Two Different Questions About Motion
A particle moves along a line, and its velocity \(v(t)\) describes both its speed and its direction. If the particle moves forward and then reverses direction, two reasonable questions have different answers: how far is its final position from its initial position, and how much ground has it covered in total?
The first question concerns displacement, or net change in position. Motion in one direction contributes positively and motion in the opposite direction contributes negatively. The second concerns total distance traveled. Every part of the motion contributes a nonnegative amount, regardless of direction.
Definitions and Formulas
Suppose \(x(t)\) is the particle’s position and \(v(t)=x'(t)\) is its velocity on the time interval \([a,b]\). The displacement from time \(a\) to time \(b\) is the change in position. By the Fundamental Theorem of Calculus, it can be calculated from velocity:
Total distance counts motion in either direction positively. Since speed is the magnitude of velocity, the distance traveled is the integral of speed:
These formulas assume that the particle’s position is differentiable with velocity \(v\), and that \(v\) is integrable on the interval. In typical AP problems, the velocity is continuous or piecewise continuous, so these integrals are defined.
When velocity is positive, the particle moves in the positive direction; when velocity is negative, it moves in the negative direction. The signed integral allows those contributions to cancel for displacement. The absolute-value integral prevents cancellation, so it records all the distance traveled.
Why the Formulas Work
The displacement formula follows from the relationship between velocity and position: \(x'(t)=v(t)\). The Fundamental Theorem of Calculus says that integrating a derivative over an interval gives the net change in the original function.
For distance, consider a very short time interval of length \(\Delta t\). During that interval, the particle’s change in position is approximately \(v(t)\Delta t\). The length of that change, which is the small distance traveled, is approximately \(|v(t)|\Delta t\). Adding these nonnegative pieces over the whole interval and taking the limit gives the integral of \(|v(t)|\).
Equivalently, speed is defined as the magnitude of velocity. Total distance is the accumulation of speed over time, so its rate of accumulation is \(|v(t)|\), not \(v(t)\).
A Reliable Calculation Method
To calculate displacement, integrate the velocity over the entire interval, retaining its sign. To calculate distance, integrate the absolute value. If velocity changes sign, find the relevant times and split the distance calculation there: on intervals where velocity is positive, \(|v|=v\); where it is negative, \(|v|=-v\). A zero of velocity is a possible direction change, but the sign on the surrounding intervals determines whether one occurs.
Use the signed integral for displacement and the absolute-value integral for total distance.
Compute \(\int_a^b v(t)\,dt\) and preserve the sign of the result.
Split at known sign changes, make each contribution nonnegative, and add them.
Displacement and distance have units of position, such as meters; do not report them in velocity units.
Worked Examples
Worked Example: A Particle That Returns to Its Starting Position
A particle has velocity \(v(t)=3t^2-6t\) meters per second for \(0\leq t\leq3\), where time is measured in seconds. Find its displacement and total distance traveled.
An antiderivative of velocity is \(F(t)=t^3-3t^2\), because \(F'(t)=3t^2-6t\). The displacement is
To find the distance, determine where the velocity changes sign. Factor it:
On \(0<t<2\), \(t\) is positive and \(t-2\) is negative, so \(v(t)<0\). On \(2<t<3\), both factors are positive, so \(v(t)>0\). Thus the distance is the sum of the magnitudes of the two signed integrals:
Check: The signed contributions are \(-4\) m and \(+4\) m, which sum to the displacement \(0\) m. Their magnitudes sum to \(8\) m, the total distance. The particle ends where it began but has traveled \(8\) m.
Worked Example: Displacement, Distance, and Final Position
A particle’s velocity is \(v(t)=4-2t\) meters per second on \(0\leq t\leq4\). Its initial position is \(x(0)=5\) meters. Find the displacement, total distance traveled, and final position.
An antiderivative is \(F(t)=4t-t^2\), since \(F'(t)=4-2t\). The displacement over the whole interval is
The velocity is zero when \(4-2t=0\), so \(t=2\). It is positive for \(0\leq t<2\) and negative for \(2<t\leq4\). Therefore, the particle travels forward during the first part and backward during the second. The distance is
Final position is initial position plus displacement:
Check: The particle moves \(4\) m in the positive direction and then \(4\) m in the negative direction. Its net change is \(4-4=0\) m, consistent with the final position of \(5\) m, while its total distance is \(4+4=8\) m.
Worked Example: A Velocity That Changes Sign Once
A particle has velocity \(v(t)=t^2-1\) meters per second on \(0\leq t\leq2\). Find its displacement and total distance traveled.
An antiderivative is \(F(t)=\frac{t^3}{3}-t\), because
The displacement is
For distance, \(t^2-1=0\) at \(t=1\). The velocity is negative on \(0\leq t<1\) and positive on \(1<t\leq2\). The two distance contributions are
Adding the nonnegative contributions gives
Check: The signed contributions are \(-\frac{2}{3}\) m and \(+\frac{4}{3}\) m. Their sum is \(\frac{2}{3}\) m, matching the displacement; their magnitudes sum to \(2\) m, matching the distance.
A Common Mistake: Taking the Absolute Value Too Late
A tempting but incorrect method is to calculate the displacement and then take its absolute value to obtain distance. The absolute value of the net change gives the magnitude of displacement, not the total distance. Earlier motion in opposite directions may cancel in the displacement calculation even though that motion still contributes to distance.
For the first worked example, the incorrect calculation would be
This is the magnitude of the displacement, but it is not the distance traveled. The particle moves \(4\) m in one direction and then \(4\) m in the other, so the correct distance is \(4+4=8\) m. Apply the absolute value to the velocity before accumulating, or split the integral at each sign change and add the magnitudes of the signed contributions.
Check Your Understanding
For each question, distinguish net change from total motion and include units where relevant.
- In words, what does \(\int_a^b v(t)\,dt\) measure, and why can it be negative?
- Why is total distance calculated using \(\int_a^b |v(t)|\,dt\) rather than the signed velocity integral?
- A particle moves \(6\) meters in the positive direction and then \(2\) meters in the negative direction. What are its displacement and total distance?
- What does the absolute value of displacement tell you, and how does that differ from total distance traveled?
- If velocity is negative throughout an interval, how can you calculate distance from the signed velocity integral?
