangle\text{ m/s}. $$

Check: The derivatives of \(t^2/2\) and \(2t-t^2/2\) are \(t\) and \(2-t\). Their endpoint differences are \(15/2\) and \(-3/2\), as calculated. Multiplying the average velocity by the elapsed time returns the displacement: \(3\langle5/2,-1/2\rangle=\langle15/2,-3/2\rangle\) meters.

Displacement Is Not Distance

A definite integral of velocity gives net displacement, because positive and negative contributions in a component can cancel. It does not, in general, give the total distance traveled. Total distance counts motion along the path without allowing reversals to cancel. For instance, in the second example, the horizontal component of displacement is zero even though the horizontal velocity is negative before \(t=1.5\) and positive after \(t=1.5\).

Average velocity also differs from average speed. Average velocity is a vector obtained by dividing displacement by elapsed time. Average speed is a scalar based on total distance traveled divided by elapsed time. Do not replace the vector integral with an integral of the magnitude of velocity when the question asks for displacement or average velocity.

A Common Mistake: Dividing by the Wrong Time

In the third example, a tempting but incorrect calculation is to divide the displacement by \(4\), because \(4\) is the ending time. That would give:

$$ \frac{\langle15/2,-3/2\rangle}{4} = \left\langle\frac{15}{8},-\frac{3}{8}\right\rangle\text{ m/s}. $$

This is wrong because the motion interval begins at \(t=1\). The elapsed time is \(4-1=3\) seconds. The correction is:

$$ \frac{\langle15/2,-3/2\rangle}{4-1} = \left\langle\frac{5}{2},-\frac{1}{2}\right\rangle\text{ m/s}. $$

A useful check is to multiply the proposed average velocity by the elapsed time. The incorrect value multiplied by \(3\) gives \(\langle45/8,-9/8\rangle\), not the displacement \(\langle15/2,-3/2\rangle\). The corrected value multiplied by \(3\) returns the displacement exactly.

How to Organize a Calculation

For a vector velocity problem, write the time interval first and identify its elapsed time. Then integrate each velocity component over the same bounds. Keep the endpoint subtraction visible, especially when the lower time is not zero. Finally, divide the displacement vector by the elapsed time only if average velocity is requested.

Takeaway: Integrating velocity gives displacement, a vector of net coordinate changes. Dividing that vector by elapsed time gives average velocity. Use the actual interval length and retain the signs of all components.

Check Your Understanding

For each question, retain component signs and include units when they are supplied.

  1. If \(\mathbf{v}(t)=\langle2,3t\rangle\) meters per second on \([0,2]\), what are displacement and average velocity?
  2. Why can one component of displacement be zero even when the object has moved in that coordinate direction?
  3. An object moves from \(t=2\) to \(t=7\). What elapsed time belongs in the average-velocity formula?
  4. What does the definite integral of velocity give: displacement or total distance? Explain briefly.
  5. What calculation can you use to check an average-velocity vector once displacement and elapsed time are known?

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