Displacement and Total Distance

Displacement is the net change in position, \int_a^b v(t)\,dt. Total distance counts all motion as positive, \int_a^b |v(t)|\,dt. To find total distance, find where the velocity changes sign, integrate v on each piece, and add the absolute values of the results.

Before this lesson: Position, Velocity and Acceleration, Fundamental Theorem of Calculus, Part 2 and Absolute Value

Displacement vs. distance

Picture an object moving back and forth on a number line. Its velocity v(t) is positive when it moves right and negative when it moves left. Displacement asks, “Where did it end up, compared with where it started?” Total distance asks, “How far did it travel in all?”

Walk 5 meters right, then 2 meters left. Your displacement is 5 − 2 = 3 meters. Your total distance is 5 + 2 = 7 meters. The integral of velocity gives displacement, because the backward part counts as negative.

Displacement and total distance formulas

For an object with velocity v(t) from t = a to t = b:

\text{displacement} = \int_a^b v(t)\,dt = s(b) - s(a)

\text{total distance} = \int_a^b |v(t)|\,dt

To find a position, add the displacement to the starting position: s(b) = s(a) + \int_a^b v(t)\,dt.

How to find total distance:

1. Solve v(t) = 0 to find where the velocity may change sign. Keep only times inside [a, b].
2. Split [a, b] at those times.
3. Integrate v(t) on each piece.
4. Add the absolute values of the pieces.

Displacement and total distance examples

Example 1: Velocity that stays positive

A particle has velocity v(t) = 3t^2 + 2 meters per second. Find its displacement and total distance from t = 0 to t = 2.

\int_0^2 (3t^2 + 2)\,dt = \Big[t^3 + 2t\Big]_0^2

= (8 + 4) - 0 = 12

Answer: the displacement is 12 meters. Since 3t² + 2 is always positive, the particle never turns around, so the total distance is also 12 meters.

Example 2: Velocity that changes sign once

A particle has velocity v(t) = 2t - 6 feet per second. Find its displacement and total distance from t = 0 to t = 5.

\int_0^5 (2t - 6)\,dt = \Big[t^2 - 6t\Big]_0^5

= (25 - 30) - 0 = -5

2t - 6 = 0

t = 3

\int_0^3 (2t - 6)\,dt = (9 - 18) - 0 = -9

\int_3^5 (2t - 6)\,dt = (25 - 30) - (9 - 18) = 4

|-9| + |4| = 9 + 4 = 13

Answer: the displacement is −5 feet, so the particle ends 5 feet left of where it started. The total distance is 13 feet: 9 feet left, then 4 feet right.

Example 3: Velocity that changes sign twice

A particle has velocity v(t) = t^2 - 4t + 3. Find its displacement and total distance from t = 0 to t = 4.

t^2 - 4t + 3 = (t - 1)(t - 3)

t = 1, \; t = 3

F(t) = \dfrac{t^3}{3} - 2t^2 + 3t

F(0) = 0, \; F(1) = \dfrac{4}{3}, \; F(3) = 0, \; F(4) = \dfrac{4}{3}

\int_0^1 v\,dt = F(1) - F(0) = \dfrac{4}{3}

\int_1^3 v\,dt = F(3) - F(1) = -\dfrac{4}{3}

\int_3^4 v\,dt = F(4) - F(3) = \dfrac{4}{3}

\dfrac{4}{3} - \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{4}{3}

\dfrac{4}{3} + \dfrac{4}{3} + \dfrac{4}{3} = 4

Answer: displacement \dfrac{4}{3}, total distance 4. These match the shaded areas in the graph above.

Example 4: Back where it started

A weight on a spring has velocity v(t) = \cos t centimeters per second. Find its displacement and total distance from t = 0 to t = π.

\int_0^{\pi} \cos t\,dt = \Big[\sin t\Big]_0^{\pi} = 0 - 0 = 0

\cos t = 0 \text{ at } t = \dfrac{\pi}{2}

\int_0^{\pi/2} \cos t\,dt = \sin\dfrac{\pi}{2} - \sin 0 = 1

\int_{\pi/2}^{\pi} \cos t\,dt = \sin\pi - \sin\dfrac{\pi}{2} = -1

|1| + |-1| = 2

Answer: the displacement is 0 cm, because the weight ends where it started. The total distance is 2 cm: 1 cm out and 1 cm back.

Example 5: A remote-control car

A remote-control car moves along a straight track with velocity v(t) = t^2 - 6t + 8 feet per second. At t = 0 it is 2 feet from the starting line. Find its position at t = 5 and the total distance it travels from t = 0 to t = 5.

t^2 - 6t + 8 = (t - 2)(t - 4)

F(t) = \dfrac{t^3}{3} - 3t^2 + 8t

F(2) = \dfrac{8}{3} - 12 + 16 = \dfrac{20}{3}

F(4) = \dfrac{64}{3} - 48 + 32 = \dfrac{16}{3}

F(5) = \dfrac{125}{3} - 75 + 40 = \dfrac{20}{3}

s(5) = 2 + \left(\dfrac{20}{3} - 0\right) = \dfrac{26}{3}

\dfrac{20}{3} + \left|\dfrac{16}{3} - \dfrac{20}{3}\right| + \left|\dfrac{20}{3} - \dfrac{16}{3}\right|

= \dfrac{20}{3} + \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{28}{3}

Answer: at t = 5 the car is \dfrac{26}{3} \approx 8.67 feet from the starting line. It traveled \dfrac{28}{3} \approx 9.33 feet in all, because it backed up for a while between t = 2 and t = 4.

Common mistakes

Related methods

Velocity from acceleration

The same idea works one level down. If a(t) is the acceleration, then v(b) = v(a) + \int_a^b a(t)\,dt. First find v(t) from a(t), then use v(t) for displacement and distance.

Displacement from a velocity graph

If you only have a graph made of lines, find each signed area with geometry (see area of triangles). Area above the axis is positive and area below is negative. Add them for displacement; add their sizes for total distance.

Net change in other settings

Displacement is one case of the net change idea: the integral of any rate is the change in the amount. Water flow in and out of a tank works the same way. The average value of a function of v is the displacement divided by the time.

Practice problems

1. For v(t) = 4t + 1 on [0, 3], find the displacement and total distance.
2. For v(t) = 3t^2 on [1, 2], find the displacement and total distance.
3. For v(t) = t - 2 on [0, 6], find the displacement and total distance.
4. For v(t) = 3 - t on [0, 4], find the displacement and total distance.
5. For v(t) = t^2 - 1 on [0, 2], find the displacement and total distance.
6. For v(t) = \sin t on [0, 2π], find the displacement and total distance.
7. A particle starts at s(0) = 4 with v(t) = 2t - 4. Find its position at t = 3.
8. For v(t) = t^2 - 5t + 6 on [0, 4], find the displacement and total distance.
9. A particle has acceleration a(t) = 6t and starting velocity v(0) = −3. Find v(t), then the total distance on [0, 2].
10. A toy train moves along a straight track with velocity v(t) = 1.5 - 0.5t meters per second. Find its displacement and total distance from t = 0 to t = 5.

\int_0^3 (4t + 1)\,dt = \Big[2t^2 + t\Big]_0^3 = 18 + 3 = 21

\int_1^2 3t^2\,dt = \Big[t^3\Big]_1^2 = 8 - 1 = 7

\int_0^6 (t - 2)\,dt = \Big[\dfrac{t^2}{2} - 2t\Big]_0^6 = 18 - 12 = 6

t - 2 = 0

\int_0^2 (t - 2)\,dt = 2 - 4 = -2

\int_2^6 (t - 2)\,dt = 6 - (-2) = 8

|-2| + |8| = 10

\int_0^4 (3 - t)\,dt = \Big[3t - \dfrac{t^2}{2}\Big]_0^4 = 12 - 8 = 4

\int_0^3 (3 - t)\,dt = 9 - 4.5 = 4.5

\int_3^4 (3 - t)\,dt = 4 - 4.5 = -0.5

4.5 + 0.5 = 5

\int_0^2 (t^2 - 1)\,dt = \Big[\dfrac{t^3}{3} - t\Big]_0^2 = \dfrac{8}{3} - 2 = \dfrac{2}{3}

\int_0^1 (t^2 - 1)\,dt = \dfrac{1}{3} - 1 = -\dfrac{2}{3}

\int_1^2 (t^2 - 1)\,dt = \dfrac{2}{3} - \left(-\dfrac{2}{3}\right) = \dfrac{4}{3}

\dfrac{2}{3} + \dfrac{4}{3} = 2

\int_0^{2\pi} \sin t\,dt = \Big[-\cos t\Big]_0^{2\pi} = -1 - (-1) = 0

\int_0^{\pi} \sin t\,dt = -\cos\pi + \cos 0 = 2

\int_{\pi}^{2\pi} \sin t\,dt = -\cos 2\pi + \cos\pi = -2

2 + 2 = 4

s(3) = 4 + \int_0^3 (2t - 4)\,dt

\int_0^3 (2t - 4)\,dt = \Big[t^2 - 4t\Big]_0^3 = 9 - 12 = -3

s(3) = 4 + (-3) = 1

t^2 - 5t + 6 = (t - 2)(t - 3)

F(t) = \dfrac{t^3}{3} - \dfrac{5t^2}{2} + 6t

F(2) = \dfrac{14}{3}, \; F(3) = \dfrac{9}{2}, \; F(4) = \dfrac{16}{3}

\dfrac{9}{2} - \dfrac{14}{3} = -\dfrac{1}{6}

\dfrac{16}{3} - \dfrac{9}{2} = \dfrac{5}{6}

\dfrac{14}{3} + \dfrac{1}{6} + \dfrac{5}{6} = \dfrac{17}{3}

v(t) = -3 + \int_0^t 6u\,du = -3 + 3t^2

-3 + 3t^2 = 0

\int_0^1 (3t^2 - 3)\,dt = \Big[t^3 - 3t\Big]_0^1 = -2

\int_1^2 (3t^2 - 3)\,dt = (8 - 6) - (-2) = 4

2 + 4 = 6

\int_0^5 (1.5 - 0.5t)\,dt = \Big[1.5t - 0.25t^2\Big]_0^5 = 7.5 - 6.25 = 1.25

1.5 - 0.5t = 0

\int_0^3 (1.5 - 0.5t)\,dt = 4.5 - 2.25 = 2.25

\int_3^5 (1.5 - 0.5t)\,dt = 1.25 - 2.25 = -1

2.25 + 1 = 3.25

Displacement and total distance FAQ

What is the difference between displacement and distance?

Displacement is how far the object ends up from its starting point, with a sign for direction. Distance is how far it traveled in all, and it is never negative.

How do you find total distance traveled with an integral?

Integrate the speed, \int_a^b |v(t)|\,dt. In practice, split the interval where v(t) = 0, integrate each piece and add the absolute values.

Can displacement be negative?

Yes. A negative displacement means the object ended up in the negative direction from where it started, for example to the left or below.

How do you find position from velocity?

Add the displacement to the starting position: s(b) = s(a) + \int_a^b v(t)\,dt.

Summary

• Displacement = \int_a^b v(t)\,dt, the net change in position.
• Total distance = \int_a^b |v(t)|\,dt, which is never negative.
• For distance, split where v(t) = 0, integrate each piece, and add the absolute values.
• Position: s(b) = s(a) + \int_a^b v(t)\,dt.
• If v never changes sign, displacement and distance have the same size.

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