Distinguish displacement from distance
Out, across and back
A particle starts at the origin and moves back and forth on a line with the position \(\langle 3\sin t,\ 4\sin t\rangle\). The teal arrow runs from the start to the particle, and it is the displacement, the change in position.
At \(t = \frac{\pi}{2}\), the particle is at \((3, 4)\), 5 units from the start. At \(t = \frac{3\pi}{2}\), it is at \((-3, -4)\), on the other side of the start.
The orange bar on the ruler adds up all the path that the particle covers, which is the distance traveled. It does not decrease, because the particle covers new path even when it turns around.
At \(t = 2\pi\), the particle is back at the start, so the teal arrow has no length and the displacement is the zero vector. The ruler reads 20, which is 5 out, 10 across and 5 back.
The shortest way and the whole route
Move the end time. The teal arrow goes straight from the start to the particle, and the orange bar on the ruler measures the path.
A straight line is the shortest way between two points, so the bar is never shorter than the arrow.
Straighten the path. If the particle moves along a line without a turn, the two lengths match. Otherwise, the distance is greater than the length of the displacement.
Notation
The displacement is a vector, the integral of the velocity, so opposite motions cancel.
\[\cb{\Delta\vec r} = \int_a^b \vec v(t)\,dt\]The speed is the length of the velocity vector, so it is never negative.
\[\text{speed} = |\cy{\vec v}(t)| = \sqrt{x'(t)^2 + y'(t)^2}\]The distance traveled is a number, the integral of the speed, so no part of the motion cancels.
\[\text{distance} = \int_a^b |\vec v(t)|\,dt\]The distance is at least the length of the displacement. The two are equal only if the particle moves along a line and does not turn around.
\[\left|\int_a^b \vec v\,dt\right| \le \int_a^b |\vec v|\,dt\]When the particle is back at the start, the displacement is the zero vector, but the ruler reads 10. A zero displacement does not mean that the particle stayed still.
When it fails
Move to the turning point. Up to there, the particle moves along the line without a turn, so the arrow and the ruler both read 5.
Let the particle turn and move 2 units back. The arrow shrinks to 3, because the particle is closer to the start, but the ruler grows to 7, because the particle covers more path.
Example 1: Half a lap
A skater has the position \(\vec r(t) = \langle 2\cos t,\ 2\sin t\rangle\) for \(0 \le t \le \pi\). Find the displacement and the distance traveled.
To find the displacement, integrate each part of the velocity \(\langle -2\sin t,\ 2\cos t\rangle\) from 0 to \(\pi\). The y-part is 0, because the skater ends at the same height.
\[\begin{aligned} \int_0^{\pi} -2\sin t\,dt &= -4 \\ \int_0^{\pi} 2\cos t\,dt &= 0 \\ \Delta\vec r &= \langle -4,\ 0\rangle \end{aligned}\]For the distance, find the speed. The two squares add to 4, because \(\sin^2 t + \cos^2 t = 1\), so the speed is 2 at every time, as the faint arrows of equal length show.
\[\begin{aligned} |\vec v| &= \sqrt{(-2\sin t)^2 + (2\cos t)^2} \\ &= 2 \end{aligned}\]Integrate the constant speed 2 over the \(\pi\) seconds.
\[\int_0^{\pi} 2\,dt = 2\pi\]As a check, \(2\pi \approx 6.28\) is more than 4, the length of the displacement, because the skater travels along the arc and not along the diameter.
Example 2: A particle that stops and turns
A particle has the velocity \(\vec v(t) = (3 - t)\langle 3,\ 4\rangle\). Find the displacement and the distance traveled from \(t = 0\) to \(t = 4\).
Integrate the velocity from 0 to 4. The constant vector \(\langle 3,\ 4\rangle\) stays outside the integral.
\[\begin{aligned} \Delta\vec r &= \left[3t - \tfrac12 t^2\right]_0^4 \langle 3,\ 4\rangle \\ &= 4\,\langle 3,\ 4\rangle = \langle 12,\ 16\rangle \end{aligned}\]The length of the displacement is 20, because \(\sqrt{12^2 + 16^2} = 20\).
A tempting wrong move: The distance traveled is the length of the displacement, which is 20.
\[\text{distance} = 20\]The particle stops at \(t = 3\) and turns around. At that time the ruler already reads 22.5, past the red 20, and the ruler never decreases.
The speed is \(5|3 - t|\), because \(\langle 3,\ 4\rangle\) has length 5. The formula for \(|3 - t|\) changes at \(t = 3\), so split the integral there.
\[\begin{aligned} \text{distance} &= \int_0^3 5(3 - t)\,dt \\ &\quad + \int_3^4 5(t - 3)\,dt \\ &= 22.5 + 2.5 = 25 \end{aligned}\]
Practice
A runner on a circular track has the velocity \(\vec v(t) = \langle -6\sin 2t,\ 6\cos 2t\rangle\) for \(0 \le t \le \pi\). Find the displacement and the distance traveled.
Find the position at both ends and compare them. Then find the speed, the length of the velocity vector. Move the time to the end to see the arrow and the ruler.
Solution
Integrate each part of the velocity from 0 to \(\pi\).
\[\begin{aligned} \Delta\vec r &= \left[\langle 3\cos 2t,\ 3\sin 2t\rangle\right]_0^{\pi} \\ &= \langle 3, 0\rangle - \langle 3, 0\rangle = \langle 0,\ 0\rangle \end{aligned}\]The displacement is the zero vector, because the runner finishes where it started.
The speed is 6, because \(\sin^2 2t + \cos^2 2t = 1\). Integrate the speed.
\[\int_0^{\pi} 6\,dt = 6\pi\]The runner travels \(6\pi\) units, but the displacement is the zero vector.
Questions
A drone flies along the sides of a triangle. It starts at \((0, 0)\), flies to \((6, 0)\), then to \((6, 8)\), and then returns to \((0, 0)\). The ruler under the figure shows the distance traveled at each corner: 6, 14 and 24.
The drone is at \((6, 8)\). What is the length of its displacement from the start?
- 10
- 14
- 8
Answer
10
The drone lands at \((0, 0)\). What are its displacement and distance traveled?
- Displacement \(\langle 0, 0\rangle\), distance 24
- Displacement \(\langle 0, 0\rangle\), distance 0
- Displacement \(\langle 6, 8\rangle\), distance 24
Answer
Displacement \(\langle 0, 0\rangle\), distance 24
A particle moves along a vertical line with the y-velocity \(t^2 - 4\) for \(0 \le t \le 3\). What is the distance traveled?
- \(\dfrac{23}{3}\)
- 3
- \(\dfrac{16}{3}\)
Answer
\(\dfrac{23}{3}\)
Summary
You find the displacement by integrating the velocity, and the distance traveled by integrating the speed.
This works because the velocity integral lets opposite motions cancel, and the speed integral does not.
You use the displacement when a problem asks for a change in position, and the distance when it asks how far the object traveled.
This is one of four skills in topic 9.5 of Which & Why: AP Calculus.