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FRQ: Vector Components and the Direction of a Resultant — A Drone’s Two-Leg Survey Flight

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A survey crew maps a flat, rectangular construction site using a drone that reports each flight leg as a displacement on the site grid. The drone lifts off from a corner marker and flies two straight legs at a constant altitude.

On the first leg the onboard sensor reports a displacement d whose magnitude is d = 48.0 m and whose direction is θ = 215°. For the second leg, flown immediately after the first, the sensor reports the displacement e directly in component form: e_x = +18.0 m and e_y = −32.0 m.

The crew needs the drone’s net displacement from the corner marker, r = d + e. Figure 1 shows the site grid on which every displacement is quoted, together with the values the sensor reported.

Figure 1. The site grid at the corner marker, with the angle convention and the sensor’s values.

Both legs are flown at the same altitude, so every displacement lies in the horizontal plane and the problem is two-dimensional. Use the site grid as the reference frame: +x is east, +y is north, and angles are measured counterclockwise from +x. Throughout, r, d and e name the three displacement vectors; the same letter in a scalar expression or a variable list names that vector’s magnitude, so d = 48.0 m is the magnitude of the first leg, and |r| names the magnitude of r.

✎ Free Response Questions

Part A

i.    On the axes of Figure 1, sketch the vector d with its tail at the origin. Then, without calculating either component, state whether d_x and d_y are positive, negative, or zero. Justify each sign using the quadrant in which d lies.

ii.   Starting from the relationship between a vector and its perpendicular components, derive an expression for the magnitude of the net displacement r.

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

iii.  Evaluate your expression from part A (ii) for the values given in the Scenario. Show d_x, d_y, r_x and r_y along the way, and state the magnitude of r with its unit.

Part B

A member of the survey crew needs θ_r, the direction of r measured counterclockwise from the +x axis, and evaluates tan⁻¹(r_y / r_x) to find it. The crew member’s calculator returns 70.3°, and the crew member records 70.3° as the direction of r.

Indicate whether θ_r is greater than, less than, or equal to 70.3° by writing one of the following.

•   θ_r > 70.3°

•   θ_r < 70.3°

•   θ_r = 70.3°

Justify your response. In your justification, state the general condition under which the value an inverse tangent returns must be corrected before it can be reported as the direction of a vector.

❖ 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

The angle θ = 215° lies between 180° and 270°, so d points into Quadrant III — down and to the left of the origin.

d_x < 0 (negative)         d_y < 0 (negative)

In Quadrant III the arrow’s projection onto the x-axis points along −x and its projection onto the y-axis points along −y, so both components are negative. Neither component is zero, because 215° is not a quadrantal angle (0°, 90°, 180°, 270°), where one component would vanish. The sign of each component is fixed by the quadrant the arrow points into, so both signs can be read off the sketch before anything is calculated.

Figure 2. d with its tail at the origin; 215° counterclockwise from +x places it in Quadrant III.

Scoring (2 points):

▸ A1 — 1 point: For a sketch showing d with its tail at the origin and its head in Quadrant III — down and to the left — on the supplied axes. This point is awarded for the drawn representation alone.

▸ A2 — 1 point: For justifying both signs from the quadrant: in Quadrant III the projection onto the x-axis points along −x and the projection onto the y-axis points along −y, so d_x < 0 and d_y < 0.

Scoring Note: A response that states both signs and marks the two projections on its own sketch, labelled along −x and along −y, has justified them from the quadrant and earns A2; the justification need not be in words. Noting that neither component is zero, because 215° is not a quadrantal angle, is not required for A2. A2 is earned on the quadrant argument the response gives; computing d_x and d_y as well does not forfeit it, but a response whose only support for the signs is a computed or calculator-quoted value has not justified them from the quadrant. A2 carries no consistency clause with A1: a response that sketches d into the wrong quadrant and then argues the signs correctly for that quadrant has justified the signs of a different leg from the one the sensor reported.

⚠︎ Common error (partial credit): The sketch is correct, but the signs are justified by quoting cos 215° = −0.819 and sin 215° = −0.574 from a calculator rather than from the quadrant. The part asks for each sign to be justified using the quadrant in which d lies, so the justification A2 requires is absent — earns A1 only, 1 of 2 points.

✗ Common error (no credit): Draws d in Quadrant II, either by swinging the extra 35° past 180° upward from the −x axis instead of downward, or by measuring 215° clockwise from +x. Both place the head up and to the left and give d_y > 0, so the drawn leg is not the one the sensor reported and neither A1 nor A2 is earned.

Part A (ii) — Model Answer

A vector is the resultant of two perpendicular components, and those components follow from its magnitude and its direction angle:

d_x = d cos θ         d_y = d sin θ

Because r = d + e, each component of r is the sum of the corresponding components of the two legs:

r_x = d cos θ + e_x         r_y = d sin θ + e_y

r_x and r_y are perpendicular, so they are the two legs of a right triangle whose hypotenuse has length |r|. The Pythagorean relationship a² + b² = c² then gives the magnitude:

|r| = √( (d cos θ + e_x)² + (d sin θ + e_y)² )

The expression takes the first leg into components, adds each axis separately, and recombines the two sums through the right triangle they form. Vectors add by components and not by magnitudes, because each axis carries its own independent one-dimensional sum.

Scoring (3 points):

▸ A3 — 1 point: For starting from the relationship between a vector and its perpendicular components — writing d_x = d cos θ and d_y = d sin θ, or stating that d is the resultant of two perpendicular components and giving the trigonometric ratios that produce them. This point is awarded for the correct starting relationship alone, before any algebra or substitution.

▸ A4 — 1 point: For adding the two legs component by component, reaching r_x = d cos θ + e_x and r_y = d sin θ + e_y. Award this point for a component sum consistent with the response’s own expressions for d_x and d_y.

▸ A5 — 1 point: For applying the Pythagorean relationship to the two perpendicular components of r, reaching |r| = √((d cos θ + e_x)² + (d sin θ + e_y)²). Award this point wherever the relationship is correctly applied to the response’s own expressions for r_x and r_y, whatever those expressions are.

Scoring Note: A correct, isolated, final expression for |r| earns points A4 and A5. A3 is a starting-relationship criterion and carries no consistency clause — it is earned or not on the relationship the response itself writes down.

⚠︎ Common error (partial credit): Exchanges the two trigonometric functions, writing d_x = d sin θ and d_y = d cos θ. The starting relationship is wrong, so A3 is not earned; the component sum and the Pythagorean step are then carried out correctly on those components, so A4 and A5 are — earns 2 of 3 points.

⚠︎ Common error (partial credit): Writes the component relationship d_x = d cos θ, d_y = d sin θ, then adds the magnitudes rather than the components, reaching |r| = d + √(e_x² + e_y²). A3 is earned; no component sum for r is formed and the Pythagorean relationship is never applied to r’s own components, so A4 and A5 are lost — earns 1 of 3 points.

✗ Common error (no credit): Writes |r| = √(d² + e_x² + e_y²), combining the first leg’s magnitude with the second leg’s components as though the three were mutually perpendicular. The direction of the first leg never enters, so there is no component relationship for d, no component sum for r, and no application of the Pythagorean relationship to two perpendicular components of r.

Part A (iii) — Model Answer

Substituting the values given in the Scenario, with the calculator set to degrees:

d_x = (48.0 m) cos 215° = −39.3 m         d_y = (48.0 m) sin 215° = −27.5 m
r_x = (−39.3 m) + (+18.0 m) = −21.3 m         r_y = (−27.5 m) + (−32.0 m) = −59.5 m
|r| = √( (−21.3 m)² + (−59.5 m)² ) = √(3994 m²) ≈ 63.2 m

The drone finishes 63.2 m from the corner marker, and both components of r are negative, so that endpoint lies south-west of it. The net magnitude 63.2 m falls well short of the 84.7 m the two leg magnitudes would give, because the legs’ x-components carry opposite signs and partly cancel.

Figure 3. The two legs head to tail and the resultant r = d + e; the components add separately.

Scoring (2 points):

▸ A6 — 1 point: For all four component values with their signs — d_x ≈ −39.3 m, d_y ≈ −27.5 m, r_x ≈ −21.3 m and r_y ≈ −59.5 m. Award this point for components consistent with the response’s own expressions from part A (ii); this point may be earned independently of the response in part A (ii), including where no expression was given there.

▸ A7 — 1 point: For the magnitude |r| ≈ 63.2 m, stated with its unit. Award this point for a magnitude consistent with the response’s own values for r_x and r_y; where the response states no values for r_x and r_y, award it for |r| ≈ 63.2 m stated with its unit.

Scoring Note: Intermediate values may be rounded to three significant figures or carried unrounded; both routes give |r| ≈ 63.2 m.

⚠︎ Common error (partial credit): Reports the four components correctly but combines them as |r| = |r_x| + |r_y| = 21.3 m + 59.5 m = 80.8 m, adding two perpendicular components arithmetically instead of through the Pythagorean relationship — earns A6 only, 1 of 2 points.

⚠︎ Common error (partial credit): Drops the signs, reporting d_x = 39.3 m and d_y = 27.5 m and then r_x = 57.3 m and r_y = 59.5 m, as though a component were a length rather than a signed projection. A6 requires the signs and is not earned; the magnitude 82.6 m that follows is consistent with those components, so A7 is — earns 1 of 2 points.

Part B — Model Answer

θ_r > 70.3°

From part A (iii), r_x ≈ −21.3 m and r_y ≈ −59.5 m. Both components are negative, so r points down and to the left of the corner marker, into Quadrant III, where directions measured counterclockwise from +x run from 180° to 270°. An angle of 70.3° points up and to the right, into Quadrant I, where both components would have to be positive. The two directions cannot describe the same displacement, and the Quadrant III direction is the larger angle.

The inverse tangent cannot tell the two apart. It is handed only the single ratio r_y / r_x, and that ratio is the same positive number for a pair of negative components as for a pair of positive ones; its principal values run only from −90° to +90°, which covers the right half of the plane, so it always answers as though the x-component were positive. Here it returns the Quadrant I angle, and 180° must be added:

θ_r = 70.3° + 180° = 250.3° ≈ 250°

Figure 4. Principal value and true direction differ by 180°; the component signs separate them.

The net displacement is 63.2 m at approximately 250° counterclockwise from +x — south-west of the corner marker, and in the same quadrant as the first leg sketched in part A (i).

General condition: the value the inverse tangent returns must be corrected by adding 180° whenever the x-component is negative, that is, in Quadrants II and III. Where the x-component is positive the returned value already gives the direction, and where that value is itself negative (Quadrant IV) adding 360° expresses the same direction as a positive angle. The condition assumes r_x ≠ 0; when r_x = 0 the ratio is undefined, there is no returned value to correct, and the direction is read off directly as 90° if r_y > 0 or 270° if r_y < 0.

A reported direction can always be tested against the component signs, because r_x = |r| cos θ_r and r_y = |r| sin θ_r with |r| > 0, so each component must carry the sign of its own trigonometric function. The component signs decide the quadrant, so an inverse tangent’s answer is only a candidate until those signs confirm it.

Scoring (3 points):

▸ B1 — 1 point: For indicating the comparison that follows from the response’s own signs for r_x and r_y in part A (iii). For the correct signs r_x < 0 and r_y < 0 that comparison is θ_r > 70.3°. Award the point wherever the indicated comparison is the one the response’s own component signs support, whichever comparison that turns out to be. This point may be earned independently of the response in part A (iii); where part A (iii) gives no values for r_x and r_y, award it for θ_r > 70.3°.

▸ B2 — 1 point: For establishing, from the response’s own signs for r_x and r_y, whether the direction 70.3° is or is not consistent with those signs. Award this point for a conclusion consistent with the response’s own values for r_x and r_y, whether that conclusion rejects 70.3° or accepts it, and award it independently of the response in part A (iii) where no values for r_x and r_y are given there.

Scoring Note: B2 is a content criterion, not a route criterion. Any argument that turns on the component signs earns it — naming the quadrant r lies in and the quadrant 70.3° points into, testing the signs of cos 70.3° and sin 70.3° against the signs of r_x and r_y, or reasoning that an inverse tangent returns only angles with a positive cosine while r_x is negative. Asserting that a correction is needed, without support from the component signs, does not earn B2.

▸ B3 — 1 point: For stating the general condition: the inverse-tangent value must be corrected by adding 180° whenever the x-component is negative, that is, in Quadrants II and III.

Scoring Note: B3 carries no consistency clause — it is earned or not on the general condition the response itself states, independently of that response’s own values for r_x and r_y. Neither the undefined r_x = 0 case nor the +360° convention for a negative returned value is required for B3.

⚠︎ Common error (partial credit): Indicates θ_r > 70.3° and reports 250.3°, but offers no justification beyond the assertion that a correction of 180° is needed here — neither establishing from its own component signs that 70.3° cannot be the direction of r, nor stating the condition under which the correction applies — earns B1 only, 1 of 3 points.

⚠︎ Common error (partial credit): Indicates θ_r > 70.3°, identifies Quadrant III from the component signs and notes that 70.3° points into Quadrant I so it cannot describe r, but states the general condition as “add 180° whenever the calculator returns a negative angle.” That confuses the sign of the returned value with the sign of r_x: in Quadrant III the returned value is positive, so every Quadrant III direction would be left uncorrected, and every Quadrant IV direction would be wrongly corrected — earns B1 and B2, 2 of 3 points.

✗ Common error (no credit): Reports r_x < 0 and r_y < 0 in part A (iii) and still indicates θ_r = 70.3°, recording the value the calculator returns as the direction of r. The magnitude found in part A (iii) still stands, but the indicated comparison contradicts the response’s own component signs, nothing in the response establishes from those signs whether 70.3° can be the direction of r, and no general condition is stated, so none of B1, B2 or B3 is earned.

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