Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Position and Displacement Vectors — The Laboratory Floor Car

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A remote-controlled car is driven across a flat laboratory floor. An overhead camera tracks the car and reports its position relative to a fixed origin at one corner of the room.

At time t_1 = 0 s the camera records the car at (1.0 m, 3.0 m). At time t_2 = 5.0 s it records the car at (5.0 m, 6.0 m). Between these two instants the car does not travel in a straight line — it follows a curved route across the floor. A student lays a flexible measuring tape along the actual route the car took and finds its length to be 8.5 m. Figure 1 shows the camera’s floor grid, the two recorded positions and the route between them.

Figure 1. The camera’s floor grid, the two recorded positions, and the measured route between them.

All positions lie in the horizontal plane of the floor, so the situation is two-dimensional. Use the camera’s fixed floor grid as the reference frame: the origin is the room corner, +x runs east along one wall, +y runs north along the adjoining wall, and angles are measured counterclockwise from the +x axis. Throughout, r_1 and r_2 name the car’s position vectors at t_1 and t_2 and Δr its displacement over that interval; a magnitude is written with vertical bars, so |Δr| is the magnitude of Δr.

✎ Free Response Questions

Part A

i.    On the axes of Figure 1, draw the car’s position vector at t_1 = 0 s, its position vector at t_2 = 5.0 s, and its displacement over the interval from t_1 to t_2.

•   Each vector must be drawn as a distinct arrow with an arrowhead at its head end.

•   Each arrow must have its tail at the correct point for the quantity it represents.

•   Label the three arrows r_1, r_2 and Δr.

 ii.   Starting from the definition of displacement, derive an expression for the magnitude of the car’s displacement over the interval from t_1 to t_2.

Express your final answer in terms of x_1, y_1, x_2 and y_2, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

iii.  Using the coordinates recorded by the camera, calculate each of the following. Show your work.

•   the magnitude of the car’s position vector at t_2 = 5.0 s

•   the magnitude of the car’s displacement over the interval from t_1 to t_2

•   the direction of that displacement, measured counterclockwise from the +x axis

Part B

After the first run, a student compares the 8.5 m route length with the displacement magnitude found in part A (iii) and claims: “Since the displacement magnitude is smaller than the length of the route, part of the motion must have cancelled out — the car must have reversed direction somewhere during the interval.”

The car is later driven a second time, from (1.0 m, 3.0 m) to (5.0 m, 6.0 m) again, along a different route. On this run the car does reverse: it drives past the finishing point, stops, and backs up to it. A tape laid along the second route measures 11.4 m. Let Δr′ be the car’s displacement on this second run.

Indicate whether |Δr′| is greater than, less than, or equal to |Δr| by writing one of the following.

•   |Δr′| > |Δr|

•   |Δr′| < |Δr|

•   |Δr′| = |Δr|

Justify your response. In your justification, evaluate the student’s claim.

❖ 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 position vector runs from the origin to where the object is at that instant, so r_1 is drawn from (0, 0) to (1.0 m, 3.0 m) and r_2 from (0, 0) to (5.0 m, 6.0 m). Displacement is the change in position, so Δr is drawn from the earlier position to the later one — tail at (1.0 m, 3.0 m), head at (5.0 m, 6.0 m), not starting at the origin.

The three arrows close a triangle, r_1 + Δr = r_2, which is the geometric statement of Δr = r_2 − r_1. A displacement arrow starts where the motion started, not at the origin — which is why it is the one arrow of the three whose tail is not at O.

Figure 2. The two position vectors drawn from the origin, and the displacement drawn between their heads.

Scoring (2 points):

▸ A1 — 1 point: For both position vectors drawn as arrows from the origin, with heads at (1.0 m, 3.0 m) and (5.0 m, 6.0 m), each carrying an arrowhead and labelled r_1 and r_2. This point is awarded for the drawn representation alone.

▸ A2 — 1 point: For the displacement vector drawn with its tail at (1.0 m, 3.0 m) and its head at (5.0 m, 6.0 m), carrying an arrowhead and labelled Δr.

Scoring Note: The arrows run between points the supplied grid already plots, so their lengths are fixed by the figure and are not scored. A1 and A2 are independent of each other: a response that draws the displacement correctly while misplacing a position vector earns A2, and the reverse earns A1. Neither A1 nor A2 carries a consistency clause — both are scored on the drawing itself, against the coordinates given in the Scenario.

⚠︎ Common error (partial credit): Draws r_1 and r_2 correctly from the origin, then draws Δr from the origin to (5.0 m, 6.0 m) as well. That third arrow is r_2 redrawn; it marks where the car is rather than how far it moved, so A2 is not earned — earns A1 only, 1 of 2 points.

⚠︎ Common error (partial credit): Draws r_1 and r_2 correctly from the origin, and places the third mark correctly between (1.0 m, 3.0 m) and (5.0 m, 6.0 m) and labels it Δr, but draws it as a plain line segment with no arrowhead. Displacement is a vector and the arrowhead is what carries its direction, so A2 is not earned — earns A1 only, 1 of 2 points.

✗ Common error (no credit): Draws a single arrow from the origin to (5.0 m, 6.0 m) and labels it Δr, with no arrow to (1.0 m, 3.0 m) anywhere. The earlier position never enters, so no pair of position vectors is drawn and no displacement vector is drawn; neither A1 nor A2 is earned.

Part A (ii) — Model Answer

Displacement is defined as the change in an object’s position:

Δr = r_2 − r_1

Vector subtraction is carried out component by component, so the subtraction is performed separately along each axis:

Δx = x_2 − x_1         Δy = y_2 − y_1

Δr is the resultant of these two perpendicular components, so they are the legs of a right triangle whose hypotenuse has length |Δr|, and the Pythagorean relationship a² + b² = c² gives the magnitude:

|Δr| = √( (x_2 − x_1)² + (y_2 − y_1)² )

Only the two recorded positions appear in the expression. The subtraction is carried out on each axis separately, and only then does the Pythagorean relationship recombine the two results.

Scoring (2 points):

▸ A3 — 1 point: For starting from the definition of displacement as the change in position — writing Δr = r_2 − r_1, or equivalently Δx = x_2 − x_1 and Δy = y_2 − y_1. This point is awarded for the correct starting definition alone, before any algebra is performed.

▸ A4 — 1 point: For the correct symbolic expression |Δr| = √((x_2 − x_1)² + (y_2 − y_1)²), or any algebraically equivalent form — the reversed differences (x_1 − x_2) and (y_1 − y_2) give the same magnitude and are equally acceptable. Award this point wherever the response applies the Pythagorean relationship, or any equivalent magnitude relationship, to its own componentwise differences of the two positions, however those differences are written. Where the response forms two components explicitly, they must be differences of the two positions; a response that writes the correct final expression without naming components earns A4 under the Scoring Note below.

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

⚠︎ Common error (partial credit): Writes Δr = r_2 − r_1, then |Δr| = √(x_2² + y_2²) − √(x_1² + y_1²), subtracting the magnitudes of the two position vectors instead of subtracting the vectors and then taking a magnitude. The starting definition is present, so A3 is earned; no componentwise difference is ever formed, so A4 is not — earns 1 of 2 points.

✗ Common error (no credit): Gives |Δr| = √((x_2 + x_1)² + (y_2 + y_1)²), adding the coordinates instead of subtracting them, with no statement of the definition anywhere. Neither the starting definition nor a componentwise difference is present, so neither A3 nor A4 is earned.

Part A (iii) — Model Answer

The components of a position vector are the coordinates themselves, so at the later instant t_2:

|r_2| = √( x_2² + y_2² ) = √( (5.0 m)² + (6.0 m)² ) = √(61 m²) ≈ 7.8 m

The displacement components are final minus initial on each axis, and the two perpendicular components recombine through the Pythagorean relationship:

Δx = 5.0 m − 1.0 m = +4.0 m         Δy = 6.0 m − 3.0 m = +3.0 m
|Δr| = √( (+4.0 m)² + (+3.0 m)² ) = √(25 m²) = 5.0 m

The direction follows from the ratio of the components:

θ = tan⁻¹( Δy / Δx ) = tan⁻¹( (+3.0 m) / (+4.0 m) ) = tan⁻¹(0.75) ≈ 37°

Both Δx and Δy are positive, so Δr points up and to the right of the starting position and the value the inverse tangent returns needs no correction. The car’s displacement is 5.0 m directed approximately 37° counterclockwise from the +x axis — that is, 37° north of east.

The two magnitudes answer different questions: |r_2| ≈ 7.8 m is how far the car is from the room corner at t_2, while |Δr| = 5.0 m is how far it ended up from where it started. They are not two values of one quantity — one locates the car, the other measures how far it moved.

Figure 3. The displacement, its components and its 37° direction, beside the position vector at t_2.

Scoring (3 points):

▸ A5 — 1 point: For the position-vector magnitude |r_2| = √((5.0 m)² + (6.0 m)²) ≈ 7.8 m, stated with its unit.

▸ A6 — 1 point: For evaluating the magnitude of the displacement from the given coordinates and stating it with its unit. The expected value is |Δr| = 5.0 m, from the component differences Δx = x_2 − x_1 = +4.0 m and Δy = y_2 − y_1 = +3.0 m. Award this point for a magnitude obtained by correctly substituting the given coordinates into the response’s own expression from part A (ii), whatever that expression is; where part A (ii) gives no expression, award it for |Δr| = 5.0 m obtained from the component differences. A single value cannot earn both A5 and A6: where the response reports the same number as the position-vector magnitude and as the displacement magnitude, it earns A5 only.

▸ A7 — 1 point: For stating the direction of the displacement as an angle with its reference direction rather than as a bare number. The expected value is θ ≈ 37° (“37° counterclockwise from the +x axis” and “37° north of east” are equally acceptable). Award this point for a direction correctly obtained from the components the response itself uses — by tan⁻¹(Δy / Δx), by sin⁻¹(Δy / |Δr|), by cos⁻¹(Δx / |Δr|), or by any equivalent route.

Scoring Note: √61 m² may be reported as 7.8 m or carried unrounded; both give the same value to two significant figures. An equivalent statement of the same direction in the Scenario’s convention earns A7: “37° north of east” names the same direction as “37° counterclockwise from the +x axis”. An angle reported from the +y axis has not been given in the convention the Scenario fixes and does not earn A7 on its own. A5 carries no consistency clause — the coordinates it uses are given in the Scenario.

⚠︎ Common error (partial credit): Uses the t_2 coordinates as the displacement components, reporting |Δr| = √((5.0 m)² + (6.0 m)²) ≈ 7.8 m and a direction of tan⁻¹(6.0 / 5.0) ≈ 50° counterclockwise from the +x axis. This is |r_2| offered as the displacement, so the same number is reported for both magnitudes and A6 is lost. A5 still stands, and A7 is earned on the components the response does use, since 50° is the angle those components give and it is stated with its reference direction — earns 2 of 3 points.

⚠︎ Common error (partial credit): Inverts the ratio, reporting the direction as tan⁻¹(Δx / Δy) = tan⁻¹(4.0 / 3.0) ≈ 53° counterclockwise from the +x axis. That ratio gives the angle measured from the +y axis, so reporting it as an angle from the +x axis names a different direction from the one the components give and A7 is not earned. The two magnitudes are unaffected — earns A5 and A6, 2 of 3 points.

Part B — Model Answer

|Δr′| = |Δr|

Displacement is the change in position, so it is fixed entirely by where the motion starts and where it ends. Both runs start at (1.0 m, 3.0 m) and end at (5.0 m, 6.0 m), so both have Δx = +4.0 m and Δy = +3.0 m and both give |Δr′| = |Δr| = 5.0 m. The expression derived in part A (ii) says the same thing in symbols: it contains only the four endpoint coordinates, and no term in it describes the route.

The student’s claim is incorrect. The second run reverses, and the reversal made its route longer, not shorter: every metre the car covers backing up is added to the distance it has already travelled, exactly as driving forward is. That is why the tape reads 11.4 m on a run that ends at the same point run 1 ended at. Distance travelled is a scalar that only ever accumulates; there is no signed cancellation for a reversal to produce. And the displacement, which is what the claim says the cancellation reduced, is identical on the two runs.

So a route longer than the displacement is not evidence of a reversal. Route length equals |Δr| only when the route is the straight segment joining the two points; any departure from straightness at all makes the route longer, whether or not the car ever turns back. The first run’s curvature is enough on its own to give 8.5 m against 5.0 m, with no reversal anywhere in it.

Figure 4. Two routes between the same two points: the distances travelled differ, the displacement does not.

Reversing adds to the distance travelled and leaves the displacement untouched, so a route longer than the displacement says only that the route was not straight.

Scoring (3 points):

▸ B1 — 1 point: For indicating |Δr′| = |Δr|.

Scoring Note: B1 carries no consistency clause with part A (iii). The two runs share both endpoints, so the displacements are equal whatever value part A (iii) reported for |Δr|, and the comparison is the same for every response.

▸ B2 — 1 point: For establishing that the displacement depends only on the initial and final positions — by appealing to the definition Δr = r_2 − r_1, by noting that the expression derived in part A (ii) contains only the four endpoint coordinates, by giving identical components for the two runs, or by any equivalent route.

Scoring Note: B2 is a content criterion, not a route criterion — any argument that establishes what the displacement depends on earns it. Giving the identical components for the two runs earns B2, because forming those components from the endpoints alone is itself the statement of what the displacement depends on. A bare assertion that the two displacements are equal, with no components and no appeal to the definition, does not.

▸ B3 — 1 point: For evaluating the student’s claim as incorrect on a supporting ground: that backing up adds to the distance travelled rather than cancelling any of it, so no motion cancels; or that route length equals |Δr| only for a straight route, so any curvature at all produces a route longer than the displacement without a reversal; or any equivalent ground showing that a route longer than the displacement is not evidence of a reversal.

⚠︎ Common error (partial credit): Indicates |Δr′| = |Δr| and argues correctly that displacement is fixed by the two endpoints, but never returns to the student’s claim — earns B1 and B2, 2 of 3 points.

⚠︎ Common error (partial credit): Indicates |Δr′| = |Δr| and rejects the claim correctly, but supports the equality by observing that the two routes “cover roughly the same ground” rather than by saying what the displacement depends on. The equality does not depend on the routes resembling one another, so B2 is not earned — earns B1 and B3, 2 of 3 points.

✗ Common error (no credit): Indicates |Δr′| < |Δr|, reasoning that the reversal on the second run cancels part of the motion and leaves a smaller net displacement. This treats distance travelled as a signed quantity that can subtract from itself, and it restates the claim the part asks the response to evaluate rather than testing it. The indicated comparison is wrong, nothing establishes what displacement depends on, and the claim is endorsed rather than evaluated, so none of B1, B2 or B3 is earned.

Keep going:

Equations & Traps deck · Mind Maps bundle