Calculate displacement
Left, then right
A particle moves along a line with velocity \(v(t) = 2t - 6\) meters per second. Positive velocity means motion to the right, and negative velocity means motion to the left. The particle starts at 0.
Drag the orange edge to \(t = 3\). The velocity is negative until then, so the area so far lies below the axis and is red: it counts as negative. Its size is 9, so the teal arrow from the start to the particle points left and shows \(-9\).
Drag the edge to \(t = 6\). The green area above the axis is also 9, and it cancels the red area. The particle is back at its start, so the arrow has length 0.
The signed area is the displacement
The numbers are gone. The teal arrow from the starting position to the present position is the displacement. It is positive when it points to the right and negative when it points to the left.
Drag the orange edge. Each green piece of area moves the particle to the right by that area, and each red piece moves it to the left. The arrows above the number line trace these moves, and the teal arrow adds them with their signs.
Move the Start slider. The particle and the arrow move together, but the arrow keeps its length and direction, because the displacement depends only on the velocity.
The displacement is the signed area under the velocity graph: green areas add and red areas subtract.
Notation
The particle is at \(s(t_1)\) at time \(t_1\) and at \(\cy{s(t_2)}\) at time \(\cy{t_2}\). The displacement is the final position minus the starting position, and it is the teal arrow.
\[\cb{s(t_2) - s(t_1)}\]Velocity is the rate of change of position, so by the Fundamental Theorem of Calculus its integral is this change in position.
\[\cb{\int_{t_1}^{t_2} v(t)\,dt} = s(t_2) - s(t_1)\]Keep the sign of \(v\), and do not take an absolute value. Area above the axis counts as positive and area below it counts as negative.
\[\cb{\int_{t_1}^{t_2} v(t)\,dt} = \cg{A_{\text{above}}} - \cr{A_{\text{below}}}\]A positive displacement means the particle ends to the right of its start, and a negative one means it ends to the left. The unit is the unit of position.
The Fundamental Theorem needs a continuous velocity. If the velocity jumps, as in Example 1, integrate each piece and add the results.
Example 1: two constant velocities
A tram moves along a straight track at 5 meters per second for the first 4 seconds. Then it moves at \(-2\) meters per second for the next 3 seconds. Find its displacement from \(t = 0\) to \(t = 7\).
The first piece is a green rectangle, 4 wide and 5 tall. Its area is how far the tram moves to the right.
\[\cg{5 \cdot 4 = 20}\]The second piece is a red rectangle, 3 wide and 2 deep below the axis. It counts as negative, because the tram moves to the left.
\[\cr{-2 \cdot 3 = -6}\]Add the two signed areas. The tram ends 14 meters to the right of its start.
\[\cb{20 - 6 = 14}\]
Example 2: a curve that crosses the axis
A particle has velocity \(v(t) = t^2 - 2t - 3\) meters per second. Find its displacement from \(t = 0\) to \(t = 5\).
Factor the velocity: \(v = (t - 3)(t + 1)\). It is negative for \(t < 3\) and positive for \(t > 3\), so the area from 0 to 3 is red and the area from 3 to 5 is green. Integrate each piece with the antiderivative \(F(t) = \tfrac{t^3}{3} - t^2 - 3t\).
\[\int_0^3 v\,dt = \cr{-9}, \qquad \int_3^5 v\,dt = \cg{\tfrac{32}{3}}\]A tempting wrong move: Add the sizes of the two areas.
\[9 + \tfrac{32}{3} = \cr{\tfrac{59}{3}}\]The red area lies under the axis, so the particle moves to the left during that time and the area subtracts. The red particle on the number line marks this answer near 19.7, but the orange particle ends near 1.7.
Count the red area as negative. The signed total is the displacement.
\[-9 + \tfrac{32}{3} = \cb{\tfrac{5}{3}}\]One integral from 0 to 5 gives the same result, \(F(5) - F(0) = \tfrac53\), because it already counts the red area as negative. The particle ends \(\tfrac53\) meters to the right of its start.
Practice
A particle has velocity \(v(t) = 12 - 3t^2\) meters per second. Find its displacement from \(t = 0\) to \(t = 3\).
The graph is above the axis until \(t = 2\) and below it after. Both areas count, each with its own sign.
Solution
An antiderivative of the velocity is \(12t - t^3\). Evaluate it at 3 and at 0.
\[\begin{aligned} \int_0^3 (12 - 3t^2)\,dt &= \Bigl[\,12t - t^3\,\Bigr]_0^3 \\ &= 36 - 27 = \cb{9} \end{aligned}\]The displacement is 9 meters to the right. As a check, the green area on \([0, 2]\) is 16 and the red area on \([2, 3]\) counts as \(-7\), so \(16 - 7 = 9\).
Questions
The figure shows the velocity of a particle on a line. It is 3 meters per second for 3 seconds, then \(-2\) meters per second for 4 seconds.
What is the displacement from \(t = 0\) to \(t = 7\)?
- 1 meter
- 17 meters
- 9 meters
Answer
1 meter
How does the red area enter the displacement?
- It subtracts: it counts as \(-8\).
- It adds: it counts as 8.
- It does not enter the displacement.
Answer
It subtracts: it counts as \(-8\).
A particle has velocity \(v(t) = 3t^2 - 6t\) meters per second. Find its displacement from \(t = 0\) to \(t = 4\).
- 16 meters
- 24 meters
- 20 meters
Answer
16 meters
Summary
You find a displacement by integrating the signed velocity over the time interval.
This works because velocity is the rate of change of position, so its integral is the final position minus the starting position.
You use it when a question asks how far from its start a particle ends, and on which side.
This is one of four skills in topic 8.2 of Which & Why: AP Calculus.