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FRQ: Vector Subtraction and Displacement — The Search-and-Rescue Base Camps

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  MID-LEVEL  |  10 points  |  25 min

▤ Scenario

A mountain search-and-rescue service runs two base camps on a wide, flat plain. Every position is quoted relative to a fixed survey marker O at the dispatch centre. Camp A lies 20.0 km due north of O, and Camp B lies 15.0 km due east of O. Figure 1 shows the layout, with the two distances drawn to one scale.

Figure 1. The two base camps and the survey marker O, with the quoted distances marked.

A helicopter stationed at Camp A is ordered to fly directly to Camp B.

Treat the plain as flat and horizontal, and neglect the curvature of Earth over these distances. All positions and displacements are measured in the ground frame, with the marker O as the origin, east as the +x direction and north as the +y direction. Every vector in this question lies in that horizontal plane.

✎ Free Response Questions

Part A

i.    On Figure 1, draw and label the position vector r_A of Camp A, the position vector r_B of Camp B, and the displacement vector Δr of the helicopter’s direct flight from Camp A to Camp B.

•   Draw the length of each arrow to represent the magnitude of each vector, consistent with the scale used in Figure 1.

•   Show clearly where the tail and the head of Δr lie.

ii.   Suppose instead that Camp A lies a distance d_A due north of the marker O and Camp B a distance d_B due east of it, the layout being otherwise unchanged. Starting from the definition of displacement as the change in position, derive an expression for the magnitude of the displacement from Camp A to Camp B, and an expression for the angle that this displacement makes with the due-east direction. State on which side of due east the displacement lies.

Express your final answer in terms of d_A and d_B, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

iii.  With the camps back at the distances given in the Scenario, calculate the magnitude of the helicopter’s direct displacement from Camp A to Camp B, and state its direction relative to a cardinal direction.

Part B

A supply aircraft based at Camp A is given a different task: it flies directly to the survey marker O, lands briefly, and then flies directly to Camp B. The dispatcher states:

“This supply flight and the direct flight have the same displacement, even though the supply flight travels the greater distance. And the route through the marker must always be the longer one: a route of two legs is always longer than a route of one.”

State whether the two flights have the same displacement, and explain why. Then use your expressions from part A (ii) to check whether the dispatcher’s second conclusion — that the route through the marker must always be the longer one — holds for any two camps placed as A and B are, one due north of the marker and the other due east of it.

Finally, indicate whether the reason she gives for that conclusion — that a route of two legs is always longer than a route of one — is sound in general, by writing one of the following.

______ Sound in general

______ Not sound in general

Justify your response. Then state the condition under which a flight’s distance travelled equals the magnitude of its displacement.

❖ 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

Position vectors are drawn from the chosen origin, so both arrows start at the marker O: r_A runs 20.0 km due north to Camp A, and r_B runs 15.0 km due east to Camp B, their lengths in the ratio 4 : 3.

Displacement is the change in position, so the arrow runs from where the flight starts to where it ends — tail at Camp A, head at Camp B. On the diagram it joins the head of r_A to the head of r_B, which is what Δr = r_B − r_A means. A displacement arrow is anchored at the starting position, not at the origin.

Figure 2. The position vectors from marker O, and the displacement Δr from Camp A to Camp B.

Reversing r_A and placing its tail at the head of r_B builds the same vector tip to tail, since r_B − r_A = r_B + (−r_A) — but that construction draws it starting from the marker, so it must then be shifted until its tail sits at Camp A.

Scoring (2 points):

▸ A1 — 1 point: Both position vectors are drawn as arrows from the marker O to their labelled camps — r_A to Camp A and r_B to Camp B — each labelled with its symbol.

▸ A2 — 1 point: The displacement is drawn as a single labelled arrow whose tail is at Camp A and whose head is at Camp B.

Scoring Note: A1 and A2 do not carry consistency credit; part A (i) is the first task in the question, so there is no upstream result for them to be consistent with.

⚠︎ Common error (partial credit): Drawing the displacement arrow from Camp B to Camp A, with the two position vectors otherwise correct — earns 1 of 2 points; that arrow is the displacement of the return flight, the negative of the one asked for.

⚠︎ Common error (partial credit): Drawing all three arrows from the marker O, so that the displacement also starts at the origin — earns 1 of 2 points; the position vectors are right, but a displacement arrow is anchored at the starting position, not at the origin.

✗ Common error (no credit): Drawing a single arrow from the marker O to Camp B and labelling it as both the position of Camp B and the displacement of the flight — earns 0 of 2 points; no position vector is drawn for Camp A, and an arrow that ends at Camp B but starts at the marker is a position, not the change in position the flight produces.

Part A (ii) — Model Answer

Displacement is the change in position, so it is the final position vector minus the initial one. The helicopter starts at Camp A and finishes at Camp B, which fixes which vector is subtracted from which:

Δr = r_B − r_A

Resolve each position into components in the stated coordinate system, with east as +x and north as +y. Camp A lies due north of the marker and Camp B due east of it, so each position vector has one zero component:

r_A = (0, d_A)         r_B = (d_B, 0)
Δr = (d_B − 0, 0 − d_A) = (d_B, −d_A)

The two components are perpendicular, so the magnitude follows from the Pythagorean relationship and the angle from the ratio of the component magnitudes:

|Δr| = √(d_A² + d_B²)         θ = tan⁻¹(d_A / d_B)

The east component is positive and the north component negative, so the displacement runs east and south of Camp A: θ is measured at Camp A from due east, turning south. The subtraction is carried out componentwise, so it is the signs of the components that fix the direction.

Scoring (3 points):

▸ A3 — 1 point: Begins from the definition of displacement as the change in position, written as a relationship between the two position vectors — Δr = r_B − r_A. This point is awarded for the correct starting relationship alone, before any components or algebra are written, and is earned even if the work that follows is incorrect. A relationship written between magnitudes or distances rather than between vectors does not earn it.

▸ A4 — 1 point: Establishes that the displacement runs d_B to the east and d_A to the south of Camp A — as the signed component pair Δr = (d_B, −d_A) in the stated coordinate system, or equivalently as the right triangle whose perpendicular legs are d_B east and d_A south.

Scoring Note: A3 and A4 do not carry consistency credit; part A (ii) is where the displacement relationship is first written. The part’s instruction to state on which side of due east the displacement lies is satisfied by the signed component pair, by a right triangle whose legs are named east and south, or by saying so in words — A4 is the criterion that carries it, and it is the side so established that A7 is judged consistent against.

▸ A5 — 1 point: Recombines the two perpendicular parts of the displacement into both a magnitude and a direction angle — |Δr| = √(d_A² + d_B²) and θ = tan⁻¹(d_A / d_B), the angle measured at Camp A from due east. The point is also earned where that recombination is correctly carried out on the response’s own perpendicular pair, provided the angle is still taken from the east–west axis and not from north or south.

Scoring Note: A correct, isolated, final pair of expressions — |Δr| = √(d_A² + d_B²) with θ = tan⁻¹(d_A / d_B) and the displacement placed south of due east — earns points A4 and A5.

⚠︎ Common error (partial credit): Reaching the correct signed components (d_B, −d_A) and the correct magnitude, but giving the angle from the wrong reference — θ = tan⁻¹(d_B / d_A), the angle from due south rather than from due east — earns 2 of 3 points; A3 and A4 stand, and A5 is not earned because the expression given is the complement of the angle the part asks for.

⚠︎ Common error (partial credit): Writing the starting relationship correctly but then executing the subtraction the other way round, reaching (−d_B, +d_A) and a direction north of west — earns 2 of 3 points; A3 stands, and A5 stands because the recombination of that pair is itself correctly carried out, but A4 is not earned: the vector obtained is the displacement from Camp B to Camp A.

⚠︎ Common error (partial credit): Adding the two position vectors instead of subtracting them, giving √(d_A² + d_B²) directed north of east — earns 1 of 3 points; only A5 stands, because that recombination is itself correctly carried out. A3 is not earned, since the relationship written is not the change in position, and A4 is not earned, since the direction established is north of east rather than east and south — the matching magnitude is a coincidence of r_A and r_B being perpendicular.

✗ Common error (no credit): Working in magnitudes throughout and never writing the vector relationship at all, so that |Δr| is taken to be d_B − d_A — earns 0 of 3 points; a vector difference is taken componentwise, and subtracting lengths discards the directions altogether.

Part A (iii) — Model Answer

Substitute d_A = 20.0 km and d_B = 15.0 km into the expressions derived in part A (ii):

Δr = (15.0 km, −20.0 km)
|Δr| = √((15.0)² + (20.0)²) = √(225 + 400) = √625 = 25.0 km
θ = tan⁻¹(20.0 / 15.0) = tan⁻¹(1.333) = 53.1°

Figure 3. The displacement resolved into east and north components, with θ from due east.

The helicopter’s direct displacement is 25.0 km, directed 53.1° south of east — equivalently, 36.9° east of south, or a compass bearing of 143.1°.

Note how the inverse tangent is used. Fed the signed ratio, tan⁻¹(−20.0 / 15.0) returns −53.1°: a calculator’s inverse tangent always names a direction in the right half of the plane, reporting the angle as though the east component were positive, whatever the true signs are. Here it genuinely is positive, so no correction is needed — the sign records only that the turn from east is toward the south. A bare angle is not a direction, so the axis it is measured from and the sense of the turn must both be named.

Scoring (2 points):

▸ A6 — 1 point: Obtains the magnitude 25.0 km with its unit. Accept any value that follows from correct substitution into the response’s own magnitude expression from part A (ii).

▸ A7 — 1 point: States the direction unambiguously against a cardinal direction: 53.1° south of east, or 36.9° east of south, or a compass bearing of 143.1°. Accept a direction that follows consistently from the response’s own angle expression and the side established in part A (ii).

⚠︎ Common error (partial credit): Giving the magnitude correctly but leaving the direction as a bare “53.1°” with no cardinal reference — earns 1 of 2 points; an angle carries no meaning until the axis it is measured from, and the sense of the turn, are both named.

⚠︎ Common error (partial credit): Reporting the direction as 53.1° north of east while the response’s own part A (ii) gave a displacement east and south — for instance by reading the negative sign returned by tan⁻¹(−20.0 / 15.0) as a turn toward the north — earns 1 of 2 points; the magnitude point stands, but that direction leads away from Camp B.

Part B — Model Answer

Take the dispatcher’s statement one conclusion at a time.

Her first conclusion is right. Displacement is the change in position, so it is fixed by the starting and finishing positions alone and carries no record of the route between them. Both flights begin at Camp A and end at Camp B, so both have the displacement found in part A (iii): 25.0 km directed 53.1° south of east. The dispatcher is right that the two distances differ — 25.0 km on the direct flight against 20.0 + 15.0 = 35.0 km by way of the marker — but distance travelled is a scalar that accumulates along the route, and it leaves the displacement untouched.

Her second conclusion is right as well, but the reason she gives for it is not. The expressions from part A (ii) settle the conclusion: the route through the marker has length d_A + d_B and the direct flight has length √(d_A² + d_B²), so comparing their squares gives

(d_A + d_B)² − (d_A² + d_B²) = 2·d_A·d_B

which is positive whenever d_A and d_B are both non-zero, so d_A + d_B > √(d_A² + d_B²) for every layout in which one camp lies genuinely north of the marker and the other genuinely east of it — the conclusion is general, not an accident of 20.0 and 15.0.

Figure 4. Both flights share one displacement; a stopover at P would add no distance.

Her reason, though, is false as stated. The distance travelled equals the magnitude of the displacement whenever the motion follows the straight line joining the two points without reversing along the way, and a two-leg route meets that condition exactly when the intermediate point lies on the straight segment between them. An aircraft that landed briefly at a point on the straight segment from Camp A to Camp B, and then flew on, would fly two legs and still cover only 25.0 km. The marker lies off the segment joining the two camps, so the supply flight is longer for its geometry and not for its number of legs.

Scoring (3 points):

▸ B1 — 1 point: States that both flights have the same displacement and gives the reason — displacement is the change in position, so it is fixed by the initial and final positions alone and carries no record of the route between them.

▸ B2 — 1 point: Tests the second conclusion against the part A (ii) expressions rather than against the two numbers, by comparing d_A + d_B with √(d_A² + d_B²) — for example by squaring, since (d_A + d_B)² − (d_A² + d_B²) = 2·d_A·d_B, which is positive whenever d_A and d_B are both non-zero. Accept a comparison of d_A + d_B against the response’s own magnitude expression from part A (ii), and award the point wherever that comparison is correctly carried out, whatever verdict it leads to; the two-leg length d_A + d_B is not carried through from part A (ii) and must be correct.

▸ B3 — 1 point: Indicates that the reason is not sound in general, and names the condition that replaces it — the distance travelled equals the magnitude of the displacement only for motion along the straight line joining the two points with no reversal, so a two-leg route is longer only when the intermediate point lies off the straight segment between them. Naming only the perpendicularity of the two legs does not earn this point: it explains this layout without addressing the blanket claim.

Scoring Note: B1 and B3 do not carry consistency credit; each is judged on its own physics and neither depends on a result from part A.

⚠︎ Common error (partial credit): Agreeing that both flights have the same displacement because displacement depends only on the endpoints and not on the route, and quoting 25.0 km against 35.0 km, but indicating “Sound in general” and offering neither the symbolic comparison nor the condition — earns 1 of 3 points; B1 stands, B2 is not earned because the check is numerical rather than symbolic, and B3 is not earned.

⚠︎ Common error (partial credit): Giving the displacement reasoning and the symbolic comparison correctly but indicating “Sound in general”, so that a route of two legs is left standing as always the longer one — earns 2 of 3 points.

⚠︎ Common error (partial credit): Giving the displacement reasoning correctly, then rejecting the dispatcher’s second conclusion along with her reason — naming the condition correctly, but concluding from it that the route through the marker need not be the longer one — earns 2 of 3 points; the reason is indeed false in general, but the marker lies off the segment joining the two camps, so the conclusion itself stands.

✗ Common error (no credit): Stating that the supply flight has the larger displacement because it covers more ground, offering no comparison of d_A + d_B with √(d_A² + d_B²), and indicating “Sound in general” with no condition named — earns 0 of 3 points; this treats displacement and distance travelled as one quantity, when the first is a vector fixed by the endpoints and the second a scalar that accumulates along the route.

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