Wemazi

Topic 8.2

Distinguish final position, displacement, and distance

Builds on calculating total distance. 7 minutes.

Three numbers for one motion

A particle starts at the \(-1\) meter mark on a line. Its velocity, in meters per second, rises to 4 at \(t = 1\), falls to 0 at \(t = 3\) and reaches \(-2\) at \(t = 5\).

Drag the orange edge to t = 3. The green area is 6, so the particle moves 6 to the right, from \(-1\) to the 5 meter mark.

Drag the edge to t = 5. The red area is 2, so the particle moves 2 to the left and ends at the 3 meter mark.

The figure shows three numbers. The final position is where the particle is, 3. The displacement is the teal arrow from the start to the particle, 4. The total distance is the length of both arrows above the line, \(6 + 2 = 8\).

Where, how far from the start, how far traveled

The numbers are gone. The ring marks where the particle starts, and the orange particle marks where it is now.

Move the Start slider. The ring, the particle and the arrows move together, but no arrow changes its length. The final position depends on the start, but the displacement and the distance do not.

Press Flip the red area up. The red area moves above the axis, where it counts as positive, so the area now measures the distance. The teal arrow does not change, because the displacement counts a move to the left as negative.

Notation

  1. The final position is the starting position plus the integral of the velocity, so it needs the start.

    \[\cy{s(b)} = s(a) + \int_a^b v(t)\,dt\]

    This needs \(v\) to be continuous on \([a, b]\).

  2. The displacement is the change in position. It is the integral of the velocity, with signs, and it does not need the start.

    \[\cb{\int_a^b v(t)\,dt = s(b) - s(a)}\]

  3. The total distance is the integral of the speed \(|v(t)|\). It counts every move as positive, so it is the total length of the arrows above the line.

    \[\cg{\int_a^b |v(t)|\,dt}\]

  4. Use the words of the question. “Where is the particle?” asks for the position. “What is the change in position?” asks for the displacement. “How far did it travel?” asks for the distance.

  5. The final position is the start plus the displacement. The distance is at least the size of the displacement, because moves in opposite directions cancel in the displacement but add in the distance.

    \[\begin{gathered} s(b) = s(a) + \text{displacement} \\ \text{distance} \ge |\text{displacement}| \end{gathered}\]

Example 1: all three numbers

A cart starts at the 3 meter mark. Its velocity is \(v(t) = 8 - 4t\) meters per second for \(0 \le t \le 3\). Find its final position, its displacement and its total distance.

  1. The velocity is 0 at \(t = 2\), so the cart turns there. The green triangle from 0 to 2 has area 8, and the red triangle from 2 to 3 has area 2.

    \[\int_0^2 v\,dt = \cg{8}, \qquad \int_2^3 v\,dt = \cr{-2}\]

  2. The displacement adds the areas with their signs.

    \[\cb{8 - 2 = 6} \text{ meters}\]

  3. The final position adds the displacement to the start, 3.

    \[s(3) = 3 + 6 = \cy{9}\]

  4. The total distance adds the sizes of the areas.

    \[8 + 2 = 10 \text{ meters}\]

Example 2: the start matters

A particle starts at the 5 meter mark. Its velocity is \(v(t) = 3t^2 - 12\) meters per second. Find its position at \(t = 3\).

  1. The velocity is 0 at \(t = 2\). The red area from 0 to 2 is 16, and the green area from 2 to 3 is 7, so the signed total is \(-9\).

    \[\begin{aligned} \int_0^3 (3t^2 - 12)\,dt &= \Bigl[\,t^3 - 12t\,\Bigr]_0^3 \\ &= \cb{-9} \end{aligned}\]

  2. A tempting wrong move: The position is the integral, \(-9\).

    \[s(3) = \cr{-9}\]

    The integral is the change in position, not the position. The red particle at \(-9\) is where the particle would end from a start at 0. It started at 5, so it ends 9 meters to the left of the ring, at \(-4\).

  3. Add the start, 5, to the displacement.

    \[s(3) = 5 + (-9) = \cy{-4}\]

  4. For the same motion, the displacement is \(-9\) meters and the total distance is \(16 + 7 = 23\) meters.

Practice

A particle starts at the \(-2\) meter mark. Its velocity is \(v(t) = t^2 - 2t\) meters per second for \(0 \le t \le 3\). Find its final position, its displacement and its total distance.

Drag the edge to 3 seconds. The velocity is negative until \(t = 2\) and positive after it, so find the red and green areas first.

Solution

  1. The velocity is 0 at \(t = 2\), so the particle turns there. The two areas have equal size and opposite signs.

    \[\begin{gathered} \int_0^2 v\,dt = \cr{-\tfrac{4}{3}} \\ \int_2^3 v\,dt = \cg{\tfrac{4}{3}} \end{gathered}\]

  2. The displacement is 0, so the final position is the start, −2. The distance adds the sizes of the areas.

    \[\begin{gathered} \text{displacement} = 0, \quad s(3) = -2 \\ \text{distance} = \tfrac{4}{3} + \tfrac{4}{3} = \tfrac{8}{3} \end{gathered}\]

Questions

A particle starts at the 4 meter mark. The graph shows its velocity in meters per second: 2 for the first 5 seconds, then \(-3\) for 4 seconds.

  1. Where is the particle at \(t = 9\)?

    • 2 meters
    • \(-2\) meters
    • 22 meters
    Answer

    2 meters

  2. At what time is the particle farthest to the right?

    • \(t = 5\)
    • \(t = 9\)
    • \(t = 0\)
    Answer

    \(t = 5\)

  3. A particle starts at the 1 meter mark. Its velocity is \(v(t) = 3t - 6\) meters per second. What is its position at \(t = 4\)?

    • 1 meter
    • 0 meters
    • 12 meters
    Answer

    1 meter

Summary

The particle ends at position 3, its displacement is the teal arrow, 4, and its distance is the two arrows above the line, \(6 + 2 = 8\).

They differ because the position depends on the start, the displacement keeps signs, and the distance counts every move as positive.

You use the words of the question to choose between where the particle is, the change in position, and how far it traveled.

This is one of four skills in topic 8.2 of Which & Why: AP Calculus.