From Position to Total Distance
A particle moving along a line has a position at each time. Its displacement over a time interval is the change from its initial position to its final position. Its total distance traveled counts the length of the path it follows, including every reversal.
When the particle keeps moving in one direction, the distance traveled is the absolute value of its displacement. When it changes direction, the net displacement can hide some of the motion: the particle may travel away from its starting point and then return. To find total distance from a position function, locate the direction changes, evaluate position at those times and at the interval endpoints, and add the absolute changes between consecutive positions.
Turning Times and Velocity Signs
Let \(x(t)\) denote a particle’s position at time \(t\). Its velocity is \(v(t)=x'(t)\). Positive velocity means position is increasing, so the particle moves in the positive direction along the line. Negative velocity means position is decreasing, so it moves in the negative direction.
An interior time at which the velocity changes sign is a turning time: the particle changes direction there. A time when \(v(t)=0\) is a rest time, but not every rest time is a turning time. For instance, velocity might be positive on both sides of a zero, so the particle pauses instantaneously without reversing. Check the velocity signs on either side rather than assuming that every zero is a direction change.
In typical AP problems, position functions are differentiable and velocity is continuous. Find the zeros of \(v(t)\) in the interval, check the signs between them, and retain the zeros where the sign changes. Include the interval’s initial and final times as well. The resulting ordered list gives the times at which to evaluate position.
A Formula for Total Distance
Suppose the time interval is \([a,b]\), and the direction changes at \(t_1,t_2,\ldots,t_n\), in increasing order. Set \(t_0=a\) and \(t_{n+1}=b\). Then the total distance is the sum of the absolute position changes between consecutive times in this list.
This formula requires that the particle move in one direction, or remain momentarily at rest without reversing, on each part of the interval. A sign chart for velocity verifies that condition. If velocity is zero at a boundary, that boundary is still included as an endpoint; it does not need to be an interior turning time.
Why the Method Works
On any time interval where velocity is nonnegative, position is nondecreasing: as time advances, position does not go down. The particle therefore travels from the position at the start of that part to the position at the end without retracing any portion of the line. Its distance is the final position minus the initial position. If velocity is nonpositive, position is nonincreasing, and distance is the initial position minus the final position. Both cases are expressed by the absolute value of the position change.
Differentiate the position function to obtain \(v(t)=x'(t)\).
Find interior zeros of velocity and check which ones have a sign change.
Find \(x(t)\) at both endpoints and at every direction-changing time.
Take the absolute difference for each pair of consecutive positions, then add the results.
The reasoning is complete because each part of the interval has motion in only one direction. Its distance is exactly the length between its endpoint positions. The entire path consists of these consecutive parts, so adding their lengths gives the total distance.
Worked Examples
Worked Example: Motion Without a Direction Change
A particle’s position, in meters, is \(x(t)=t^2+2t\), where \(t\) is measured in seconds. Find its total distance traveled on \(0\le t\le3\).
Step 1: Find velocity.
For \(0\le t\le3\), \(2t+2\ge2>0\). Thus position is increasing throughout the interval, and there is no direction change.
Step 2: Evaluate position at the endpoints.
Step 3: Find distance and check. There is only one part of the motion, so its distance is
The positive velocity confirms that the particle moves from \(0\) m to \(15\) m without reversing. The calculated distance agrees with that change in position.
Worked Example: Motion With a Direction Change
A particle’s position, in meters, is \(x(t)=(t-4)^2\), with time \(t\) in seconds. Find its total distance traveled on \(1\le t\le7\).
Step 1: Find velocity and its sign.
For \(1\le t<4\), \(t-4<0\), so \(v(t)<0\). For \(4<t\le7\), \(t-4>0\), so \(v(t)>0\). The velocity changes from negative to positive at \(t=4\), so this is a turning time.
Step 2: Evaluate position at the endpoints and turning time.
Step 3: Add the absolute changes.
The check is that the particle moves \(9\) m in the negative direction, then \(9\) m in the positive direction. Its final position equals its initial position, so its displacement is \(9-9=0\) m, while its total distance is \(18\) m.
Worked Example: An Endpoint at Rest
A particle’s position, in meters, is \(x(t)=\frac{1}{3}t^3-2t^2+3t\), with time \(t\) in seconds. Find its total distance on \(0\le t\le3\).
Step 1: Differentiate and factor velocity.
The velocity is positive on \(0<t<1\), because both factors are negative there. It is negative on \(1<t<3\), because \(t-1>0\) and \(t-3<0\). Thus \(t=1\) is a direction-changing time. At \(t=3\), velocity is zero, but this is the endpoint of the interval, so there is no interval to its right on which to check for a reversal.
Step 2: Evaluate position at \(t=0,1,3\).
Step 3: Add the changes in position as distances.
Differentiate the stated position function to check the velocity used: \(\frac{d}{dt}\left(\frac{1}{3}t^3-2t^2+3t\right)=t^2-4t+3\), as required. The first segment moves from \(0\) m to \(\frac{4}{3}\) m, and the second returns to \(0\) m. The displacement is \(0\) m, but the sum of the two path lengths is \(\frac{8}{3}\) m.
A Common Mistake: Using Only the Net Change
For the position function \(x(t)=(t-4)^2\) on \(1\le t\le7\), a tempting but incorrect calculation is to use the absolute difference between the final and initial positions as the total distance:
This gives the magnitude of the displacement, not the total distance. It ignores the motion from \(9\) m to \(0\) m and back to \(9\) m. The correction is to split at the direction change \(t=4\):
Check Your Understanding
For each question, use velocity signs to decide whether a direction change occurs before adding position changes.
- A particle has \(x(t)=t^2+4t\) on \(0\le t\le2\). Find its total distance traveled.
- A particle has \(x(t)=(t-2)^2\) on \(0\le t\le5\). Find its turning time and total distance.
- What is the difference between a rest time and a turning time?
- A particle starts at position \(3\) m, reaches position \(-2\) m at a turning time, and ends at position \(4\) m. What is its total distance over these two parts?
- Why does the absolute difference between the initial and final positions fail to give total distance when a particle returns to its starting position?
