Same Displacement, Different Distance? Two Hikers FRQ
Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.
FRQ: Displacement and Distance in One Dimension — Two Hikers on a Mountain Trail
Assessments aligned to 2026 AP Physics 1 standards
Mathematical Routines (MR) | MID-LEVEL | 10 points | 25 min
▤ Scenario
Two hikers, Priya and Sam, begin at the same trailhead on a straight north–south mountain trail. The trailhead is defined as the origin (x = 0 m), and north is defined as the positive direction.
Priya hikes north to a viewpoint at x = +3200 m, then south to a rest stop at x = +800 m, and finally north to the summit at x = +4400 m. Sam hikes directly north from the trailhead to the summit at x = +4400 m, without stopping at the viewpoint or the rest stop.
All motion is one-dimensional along the trail. Use the ground as the reference frame, and treat each hiker as a single point object.
✎ Free Response Questions
Part A
i. On the position axis shown in Figure 1, represent Priya’s journey from the trailhead to the summit.
• Label the trailhead, the rest stop, the viewpoint, and the summit with their position coordinates.
• Draw an arrow for each of the three legs of Priya’s journey, from the position where the leg begins to the position where it ends, and label each arrow with the sign of that leg’s displacement.
Figure 1 — Position axis for the trail, with x measured from the trailhead.
ii. Calculate Priya’s total displacement and her total distance travelled from the trailhead to the summit.
iii. Consider a third hiker on the same trail who starts at the trailhead (x = 0 m) and walks directly to a position x_1, then directly to a position x_2, then directly to a position x_3, where 0 ≤ x_2 ≤ x_1 and x_2 < x_3. Working in symbols throughout, derive an expression for the total distance d travelled by that hiker, and use it to determine the condition under which d is equal to the magnitude of that hiker’s displacement. Express your final answer in terms of x_1, x_2, x_3, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
Part B
A park ranger says: “Priya and Sam both started at the trailhead and ended at the summit. Priya’s displacement is therefore the same as Sam’s, and her total distance travelled is the same as his.”
Let d_Priya and d_Sam be the total distances travelled by Priya and Sam, respectively. Indicate whether d_Priya is greater than, less than, or equal to d_Sam by writing one of the following.
• d_Priya > d_Sam
• d_Priya < d_Sam
• d_Priya = d_Sam
Justify your answer. In your justification, include qualitative reasoning beyond mathematical derivations or expressions.
State whether the ranger’s claim is fully correct, partially correct, or fully incorrect, and explain why the part of the claim about displacement does or does not hold.
❖ Answer Key & Scoring Guide
▸ earns credit ⚠︎ common error, partial credit ✗ common error, no credit
Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.
Part A (i) — Model Answer
A correct diagram places the four positions on the axis — trailhead 0 m, rest stop +800 m, viewpoint +3200 m and summit +4400 m — and shows Priya’s journey as three arrows, one per leg, each drawn from the position where the leg begins to the position where it ends:
• Leg 1: from 0 m to +3200 m, pointing north — positive displacement
• Leg 2: from +3200 m to +800 m, pointing south — negative displacement
• Leg 3: from +800 m to +4400 m, pointing north — positive displacement
Each arrow is that leg’s displacement: it points the way the hiker walked, and its length is the magnitude of the displacement, equal to the leg’s length because each leg is walked in one direction. In one dimension the sign of a displacement completely describes its direction, so an arrow and a sign are a complete representation — no vector notation is required.
Figure 2 — Priya’s journey. Legs 1 and 3 have positive displacements; leg 2 has a negative one.
Scoring (2 points):
▸ A1 — 1 point: For placing the trailhead, the rest stop, the viewpoint and the summit on the axis at 0 m, +800 m, +3200 m and +4400 m, each identified by name and labelled with its coordinate (the 0 printed on the axis serves for the trailhead).
▸ A2 — 1 point: For three arrows, one per leg, each drawn from the position where the leg begins to the position where it ends — 0 m to +3200 m, +3200 m to +800 m, +800 m to +4400 m — and labelled with the sign of that leg’s displacement: +, −, +.
⚠︎ Common error (partial credit): Places and labels all four positions correctly but draws the arrow for leg 2 pointing north, with its tail at the rest stop and its head at the viewpoint — earns 1 of 2 points, A1.
⚠︎ Common error (partial credit): Places and labels all four positions correctly but draws three northward arrows between neighbouring landmarks — trailhead to rest stop, rest stop to viewpoint, viewpoint to summit — which trace the trail, not Priya’s journey — earns 1 of 2 points, A1.
✗ Common error (no credit): Draws a single arrow from the trailhead to the summit and marks only those two positions — neither the rest stop and the viewpoint nor the three legs appear, so neither A1 nor A2 is earned.
Part A (ii) — Model Answer
Displacement depends only on the initial and final positions:
Δx = x_f − x_i = (+4400 m) − (0 m) = +4400 m
Priya’s displacement is 4400 m north. Distance is the total length of the route: the sum of the lengths of the three legs. Each leg is walked in one direction, so its length is the magnitude of its change in position.
leg 1: |(+3200 m) − (0 m)| = 3200 m
leg 2: |(+800 m) − (+3200 m)| = 2400 m
leg 3: |(+4400 m) − (+800 m)| = 3600 m
d_Priya = 3200 m + 2400 m + 3600 m = 9200 m
Priya’s total distance travelled is 9200 m. Distance is a scalar and carries no direction, while displacement carries both a magnitude and a sign.
Scoring (2 points):
▸ A3 — 1 point: For a displacement of +4400 m (4400 m, or 4400 m north).
▸ A4 — 1 point: For a total distance travelled of 9200 m.
Scoring Note: A4 accepts any correct method — the sum of the three leg lengths, or the magnitude of the displacement, 4400 m, plus twice the 2400 m stretch that Priya walks back over. A3 and A4 each require units; A3 is not earned for −4400 m or 4400 m south. A3 and A4 do not depend on the response in part A (i).
⚠︎ Common error (partial credit): Gives the displacement correctly, then adds the signed leg displacements, (+3200 m) + (−2400 m) + (+3600 m), and reports 4400 m as the distance too — earns 1 of 2 points, A3. Adding signed changes in position gives the displacement again; the distance adds their magnitudes.
⚠︎ Common error (partial credit): Subtracts the positions the wrong way round, x_i − x_f, and gives the displacement as −4400 m, with the distance correct — earns 1 of 2 points, A4. Displacement is the final position minus the initial position, and Priya ends north of where she started.
✗ Common error (no credit): Reports the displacement as 9200 m and the distance as 4400 m — the two definitions are swapped, so neither A3 nor A4 is earned.
Part A (iii) — Model Answer
Start from the definition of displacement:
Δx = x_f − x_i
Each leg is walked directly, in one direction only, so its length is the magnitude of its displacement. With 0 ≤ x_2 ≤ x_1 and x_2 < x_3:
d_1 = |x_1 − 0| = x_1
d_2 = |x_2 − x_1| = x_1 − x_2
d_3 = |x_3 − x_2| = x_3 − x_2
Adding the three leg lengths gives the total distance:
d = x_1 + (x_1 − x_2) + (x_3 − x_2) = 2x_1 − 2x_2 + x_3
The hiker’s displacement is Δx = x_3 − 0 = x_3, and |Δx| = x_3 because x_3 > x_2 ≥ 0. Comparing the two:
d − |Δx| = 2x_1 − 2x_2 = 2(x_1 − x_2)
Because x_1 ≥ x_2, that difference is never negative, so d ≥ |Δx|. The extra distance is twice the length of the leg walked back, so d equals |Δx| only when x_1 = x_2 — only when the hiker never turns around.
Scoring (3 points):
▸ A5 — 1 point: For including the length of a leg as the magnitude of its change in position — stated generally for a single leg walked in one direction, written with absolute-value bars for any leg, or applied to leg 2, the leg walked back, as d_2 = x_1 − x_2. This point is earned for the starting relationship alone, even if the steps that follow are incorrect.
Scoring Note: A term for leg 1 or leg 3 written without bars (x_1, or x_3 − x_2) does not meet A5 on its own: the change in position along those legs is never negative, so it does not show that the magnitude was taken. Distance = |Δx| stated for the whole trip does not meet A5.
▸ A6 — 1 point: For a correct expression for d, such as d = 2x_1 − 2x_2 + x_3.
Scoring Note: A correct, isolated, final expression for d earns points A5 and A6. Any correct form earns A6, including the unsimplified x_1 + |x_2 − x_1| + |x_3 − x_2|.
▸ A7 — 1 point: For the condition under which d equals |Δx| = x_3, consistent with the response’s expression for d — x_1 = x_2 for the correct expression, or in words, that the hiker never turns around; for an expression that reduces to d = x_3, the consistent statement is that the two are always equal.
Scoring Note: A7 requires the condition reduced — to a relation between the positions, to words (the hiker never turns around), or to “always equal” where the response’s expression gives that; 2x_1 − 2x_2 + x_3 = x_3 left unreduced does not meet it. A5 and A6 carry no consistency credit, and A5–A7 do not depend on parts A (i) and A (ii).
⚠︎ Common error (partial credit): Derives d correctly but states no condition, or gives only a numerical case in which d and |Δx| differ — earns 2 of 3 points, A5 and A6.
⚠︎ Common error (partial credit): Writes d = x_1 + x_2 + x_3, taking each position as the length of the leg that ends there, with no statement of how a leg’s length relates to its change in position, and concludes that d equals |Δx| only when x_1 = x_2 = 0 — earns 1 of 3 points, A7. A position is measured from the trailhead; a leg’s length is the magnitude of the change in position along it.
⚠︎ Common error (partial credit): Adds the signed changes in position, x_1 + (x_2 − x_1) + (x_3 − x_2), with no absolute-value bars, so that d = x_3, and concludes that d always equals |Δx| — earns 1 of 3 points, A7, as consistency credit only: in fact d equals |Δx| only when x_1 = x_2. Adding signed changes gives the displacement; the distance adds their magnitudes.
Part B — Model Answer
d_Priya > d_Sam
Both hikers begin at x_i = 0 m and end at x_f = +4400 m, so applying Δx = x_f − x_i to each gives the same displacement, +4400 m. The ranger is right about that part of the claim.
Figure 3 — Same displacement, different distance: Priya covers 9200 m, Sam 4400 m.
Distance, however, is the total length of the route. Sam never turned around, so his distance equals the magnitude of his displacement, 4400 m. Priya turned around at the viewpoint, so she walked the 2400 m stretch between the rest stop and the viewpoint three times — north, south, then north again — where Sam walked it once. Her distance exceeds his by 2 × 2400 m = 4800 m, the 2(x_1 − x_2) of part A (iii). Equal endpoints fix the displacement but set only a lower limit on the distance, which is why the ranger’s claim is only half right.
The ranger’s claim is therefore partially correct: right about the displacement, which depends only on the initial and final positions, and wrong about the distance.
Scoring (3 points):
▸ B1 — 1 point: For indicating d_Priya > d_Sam.
▸ B2 — 1 point: For a justification that indicates that Priya’s reversal makes her walk part of the trail more than once.
▸ B3 — 1 point: For stating that the ranger’s claim is partially correct, with the reason the displacement part holds: displacement depends only on the initial and final positions.
Scoring Note: B1, B2 and B3 are scored independently of part A and of one another. A statement that distance depends on the route supports the reversal argument but does not meet B2 on its own. B3 requires the principle that displacement depends only on the initial and final positions — in words, or as Δx = x_f − x_i applied to both hikers; restating the shared endpoints alone, as the ranger does, does not meet it.
⚠︎ Common error (partial credit): Indicates d_Priya > d_Sam and gives the reversal argument, but calls the claim fully incorrect, arguing that Priya’s longer route also gives her the larger displacement — earns 2 of 3 points, B1 and B2. Displacement depends only on where a hiker starts and ends, not on the route.
⚠︎ Common error (partial credit): Indicates d_Priya > d_Sam because Priya stopped at the viewpoint and the rest stop and Sam did not, and calls the claim partially correct because displacement depends only on where each hiker starts and ends — earns 2 of 3 points, B1 and B3. A stop adds no distance; the extra distance comes from walking back over the same ground.
⚠︎ Common error (partial credit): Indicates d_Priya > d_Sam and supports it only with the values 9200 m and 4400 m, with no reference to the reversal and no verdict on the claim — earns 1 of 3 points, B1.
✗ Common error (no credit): Indicates d_Priya = d_Sam because both hikers covered the same trail between the same two points, and calls the claim fully correct — displacement and distance are treated as one quantity, so no criterion is earned.
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