Distance and Displacement Problems

Distance is the easy one: add up every leg, ignore direction

To find total distance you simply add the length of every part of the journey, direction never enters into it, so the answer just keeps growing.

If a trip has legs of $8\ \text{km}$, $6\ \text{km}$ and $3\ \text{km}$, the total distance is $8 + 6 + 3 = 17\ \text{km}$, regardless of how the legs are oriented or whether the traveller doubles back. Distance is a scalar, so every leg counts as a positive addition.

Total distance is the arithmetic sum of all the path lengths in a journey, ignoring direction. It is a scalar and always increases as more ground is covered, even if the traveller returns to the start.

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When motion reverses, direction (sign) decides the answer

We just saw that distance simply sums the legs. That raises a question: what happens to displacement when a journey doubles back on itself? This card answers it → choose a positive direction and let opposite legs subtract.

For motion along one line, pick one direction as positive; legs going the other way are negative, and the displacement is the signed total from start to finish.

Set a sign convention (say east $= +$, west $= -$). Then add the signed legs to get the net displacement, keeping the sign to show direction.

For straight-line motion, choose one direction positive and add the signed legs: $s = \sum (\pm \text{leg})$. The sign of the result gives the direction. Distance still sums all legs as positives.

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When legs are at right angles, combine them with Pythagoras

We just learned to handle reversals along a straight line. That raises a question: what if the legs point in different, perpendicular directions? This card answers it → resolve into east–west and north–south totals, then combine with Pythagoras.

For a 2D journey, add up all the east–west motion and all the north–south motion separately, then join those two totals into one displacement using the right-triangle rule.

Total the $x$ (east–west) legs and the $y$ (north–south) legs with signs. The displacement magnitude is $s = \sqrt{s_x^2 + s_y^2}$ and its direction is $\theta = \tan^{-1}(s_y / s_x)$. For our hook, $8\ \text{km}$ east and $6\ \text{km}$ north give $s = \sqrt{8^2 + 6^2} = 10\ \text{km}$, while the distance is $14\ \text{km}$.

In 2D, total the perpendicular legs separately, then combine: displacement $s = \sqrt{s_x^2 + s_y^2}$ at direction $\theta = \tan^{-1}(s_y/s_x)$. Distance is still the simple sum of all leg lengths.

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